0$. Taking $r=p, q=1$ gives \[ \|I^{\alpha}u\|_{L^p}\le \frac{T^{\alpha}}{\Gamma(\alpha+1)}\|u\|_{L^p}. \] The case $p=\infty$ is simple: for $u \in L^{\infty}$ the integral for $I^{\alpha}u$ is well defined and we have \begin{align*} I^{\alpha}u(t) &=\frac{1}{\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}u(s)\;ds\\ &\le \frac{1}{\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}\|u\|_{\infty}\;ds\\ &= \frac{1}{\Gamma(\alpha)}\frac{t^{\alpha}}{\alpha}\|u\|_{\infty}, \end{align*} hence $\|I^{\alpha}u\|_{\infty} \le \dfrac{T^{\alpha}}{\Gamma(\alpha+1)}\|u\|_{\infty}$. \eqref{rlint2} We only give the case $r

-1$, that is $q<1/(1-\alpha)$ and hence $1/r>1/p-\alpha$, that is $r

-1$. Therefore by H\"{o}lder's inequality we have \begin{align*} | I^{\alpha}u(t)| &\le \frac{1}{\Gamma({\alpha})}\Bigl(\int_0^t (t-s)^{(\alpha-1)q}\,ds\Bigr)^{1/q}\Bigl(\int_0^t |u|^p(s)\,ds\Bigr)^{1/p},\\ &=\frac{1}{\Gamma({\alpha})}\Bigl(\frac{t^{(\alpha-1)q+1}}{(\alpha-1)q+1}\Bigr)^{1/q}\Bigl(\int_0^t |u|^p(s)\,ds\Bigr)^{1/p}, \end{align*} and both terms in the product have a zero limit as $t \to 0+$. \eqref{rlint4} For $u \in C_{-\gamma}$ we have $u(t)=t^{-\gamma}v(t)$ for some function $v \in C[0,T]$ and $\|u\|_{-\gamma}=\|v\|_{\infty}$. So we have \begin{equation} \begin{split}\label{eqcgamemb} I^{\alpha}u(t)&= \frac{1}{\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}s^{-\gamma}v(s)\,ds,\\ &=\frac{1}{\Gamma(\alpha)} t^{\alpha-\gamma}\int_0^1 (1-\sigma)^{\alpha-1}\sigma^{-\gamma}v(t\sigma)\,d{\sigma}. \end{split} \end{equation} Since $v$ is continuous on $[0,T]$ it is bounded, say $\|v\|_{\infty} = M$, and we have \[ \int_0^1 (1-\sigma)^{\alpha-1}\sigma^{-\gamma}v(t\sigma)\,d{\sigma} \le M B(\alpha,1-\gamma) \] Therefore $\int_0^1 (1-\sigma)^{\alpha-1}\sigma^{-\gamma}v(t\sigma)\,d{\sigma}$ is a continuous function of $t$ by the dominated convergence theorem, thus $I^{\alpha}u \in C_{\alpha-\gamma}[0,T]$. Moreover, \[ \|I^{\alpha}u\|_{\alpha-\gamma} \le \frac{1}{\Gamma(\alpha)}\int_0^1 (1-\sigma)^{\alpha-1}\sigma^{-\gamma}\,d{\sigma}\|v\|_{\infty} =\frac{\Gamma(1-\gamma)}{\Gamma(1+\alpha-\gamma)}\|u\|_{-\gamma}. \] \eqref{rlint5} Let $0 \le \gamma \le \alpha $. By part \eqref{rlint3}, $I^{\alpha}$ maps $C_{-\gamma}$ into $C_{\alpha-\gamma} \subset C[0,T]$ and from \eqref{eqcgamemb} we obtain $\|I^{\alpha}u\|_{\infty} \le T^{\alpha-\gamma} \frac{\Gamma(1-\gamma)}{\Gamma(1+\alpha-\gamma)}\|u\|_{-\gamma}$. Also from \eqref{eqcgamemb} we obtain $\lim_{t \to 0+}I^{\alpha}u(t)=0$ if $\gamma <\alpha$. By the dominated convergence theorem this also holds if $\gamma=\alpha$ and $v(0)=0$. \eqref{rlint6} For $u \in AC$, $u' \in L^1$ exists a.e.\ and $u(t)-u(0)=Iu'(t)$ for all $t$. Then \begin{equation}\label{eqI-AC} I^{\alpha}u(t)=I^{\alpha}\,Iu'(t)+I^{\alpha}u(0)=I\,I^{\alpha}u'(t)+u(0)t^{\alpha}/\Gamma(\alpha+1) \end{equation} where $I^{\alpha}u' \in L^1$ so the first term is in $AC$, and the second term is also in $AC$ since $\alpha>0$. \eqref{rlint6b} When $u' \in C_{-\gamma}$, we have $I^{\alpha}u' \in C_{\alpha-\gamma} \subset C$ by part~\eqref{rlint4}, hence $I(I^{\alpha}u') \in C^1$ and from \eqref{eqI-AC} we see that $I^{\alpha}u \in C^1[0,T]$ if and only if $u(0)=0$. For the last part, taking $v(t) \equiv 1$ we have $v \in C^{\infty}$ and $I^{\alpha}v=t^{\alpha}/\Gamma(\alpha+1)$ is an $AC$ function but is not in $C^1$ for $\alpha<1$. \par \eqref{rlint7} This follows at once since $I^{m+\alpha}u=I^{m}I^{\alpha}u$ where $I^{\alpha}u\in C$ by part \eqref{rlint5}. \eqref{rlint8} Let $u \in C_{1,-\gamma}[0,1]$ so that $u(t)=t^{-\gamma}v(t)$ for some $v \in C^1[0,1]$. Then we have as above \[ I^{\alpha}u(t)=\frac{1}{\Gamma(\alpha)} t^{\alpha-\gamma}\int_0^1 (1-\sigma)^{\alpha-1}\sigma^{-\gamma}v(t\sigma)\,d{\sigma}, \] where, by differentiation under the integral sign, \[ \frac{d}{dt}\Bigl(\int_0^1 (1-\sigma)^{\alpha-1}\sigma^{-\gamma}v(t\sigma)\,d{\sigma}\Bigr) = \int_0^1 (1-\sigma)^{\alpha-1}\sigma^{1-\gamma}v'(t\sigma)\,d{\sigma}. \] Since $v'$ is continuous the integral on the right is a continuous function of $t$ by the dominated convergence theorem. Hence \[ \int_0^1 (1-\sigma)^{\alpha-1}\sigma^{-\gamma}v(t\sigma)\,d{\sigma} \] is in $C^1[0,1]$,that is $I^{\alpha}u \in C_{1,\alpha-\gamma}$; the example $v(t) \equiv 1$ shows this is optimal. Also, if $\alpha \ge \gamma$, $I^{\alpha}u$ is $t^{\alpha-\gamma}$ multiplied by a $C^1$ function so is in $AC$. \eqref{rlint9} We have \begin{align*} I^{\alpha}u(t)&=\frac{1}{\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}u(s)\,ds\\ &=\frac{1}{\Gamma(\alpha)} t^{\alpha} \int_0^1 (1-\sigma)^{\alpha-1}u(t\sigma)d{\sigma}, \end{align*} and this is a non-decreasing function of $t$ when $u$ is non-decreasing. \end{proof} \begin{remark} \rm Part \eqref{rlint1} is stated in \cite[Lemma 2.1 a]{kst} and Theorem 2.6 of \cite{skm} states: ``may be verified by simple operations using the generalized Minkowski's inequality''. Part \eqref{rlint2} The first part for $p \ge 1$ is stated in \cite[Lemma 2.1 b]{kst} and is proved in \cite[Theorem 3.5]{skm} by another method. The fact that $I^{\alpha}$ is bounded from $L^{p}$ (with $p>1$) into $L^{r}$ with the equality $r = p/(1-{\alpha}p)$ was proved by Hardy and Littlewood~\cite[Theorem 4]{Hardy-Lit}. Hardy and Littlewood show that the result does not hold if $p=1$ for any $\alpha \in (0,1)$, and they also show that if $0<\alpha<1$ and $p=1/\alpha$ the potentially plausible result is false, that is, $I^{\alpha}u$ is not necessarily bounded. Part \eqref{rlint3} This was proved by Hardy and Littlewood in \cite[Theorem 12]{Hardy-Lit}. The proof can be found in \cite[Theorem 2.6 ]{diethelm-bk} and in \cite[Theorem 3.6 ]{skm}. The H\"{o}lder space $C^{0,\lambda}$ has other notations in these references. Hardy and Littlewood point out that the result is not true, in the cases $p>1$, $\alpha=1/p$, and $\alpha=1+1/p$. The continuity of $I^{\alpha}u$ is proved in \cite[Lemma 2.2]{Ag-OR-St} by a direct method using H\"{o}lder's inequality and the last part is also proved in \cite[Lemma 2.1]{Ag-OR-St} as in the given proof. The result is also a consequence of Corollary to Theorem 3.6 of \cite{skm}. Part \eqref{rlint4} is stated in \cite[Lemma 2.8]{kst} with $\alpha, \gamma$ complex numbers, the proof is referred to \cite{kbt}, a paper in Russian. Part \eqref{rlint5}, the first part is stated in \cite[Lemma 2.8]{kst}; the last part may be novel. Part \eqref{rlint6}, this is proved in \cite[Lemma 2.1]{skm}, with a different notation, by a longhand version of the same proof, and is proved in \cite[Lemma 2.3]{lan-lin-caa} by using Proposition~\ref{propRLexist} below. \par Part \eqref{rlint6b} seems to be new for $\gamma>0$, the case $\gamma=0$ is known, see \cite[Theorem 6.26]{diethelm-bk}. The last part is well known and is pointed out in \cite[Example 6.4]{diethelm-bk}. Part \eqref{rlint7} should be known but we do not have a reference, it improves \cite[Lemma 2.4]{eloe-masthay} which has the case $\gamma=0$ and a different longer proof. Part \eqref{rlint8} This seems to be new when ${\gamma} \ne 0$. Part \eqref{rlint9} is presumably well known but we do not know a reference. \end{remark} Interchanging the order of integration, using Fubini's theorem, shows that these fractional integral operators satisfy a semigroup property as follows. \begin{lemma}\label{RLintsgp} Let $\alpha, \beta>0$ and $u \in L^1[0,T]$. Then $I^{\alpha}I^{\beta}(u)=I^{\alpha+\beta}(u)$ as $L^1$ functions, thus, $I^{\alpha}I^{\beta}(u)(t)=I^{\alpha+\beta}(u)(t)$ for a.e.\ $t \in [0,T]$. If $u$ is continuous, or if $u \in C_{-\gamma}$ and $\alpha+\beta\ge \gamma$, this holds for all $t\in [0,T]$. If $u \in L^1$ and $\alpha+\beta \ge 1$ equality again holds for all $t\in [0,T]$. \end{lemma} \begin{proof} For $u \in L^1$ the fractional integrals $I^{\beta}u$ and $I^{\alpha+\beta}u$ exist as $L^1$ functions so are finite almost everywhere. For each $t$ for which $I^{\alpha+\beta}|u|(t)$ exists (finite), that is a.e.\ $t$, we have \begin{align*} \Gamma(\alpha)\Gamma(\beta)&I^{\alpha}(I^{\beta}u)(t)\\ =&\int_0^t (t-s)^{\alpha-1} \Bigl(\int_0^s (s-\tau)^{\beta-1} u(\tau) \,d{\tau}\Bigr)\,ds\\ =&\int_0^{t} u(\tau) \Bigl(\int_{\tau}^{t} (t-s)^{\alpha-1}(s-\tau)^{\beta -1}\,ds\Bigr)\,d{\tau}, \quad\text{by Fubini's theorem},\\ =&\int_0^t (t-\tau)^{\alpha+\beta-1} u(\tau)B(\alpha, \beta)\,d{\tau}, \quad \text{by Lemma~\ref{lbeta}}, \end{align*} which proves the first part by the relationship between the Beta and Gamma functions stated earlier. When $u$ is continuous all terms are continuous, see Proposition~\ref{propfintprop} part~\eqref{rlint5}, so equality holds for all $t$. For $u \in C_{-\gamma}$, by Proposition~\ref{propfintprop} part~\eqref{rlint4}, $I^{\beta}u \in C_{\beta-\gamma}$ and $I^{\alpha}(I^{\beta}(u)) \in C_{\alpha+\beta-\gamma} \subset C^{0}$ when $\alpha+\beta \ge \gamma$, so both sides are continuous functions and equality holds for all $t$. For the last part if $\beta\ge 1$ or $\alpha \ge 1$ then the terms on both sides are continuous so the only case to consider is $0<\alpha,\beta<1$. When $\alpha+\beta \ge 1$ we have $I^{\alpha+\beta}(u)(t)=I(I^{\alpha+\beta-1}u)(t)$ and the right side of this equation is $AC$ so $I^{\alpha+\beta}u(t)$ exists for every $t$ and equals $I^{\alpha}I^{\beta}u(t)$ by the first part. \end{proof} \begin{remark} \rm Some of this is proved in \cite[Theorem 2.2] {diethelm-bk} and in \cite[(2.21)]{skm}. The part concerning $u \in C_{-\gamma}$ seems to be new. Also we make the observation that when $\alpha+\beta>1$ and $u \in L^1$ both sides are also H\"older continuous by using Proposition~\ref{propfintprop} parts~\eqref{rlint2} and \eqref{rlint3}. Note that there are functions that are H\"older continuous but not $AC$ and $AC$ functions that are not H\"older continuous. \end{remark} For $\alpha \in (0,1)$ we will see that in discussing fractional differential equations via the corresponding Volterra integral equation it is necessary to have $I^{1-\alpha}u\in AC$. The following result gives conditions for this to hold; the result is known, it is contained in the proof of \cite[Theorem 2.1]{skm}. \begin{proposition}\label{propRLexist} Let $u \in L^1[0,T]$ and $\alpha \in (0,1)$. Then $I^{1-\alpha}u \in AC$ and $I^{1-\alpha}u(0)=0$ if and only if there exists $f \in L^1$ such that $u=I^{\alpha}f$. \end{proposition} \begin{proof} Suppose that there exists $f \in L^1$ such that $u=I^{\alpha}f$. Then, by Lemma~\ref{RLintsgp} \[ I^{1-\alpha}u=I^{1-\alpha}I^{\alpha}f=If \in AC, \] and $I^{1-\alpha}u(0)=\lim_{t \to 0+} \int_0^t f(s)\,ds=0$, since $f \in L^1$. Conversely suppose that $I^{1-\alpha}u \in AC$ and $I^{1-\alpha}u(0)=0$. Let $F(t):=I^{1-\alpha}u(t)$ so that $F \in AC$ and $F(0)=0$. Then $f:=F'$ exists for a.e.\ $t$ with $f \in L^1$, and $F(t)=If=\int_0^{t}f(s)\,ds$. From $F(t):=I^{1-\alpha}u(t)$ we have $I^{\alpha}F=Iu$ that is $Iu=I^{\alpha}\,If=I\,I^{\alpha}f$. By the definition (see Proposition~\ref{propfintprop} \eqref{rlint1}, \eqref{rlint2} we have $I^{\alpha}f \in L^1$ and since $u \in L^1$ both $Iu$ and $I\,I^{\alpha}f$ are absolutely continuous so their derivatives exist a.e.\ as $L^1$ functions and are equal, that is $u=I^{\alpha}f$. \end{proof} \section{Fractional derivatives of order $\alpha \in (0,1)$}\label{s-FD-ord1} Let $D$ denote the usual differentiation operator. The Riemann-Liouville (R-L) fractional derivative of order $\alpha \in (0,1)$ is defined as follows. \begin{definition}\label{defRLder} \rm For $\alpha \in (0,1)$ and $u \in L^1$ the R-L fractional derivative $D^{\alpha}u$ is defined when $I^{1-\alpha}u\in AC$ by \[ D^{\alpha}u(t):=D\,I^{1-\alpha}u(t), \;\text{a.e.} \; t \in [0,T]. \] \end{definition} For $D\,I^{1-\alpha}u(t)$ to be defined for a.e.\ $t$, it is necessary that $I^{1-\alpha}u$ should be differentiable a.e., but we do not believe that alone is sufficient, and our discussions below show that it is necessary to always have $I^{1-\alpha}u\in AC$ in considering IVPs for R-L fractional differential equations via a Volterra integral equation, thus we make this requirement. This has been noted in the monograph \cite{skm}, see \cite[Definition 2.4]{skm} and the related comments in the `Notes to \S2.6'. Clearly $DI^{1-\alpha}u$ exists a.e.\ if $u=I^{\alpha}f$ for some $f \in L^1$, using Lemma~\ref{RLintsgp}, but Proposition~\ref{propRLexist} already implies that $I^{1-\alpha}u\in AC$. It follows using Lemma~\ref{RLintsgp} that the R-L derivative $D^{\alpha}$ is the left inverse of $I^{\alpha}$, as shown, for example, in \cite[Theorem 2.14] {diethelm-bk}. \begin{lemma} \label{lemRLinverse} Let $0<\alpha \le 1$. Then, for every $h \in L^1$, $D^{\alpha}I^{\alpha}h(t)=h(t)$ for almost every $t$. \end{lemma} \begin{proof} Since $Ih \in AC$ we have \[ D^{\alpha}I^{\alpha}h(t)=D\,I^{1-\alpha}\, I^{\alpha}h(t) =D\,Ih(t)=h(t), \quad \text{for a.e. } t. \qedhere \] \end{proof} In general fractional derivatives do not commute, see \cite[Examples 2.6, 2.7 ] {diethelm-bk}. The Caputo fractional derivative is defined with the derivative and fractional integral taken in the reverse order to that of the R-L derivative. \begin{definition}\label{defCder} \rm For $\alpha \in (0,1)$ and $u \in AC$ the Caputo fractional derivative $D_{C}^{\alpha}u$ is defined for a.e.\ $t$ by \[ D_{C}^{\alpha}u(t):=I^{1-\alpha}Du(t). \] For $u \in AC$, $Du \in L^1$ and so $D_{C}^{\alpha}u=I^{1-\alpha}(Du)$ is defined as an $L^1$ function. The modified Caputo derivative is defined by $D_{*}^{\alpha}u:=D^{\alpha}(u-u(0))$ whenever this R-L derivative exists, that is when $u(0)$ exists and $I^{1-\alpha}u \in AC$. \end{definition} There is a connection between the R-L and Caputo derivatives for functions with some regularity. \begin{proposition}\label{propcap=rl1} Let $u \in AC$ and let $u_0$ denote the constant function with value $u(0)$. For $0<\alpha < 1$ we have \begin{equation}\label{eqRL=C} D_{*}^{\alpha}u(t) = D^{\alpha}(u-u_0)(t)=D_{C}^{\alpha}u(t), \quad \text{for a.e. } t. \end{equation} \end{proposition} \begin{proof} Since $u \in AC$, by Proposition~\ref{propfintprop}\,\eqref{rlint6}, $I^{1-\alpha}u \in AC$ so the R-L derivative exists and we have \begin{align*} D_{*}^{\alpha}u= D^{\alpha}(u-u_0)&=DI^{1-\alpha}(u-u_0)\\ &=DI^{1-\alpha}Iu', \quad\text{since} \;u \in AC,\\ &=DI\,I^{1-\alpha}u', \quad\text{by Lemma~\ref{RLintsgp}} \;\text{as} \;u' \in L^1,\\ &=I^{1-\alpha}u',\quad\text{since} \;I^{1-\alpha}u' \in L^1,\\ &=D_{C}^{\alpha}u. \qedhere \end{align*} \end{proof} The result is a special case of the result given for fractional derivatives of all orders in \cite[Theorem 3.1]{diethelm-bk}, and which is proved below in Lemma~\ref{lemhigherorder}. The modified Caputo derivative $D^{\alpha}_{*}$ is the left inverse of $I^{\alpha}$ for \textbf{continuous} functions; for the higher order case see \cite[Theorem 3.7] {diethelm-bk}. \begin{lemma} \label{lemCinvRLint} Let $\alpha \in (0,1)$ and let $u$ be continuous, then $D^{\alpha}_{*}I^{\alpha}u(t)=u(t)$ for $t \in [0,T]$. \end{lemma} \begin{proof} By Proposition~\ref{propfintprop}\,\eqref{rlint5} with $\gamma=0$, $I^{\alpha}u$ is continuous and $I^{\alpha}u(0)=0$, thus \[ D^{\alpha}_{*}I^{\alpha}u(t)=D^{\alpha}(I^{\alpha}u-0)(t)=DI^{1-\alpha}(I^{\alpha}u)(t)=D(Iu)(t)=u(t), \] which is valid for every $t$ since $u\in C$. \end{proof} The Caputo derivatives do not commute in general but Diethelm has a positive result. \begin{lemma}[{\cite[Lemma 3.13] {diethelm-bk}}] Let $f \in C^{k}[0,T]$ for some $k \in \mathbb{N}$. Moreover let $\alpha,\beta>0$ be such that there exists some $\ell\in \mathbb{N}$ with $\ell \le k$ and $\alpha, \alpha+\beta \in [\ell-1,\ell]$. Then, \[ D^{\alpha}_{*}D^{\beta}_{*}f =D^{\alpha+\beta}_{*}f. \] \end{lemma} \begin{remark} \rm Diethelm \cite {diethelm-bk} notes that existence of $\ell$ is important, and gives the example $f(t)=t$, $\alpha=7/10$, $\beta=7/10$ to show that it can fail when this condition is not satisfied. Also he notes that such a result cannot be expected to hold in general for Riemann-Liouville derivatives. \end{remark} \subsection{Fractional derivatives of higher order} For higher order derivatives the definitions are as follows. For a positive integer $k$ let $D^{k}$ denote the ordinary derivative operator of order $k$, and for $n \in \mathbb{N}$ and a function $u$ such that $D^{k}u(0)$ exists for $k=0, \dots, n$ let $T_{n}u(t):=\sum_{k=0}^{n}\frac{t^{k}D^{k}u(0)}{k!}$ be the Taylor polynomial of degree $n$ and define $T_0u(t)=u(0)$. \begin{definition}\label{RL-Cap-higher-order} \rm Let $\beta \in \mathbb{R_{+}}$ and let $n =\lceil \beta \rceil$ be the smallest integer greater than or equal to $\beta$ (the ceiling function acting on $\beta$). The Riemann-Liouville fractional differential operator of order $\beta$ is defined when $D^{n-1}(I^{n-\beta}u) \in AC$, that is $I^{n-\beta}u \in AC^{n-1}$, by \[ D^{\beta}u := D^{n}I^{n-\beta}u. \] The Caputo derivative is defined for $u \in AC^{n-1}$, by the equation \[ D^{\beta}_{C}u:=I^{n-\beta}D^{n}u. \] The modified Caputo derivative is defined when $I^{n-\beta}u \in AC^{n-1}$ and $T_{n-1}u$ exists by \[ D^{\beta}_{*}u=D^{\beta}(u-T_{n-1}u), \] where $T_{n-1}u$ is the Taylor polynomial of degree $n-1$.\\ Under the given conditions each fractional derivative exists a.e. \end{definition} \begin{remark} \rm Diethelm~\cite[Definition 3.2] {diethelm-bk} calls $D^{\beta}_{*}$ the \emph{Caputo differential operator of order ${\beta}$} and thereafter uses that definition. \end{remark} In the following results $m$ always denotes a positive integer. \begin{lemma} \label{lemhigherorder} If $D^{m}u \in AC$ then for $0<\alpha<1$, $D^{m+\alpha}_{C}u$ exists, and we have \begin{equation}\label{eqcapcommhord} D^{m+\alpha}_{C}u(t)=D^{\alpha}_{C}D^{m}u(t), \quad\text{for a.e. } t \in [0,T]. \end{equation} \end{lemma} \begin{proof} By definition of Caputo derivative and the fact that $D^{m}u \in AC$ we have \[ D^{m+\alpha}_{C}u=I^{1-\alpha}D^{m+1}u=I^{1-\alpha}D(D^{m}u)=D^{\alpha}_{C}D^{m}u. \] \end{proof} \begin{lemma}\label{lem-high-ord-star} For $0<\alpha<1$ if $u \in AC[0,T]$ then $D^{m+\alpha}_{*}u(t)=D^{m+\alpha-1}_{*}u'(t)$ whenever both fractional derivatives exist. \end{lemma} \begin{proof} By definition, \begin{align*} D^{m+\alpha}_{*}u(t)&=D^{m+1}(I^{1-\alpha}(u-T_{m}u))(t)\\ &=D^{m+\alpha}(I^{1-\alpha}I(u'-T_{m-1}u))(t)\\ &=D^{m+\alpha}(I I^{1-\alpha}(u'-T_{m-1}u))(t)\\ &=D^{m+\alpha-1}(I^{1-\alpha}(u'-T_{m-1}u))(t)\\ &=D^{m+\alpha-1}_{*}u'(t). \qedhere \end{align*} \end{proof} The following result is proved in Diethelm ~\cite[Theorem 3.1] {diethelm-bk} with a different proof using integration by parts. \begin{lemma} \label{lemmthord} Let $D^{m}u \in AC$ and $0<\alpha<1$. Then \begin{equation} \label{eqcap=rlhord} D^{m+\alpha}_{*}u=D^{m+\alpha}_{C}u. \end{equation} \end{lemma} \begin{proof} Since $D^{m}u \in AC$, $u$ and derivatives of order up to $m$ are absolutely continuous and all required fractional derivatives exist. By repeated application of Lemma~\ref{lem-high-ord-star} \begin{align*} D^{m+\alpha}_{*}u &= D^{m+\alpha-1}(u')=D^{m+\alpha-2}_{*}u'' = \dots\\ &=D^{\alpha}_{*}(D^{m}u)=D^{\alpha}_{C}(D^{m}u), \;\text{by Proposition~\ref{propcap=rl1}},\\ &=D^{m+\alpha}_{C}u, \text{by Lemma~\ref{lemhigherorder}}. \qedhere \end{align*} \end{proof} \section{IVP for Caputo derivative of all orders}\label{s-IVP-Cap} We now turn to initial value problems for Caputo derivatives. For Caputo fractional differential equations with a singular nonlinearity we have the following relationships with a Volterra integral equation with a doubly singular kernel. \begin{theorem} \label{ivporderm} Let $f$ be continuous on $[0,T] \times \mathbb{R}$, let $0<\alpha<1$ and let $0 \le \gamma < \alpha$. For $m \in \mathbb{N}$, if a function $u$ with $D^m u \in AC$ satisfies the Caputo fractional initial value problem \begin{equation}\label{eqCapDEm} \begin{gathered} D^{m+\alpha}_{C} u(t)=t^{-\gamma}f(t,u(t)), \; \text{a.e.}\; t \in (0,T], \\ u(0)=u_0, \;u'(0)=u_1, \dots, u^{(m)}(0)=u_{m}, \end{gathered} \end{equation} then $u$ satisfies the Volterra integral equation \begin{equation}\label{eqCapVIEm} u(t)=\sum_{k=0}^{m}\frac{t^{k}}{k!}u_{k} + \frac{1}{\Gamma(m+\alpha)}\int_0^t (t-s)^{m+\alpha-1}s^{-\gamma}f(s, u(s))\,ds, \; t \in [0,T]. \end{equation} Secondly, if $u \in C[0,T]$ satisfies \eqref{eqCapVIEm} then $u \in C^{m}[0,T]$, and $D^{m+\alpha}_{*}u$ exists a.e.\ and satisfies \begin{equation}\label{eqmodCapDE} \begin{gathered} D^{m+\alpha}_{*}u(t)=t^{-\gamma}f(t,u(t)) \quad \text{for a.e. }t, \\ u(0)=u_0, u'(0)=u_1, \dots, u^{(m)}(0)=u_{m}. \end{gathered} \end{equation} Moreover, $I^{1-\alpha}(u-T_{m}(u)) \in AC^{m}$. Thirdly, if $u \in C^{m}[0,T]$ and $I^{1-\alpha}(u-T_{m}(u)) \in AC^{m}$ and if $u$ satisfies \eqref{eqmodCapDE}, then $u$ satisfies \eqref{eqCapVIEm}. \end{theorem} \begin{proof} Let $g(t):=t^{-\gamma}f(t,u(t))$, $t\in (0,T]$, and note that $g \in L^1$ since $0 \le \gamma \le \alpha <1$. First we show the result for the special case $m=0$. So suppose that $u \in AC$ and $D^{\alpha}_{C} u(t)=t^{-\gamma}f(t,u(t))=g(t)$, for a.e.\ $t\in (0,T]$. Then, by the definition of Caputo derivative, we have $I^{1-\alpha}(Du)=g$ where $Du \in L^1$ since $u \in AC$. This yields $I^{\alpha}I^{1-\alpha}(Du)=I^{\alpha}g$, hence, by Lemma~\ref{RLintsgp}, $I(Du)=I^{\alpha}g$, that is $u(t)-u(0)=I^{\alpha}g$ since $u \in AC$. Now we consider the case $m>0$. Let $u$ with $D^{m}u \in AC$ satisfy \eqref{eqCapDEm}. Then using Lemma~\ref{lemhigherorder} we have a.e., \begin{align*} D^{m+\alpha}_{C}u=g &\implies D^{\alpha}_{C}(D^{m}u)=g, \\ &\implies D^{m}u=u_{m}+I^{\alpha}g, \quad\text{by the special case just proved}. \end{align*} Integrating $m$ times gives $u=\sum_{k=0}^{m}\frac{t^{k}}{k!}u_{k} + I^{m+\alpha}g$. Secondly, let $u$ be continuous and suppose that $u(t)=\sum_{k=0}^{m}\frac{t^{k}}{k!}u_{k} + I^{m+\alpha}g(t)$ with $g(t)=t^{-\gamma}f(t,u(t))$. To verify the initial conditions we observe that, for every $\beta \ge \alpha>\gamma$, setting $s=t{\sigma}$, \begin{align*} I^{\beta}g(t) &= \frac{1}{\Gamma(\beta)}\int_0^{t} (t-s)^{\beta-1}s^{-\gamma}f(s,u(s))\,ds,\\ &=\frac{1}{\Gamma(\beta)}\int_0^{1} t^{\beta-\gamma}(1-\sigma)^{\beta-1} {\sigma}^{-\gamma}f(t\sigma,u(t\sigma))\,d{\sigma} \end{align*} exists for every $t$ since $\int_0^1 (1-\sigma)^{\beta-1}{\sigma}^{-\gamma}\,d{\sigma}=B(\beta,1-\gamma)$, and $f(t\sigma,u(t\sigma))$ is bounded by continuity of $u$ and $f$. We also see that $I^{\beta}g(t)$ is a continuous function of $t \in [0,T]$. In particular, $I^{\alpha}g(t)$ is continuous and therefore $I^{m+\alpha}g \in C^{m}$, hence also $u \in C^{m}$. Furthermore for $\beta>\gamma$ we have \[ I^{\beta}g(0)=\lim_{t \to 0+} I^{\beta}g(t) = \frac{1}{\Gamma(\beta)}\lim_{t \to 0+} t^{\beta-\gamma} \int_0^{1} (1-\sigma)^{\beta-1}{\sigma}^{-\gamma}f(t\sigma,u(t\sigma))\,d{\sigma}=0, \] and then taking $\beta=m+\alpha$ in the equation $u(t)=\sum_{k=0}^{m}\frac{t^{k}}{k!}u_{k} + I^{m+\alpha}g(t)$ we first obtain $u(0)=u_0$. By differentiation we have \[ u'(t)=\sum_{k=0}^{m-1}\frac{t^{k}}{k!}u_{k+1}+I^{m-1+\alpha}g(t), \] and taking $\beta=m-1+\alpha$ we obtain $u'(0)=u_1$. Similarly, differentiating and evaluating we obtain $D^{n}u(0)=u_{n}$ for $n=1, \dots, m$. Then we have \[ D^{m+\alpha}_{*}u=D^{m+\alpha}(u-T_{m}u) =D^{m+\alpha}\bigl(u-\sum_{k=0}^{m}\frac{t^{k}}{k!}u_{k}\bigr) =D^{m+\alpha}I^{m+\alpha}g=g. \] This shows that $D^{m+1}(I^{1-\alpha}(u-T_{m}(u)))=g$ a.e.. Since we have \[ I^{1-\alpha}(u-T_{m}(u))=I^{1-\alpha}I^{m+\alpha}g=I^{m+1}g=I^{m}Ig \] it follows that $D^{m}(I^{1-\alpha}(u-T_{m}(u))) =Ig\in AC$, that is, $I^{1-\alpha}(u-T_{m}(u)) \in AC^{m}$. Thirdly, for $g(t)=t^{-\gamma}f(t,u(t))$, $g \in L^1$, since $I^{1-\alpha}(u-T_{m}(u)) \in AC^{m}$ the expression $D^{m+1}(I^{1-\alpha}(u-T_m(u)))(t)=g(t)$ can be integrated $m+1$ times to get \[ I^{1-\alpha}(u-T_m(u))(t)=I^{m+1}g(t)+a_0+a_1t+ \dots +a_{m}t^{m}, \text{ for constants } a_{i}. \] Applying $I^{\alpha}$ yields \[ I(u-T_m(u))=I^{m+1+\alpha}g+b_0t^{\alpha}+b_1t^{1+\alpha}+ \dots +b_{m}t^{m+\alpha}, \] for constants $b_{i}$ whose precise values are not important here. Since $u \in C^m$ we have $I(u-T_m(u)) \in C^{m+1}$ and $I^{m+1+\alpha}g=I^{m+\alpha}Ig \in C^{m+1}$ and therefore we must have $b_{i}=0$ for every $i$. Then, we may differentiate to get $u-T_m(u)=I^{m+\alpha}g$. \end{proof} \begin{remark}\label{remmodCequiv} \rm It has often been asserted (when $\gamma=0$) that \eqref{eqCapDEm} is equivalent to \eqref{eqCapVIEm}, but it seems that the absolute continuity of solutions $u$ of \eqref{eqCapVIEm} has never been shown when $f$ is at best continuous. The proved equivalence is: \begin{quote} if $u \in C^{m}[0,T]$ and $I^{1-\alpha}(u-T_{m}(u)) \in AC^{m}$ then $u$ satisfies \eqref{eqmodCapDE} if and only if $u$ satisfies \eqref{eqCapVIEm}. \end{quote} Some of these issues are discussed in the paper \cite{lan-lin-caa} where some positive results for boundary value problems involving the fractional derivative $D^{1+\alpha}_{C}u$ are obtained under a Lipschitz condition on $f$. For some more positive results see \S\ref{s-reg} below. It is important to have $I^{1-\alpha}(u-T_m(u)) \in AC^{m}$ in the third part of the theorem otherwise the integration in the last part is not valid. This is often implicit in the, often undefined, notion of `solution' for the problem, for if $D^{m+\alpha}_{*} u$ exists and $D^{m+\alpha}_{*} u(t)=t^{-\gamma}f(t,u(t))$ when $u,f$ are continuous then $t^{-\gamma}f(t,u(t)) \in L^1$ so that $D^{m+1}I^{1-\alpha}(u-T_m(u)) \in L^1$ and therefore $D^{m}I^{1-\alpha}(u-T_m(u)) \in AC$. When the term $t^{-\gamma}$ is absent and \eqref{eqmodCapDE} is satisfied then $D^{m+1}I^{1-\alpha}(u-T_m(u))$ is continuous so $I^{1-\alpha}(u-T_m(u)) \in C^{m+1}$. \end{remark} \begin{remark} \rm The case $m=0$ is proved in the recent paper \cite{jw-jmaa}. When there is no singular term $t^{-\gamma}$ and $f$ is continuous, Diethelm \cite[Lemma 6.2] {diethelm-bk} proves the equivalence in the form that $u \in C[0,h]$ is solution of $D^{m+\alpha}_{*}u(t)=f(t,u(t))$ with initial conditions $D^{k}u(0)=u_{k}, k=0, 1,\dots, m$, if and only if $u$ is solution of \[ u(t)=\sum_{k=0}^{m} \frac{t^k}{k!}u_{k}+I^{m+\alpha}f(t). \] There it is implicit that a solution of the fractional differential equation is a function for which $D^{m+\alpha}_{*}u$ exists, which requires more than continuity of $u$. \end{remark} \section{Initial value problems for the Riemann-Liouville fractional derivative}\label{s-IVP-RL} It has often been stated imprecisely that the fractional differential equation with the Riemann-Liouville fractional derivative is equivalent to an integral equation. One such statement is as follows. Assume that $\alpha > 0$. Then $u$ satisfies $D^{\alpha}u=f$ if and only if \[ u(t) = I^{\alpha}f(t) + c_1t^{\alpha-1} + c_2t^{\alpha-2} + \dots + c_nt^{\alpha-n},\quad c_i \in \mathbb{R}, \; i = 1, 2, \dots n. \] where $n =\lceil \alpha \rceil$, the smallest integer greater than or equal to $\alpha$. The difficulty is that it is not stated in what class of functions the solution is sought, one can consider $f \in L^1$ and seek solutions $u \in L^1$ or seek solutions in the weighted space $C_{\alpha-n}$, where the space $C_{\gamma}$ was defined in \S\ref{s-tRLInt}. Seeking solutions in the space $C[0,T]$ can only be done for special cases since $c_{n}=0$ is then necessary irrespective of any condition on $u(0)$, and then only $u(0)=0$ is a consistent value. \subsection{R-L derivative of order $0<\alpha<1$} For $0<\alpha<1$ a well posed initial value problem for the equation $D^{\alpha}u=f$ with $f \in L^1$ is given by the following Proposition. \begin{proposition}\label{proprlex} Let $f \in L^1[0,T]$. Then a function $u \in L^1$ such that $I^{1-\alpha}u \in AC$ satisfies $D^{\alpha}u(t)=f(t) \;\text{a.e.}$ and $I^{1-\alpha}u(0)=c\Gamma(\alpha)$ if and only if $u(t)=ct^{\alpha-1}+I^{\alpha}f(t)$ a.e., where $c=I^{1-\alpha}u(0)/\Gamma(\alpha)$. \end{proposition} \begin{proof} Let $u\in L^1$ and suppose that $I^{1-\alpha}u \in AC$ and $D^{\alpha}u(t)=f(t)$ a.e. Thus $D(I^{1-\alpha}u)(t)=f(t)$ a.e.\ and since $I^{1-\alpha}u \in AC$ and $f \in L^1$ we may integrate to get $I^{1-\alpha}u(t)=a+If$ where $a=I^{1-\alpha}u(0)$. Applying the operator $I^{\alpha}$ gives $Iu(t)=at^{\alpha}/\Gamma(1+\alpha)+I^{1+\alpha}f(t)$. Differentiating these $AC$ functions gives $u(t)=ct^{\alpha-1}+I^{\alpha}f(t)$ a.e.\ where $c=I^{1-\alpha}u(0)/\Gamma(\alpha)$. Conversely, if $u(t)=ct^{\alpha-1}+I^{\alpha}f(t)$ a.e.\ then $u \in L^1$ and $I^{1-\alpha}u(t)=c\Gamma(\alpha)+If(t)$ so $I^{1-\alpha}u \in AC$ and $I^{1-\alpha}u(0)=c\Gamma(\alpha)$. Moreover $D(I^{1-\alpha}u)(t)=f(t)$ a.e. \end{proof} Proposition~\ref{proprlex} is stated in an equivalent form, also for the higher order case, as \cite[Lemma 2.5(b)]{kst} and was proved in \cite[Theorem 2.4]{skm}. it is also proved, under some slightly different hypotheses, in \cite[Theorems 4.10 and 5.1]{BBPejqtde2015}, see Proposition~\ref{propBBP} below. When $f$ depends also on $u$ the result takes the following form. \begin{theorem}\label{thmrlordfu} Let $u \in L^1$ be such that $I^{1-\alpha}u \in AC$ and suppose that $t \mapsto f(t,u(t)) \in L^1$. Then $D^{\alpha}u(t)=f(t,u(t))$ a.e.\ and $I^{1-\alpha}u(0)=c\Gamma(\alpha)$ if and only if $u \in L^1$ with $t \mapsto f(t,u(t))\in L^1$ satisfies $u(t)=ct^{\alpha-1}+I^{\alpha}f(t,u(t))$ a.e., where $c=I^{1-\alpha}u(0)/\Gamma(\alpha)$. \end{theorem} The `initial condition' is often given in terms of $\lim_{t\to 0+} u(t)t^{1-\alpha}$, when this exists. The result is proved for $\alpha \in \mathbb{C}$ with $0<\operatorname{Re}(\alpha)<1$ in \cite[Lemma 3.2]{kst}. \begin{lemma}\label{limimplim} Let $0<\alpha<1$ and suppose that $u \in L^1$. Then \[ \lim_{t\to 0+} u(t)t^{1-\alpha}=c \;\text{implies that} \; \lim_{t\to 0+}I^{1-\alpha}u(t)=c\Gamma(\alpha). \] \end{lemma} \begin{proof} For $\varepsilon>0$ there exists $\delta>0$ such that $|u(t)t^{1-\alpha}-c|<\varepsilon$ for $|t|<\delta$. Then for $|t|<\delta$ we have \[ I^{1-\alpha}u(t)-c\Gamma(\alpha) =\frac{1}{\Gamma(1-\alpha)}\int_0^t (t-s)^{-\alpha}(u(s)-s^{\alpha-1}c)\,ds, \] hence \begin{align*} |I^{1-\alpha}u(t)-c\Gamma(\alpha)| & \le\frac{1}{\Gamma(1-\alpha)}\int_0^t (t-s)^{-\alpha}s^{\alpha-1}\varepsilon\,ds\\ &=\frac{\varepsilon}{\Gamma(1-\alpha)}\int_0^1 (1-\sigma)^{-\alpha}\sigma^{\alpha-1}\,d\sigma=\Gamma(\alpha)\varepsilon. \end{align*} Since $\varepsilon>0$ is arbitrary this proves that $\lim_{t\to 0+}I^{1-\alpha}u(t)=c\Gamma(\alpha)$. \end{proof} The following Proposition is given in \cite[Theorem 6.2]{BBPejqtde2015}. \begin{proposition}\label{propBBP} Let $0 < \alpha < 1$, and let $f$ be continuous on $(0,T] \times J$ where $J\subset \mathbb{R}$ is an unbounded interval. If $u$ is continuous on $(0,T]$ and $u$, $t \mapsto f(t,u(t))$ belong to $L^1[0,T]$, then $u$ satisfies the initial value problem, \begin{equation}\label{eqBBPde} D^{\alpha}u(t) = f (t, u(t)), \;\; t \in (0,T], \;\lim_{t \to 0+}t^{1-\alpha} u(t) = u^{0}, \end{equation} if and only if it satisfies the Volterra integral equation \begin{equation}\label{eqBBPint} u(t) = u^{0}t^{\alpha-1} +\frac{1}{\Gamma(\alpha)} \int_0^{t} (t-s)^{\alpha-1} f(s, u(s)) \,ds, \;t \in (0, T]. \end{equation} \end{proposition} \begin{remark} \rm This is somewhat different to Theorem~\ref{thmrlordfu} since in Proposition~\ref{propBBP} it is assumed that $D^{\alpha}u(t)$ exists for every $t \in (0,T]$ and functions that are in $L^1 \cap C(0,T]$ are considered, as opposed to supposing that functions are in $L^1$ and $D^{\alpha}u(t)$ exists a.e.\ in Theorem~\ref{thmrlordfu}. The converse of Lemma \ref{limimplim} is claimed in \cite[Theorem 6.1]{BBPejqtde2015} but this is not clear as the proof uses L'H\^{o}pital's rule which assumes the limit exists whose existence is to be shown. Note that if $\lim_{t\to 0+}I^{1-\alpha}u(t)=c\Gamma(\alpha)$ and $c \ne 0$ then it is necessary that $u \not \in L^{p}$ for every $p>1/(1-\alpha)$ by Proposition~\ref{propfintprop} part \eqref{rlint3}. However, in \cite{BBPejqtde2015} the result needed for solutions of \eqref{eqBBPint} does hold since $I^{1-\alpha}I^{\alpha}f(t)=If(t) \to 0$ as $t \to 0+$ for $f \in L^1$. \end{remark} \begin{remark} \rm Comparing Proposition~\ref{propBBP} with Theorem~\ref{thmrlordfu} implies that the hypotheses of Proposition~\ref{propBBP} should imply that $I^{1-\alpha}u \in AC$. In fact, if $u \in L^1$ and is a solution of \eqref{eqBBPint} then writing $g(t)=f(t,u(t))$ we have $g \in L^1$ and \[ I^{1-\alpha}u(t)= I^{1-\alpha}\bigl(u^{0}t^{\alpha-1} + I^{\alpha}g(t)\bigr)=\Gamma(\alpha)u^{0}+Ig(t) \in AC. \] If $u$ satisfies \eqref{eqBBPde} then $D(I^{1-\alpha}u)$ is continuous on $(0,T]$, and by Lemma~\ref{limimplim}, $\lim_{t \to 0+}I^{1-\alpha}u(t)=u^{0}\Gamma(\alpha)$ so $I^{1-\alpha}u$ satisfies the hypotheses of Proposition~\ref{propac} and is therefore $AC$. \end{remark} \begin{remark} \rm An early paper on the Riemann-Liouville IVP $D^{\alpha}u =f (t, u)$ is that of Delbosco and Rodino \cite{delbosco-rodino} who studied continuous solutions. In some cases it is implicit, but not explicit, that $u(0)=0$. They also study the problem when $f$ is replaced by $t^{-\gamma}f$ under a Lipschitz condition on $f$. \end{remark} \subsection{R-L derivative of order $1<\beta<2$} It is obvious that if $\beta=1+\alpha$ with $0<\alpha<1$ and $D^{\beta}u=D^{1+\alpha}u$ exists then $D^{1+\alpha}u=D(D^{\alpha}u)$. We have the following result. \begin{theorem}\label{rlequord2} Let $f \in L^1[0,T]$. Then $u \in L^1$ such that $D^{\alpha}u=D(I^{1-\alpha}u) \in AC$ satisfies $D^{1+\alpha}u(t)=f(t)$ a.e.\ and $I^{1-\alpha}u(0)=c_1\Gamma(\alpha)$ and $D^{\alpha}u(0)=c_2$ if and only if $u(t)=c_1t^{\alpha-1}+c_2t^{\alpha}+I^{1+\alpha}f(t)$ a.e., where $c_1=I^{1-\alpha}u(0)/\Gamma(\alpha)$ and $c_2=D^{\alpha}u(0)/\Gamma(1+\alpha)$. \end{theorem} \begin{proof} Let $u\in L^1$ and suppose that $D^{\alpha}u \in AC$ and $D^{1+\alpha}u(t)=f(t)$ a.e. Thus $D(D^{\alpha}u)(t)=f(t)$ a.e.\ and since $D^{\alpha}u \in AC$ and $f \in L^1$ we may integrate to get $D^{\alpha}u(t)=a_2+If(t)$ for all $t$, where $a_2=D^{\alpha}u(0)$. Integrating again gives \[ I^{1-\alpha}u(t)=a_1+a_2t+I^2f(t), \;\text{where} \;a_1=I^{1-\alpha}u(0). \] Applying the operator $I^{\alpha}$ gives \[ Iu(t)=a_1t^{\alpha}/\Gamma(1+\alpha)+a_2t^{1+\alpha}/\Gamma(2+\alpha)+I^{2+\alpha}f(t). \] Differentiating these $AC$ functions gives \[ u(t)=a_1t^{\alpha-1}/\Gamma(\alpha)+a_2t^{\alpha}/\Gamma(1+\alpha)+I^{1+\alpha}f(t), \quad \text{for a.e. } t. \] Conversely, if $u(t)=a_1t^{\alpha-1}/\Gamma(\alpha)+a_2t^{\alpha}/ \Gamma(1+\alpha)+I^{1+\alpha}f(t)$ a.e.\ then $u \in L^1$ and $I^{1-\alpha}u(t)=a_1+a_2t+I^2f(t)$ so $I^{1-\alpha}u(0)=a_1$ and $D^{\alpha}u=D(I^{1-\alpha}u)=a_2+If \in AC$ so that $D(I^{1-\alpha}u)(0)=D^{\alpha}u(0)=a_2$. Moreover $D^{1+\alpha}u=D(D^{\alpha}u)=f$ a.e. \end{proof} When $f$ depends on $u$ the result is as follows. \begin{theorem}\label{rleqord2fu} Let $u \in L^1$ be such $D^{\alpha}u \in AC$ and that $t \mapsto f(t,u(t)) \in L^1$. Then $D^{1+\alpha}u(t)=f(t,u(t)) \;\text{a.e.}$, $I^{1-\alpha}u(0)=c_1\Gamma(\alpha)$ and $D^{\alpha}u(0)=c_2$ if and only if $u(t)=c_1t^{\alpha-1}+c_2t^{\alpha}+I^{1+\alpha}f(t,u(t))$ a.e., where $c_1=I^{1-\alpha}u(0)/\Gamma(\alpha)$ and $c_2=D^{1-\alpha}u(0)$. \end{theorem} We note that to prove this equivalence it is required that $D^{\alpha}u \in AC$, equivalently $I^{1-\alpha}u \in AC^1$. Also if we ask that $t \mapsto f(t,u(t)) \in L^1$ for \emph{every} $u \in L^1$ then it is known that a necessary and sufficient condition is $|f(t,u)| \le a(t)+b|u|$ for all $(t,u)$ for some $a \in L^1$ and some constant $b\ge 0$. \subsection{RL derivative of higher order} For the RL derivative of order $m+\alpha$ where $m \in\mathbb{N}$ and $0<\alpha<1$ the result corresponding to Theorem ~\ref{rleqord2fu} is the following and can be proved in the same way. \begin{theorem}\label{rleqordmfu} Let $u \in L^1$ be such $D^{m-1+\alpha}u=D^{m}(I^{1-\alpha}u) \in AC$ (that is $I^{1-\alpha}u \in AC^{m}$) and suppose that $t \mapsto f(t,u(t)) \in L^1$. Then $u$ satisfies \begin{gather*} D^{m+\alpha}u(t)=f(t,u(t)) \;\text{a.e.},\\ I^{1-\alpha}u(0)=b_{m+1}\Gamma(\alpha), \quad\text{and}\quad D^{m+\alpha-k}u(0)=b_{k}, \; k=1, \dots, m, \end{gather*} if and only if \[ u(t)=\sum_{k=1}^m b_{k}t^{m+\alpha-k}+I^{m+\alpha}f(t,u(t)) \; \text{a.e.}, \] where $b_{m+1}=I^{1-\alpha}u(0)/\Gamma(\alpha)$ and $ b_{k}=D^{m+\alpha-k}u(0)$. \end{theorem} This result is essentially given in Theorem 3.1 of \cite{kst} where it is assumed that $f$ is continuous on $(0,T) \times G$ ($G$ an open set in $\mathbb{R}$) satisfying $f(t,u) \in L^1$ for every $u \in G$. The proof claims that the necessary $AC$ property holds but the results cross referenced seem to be the wrong ones as they do not seem to prove this. Under a Lipschitz condition on $f$ it is also essentially given in \cite[Lemma 5.2]{diethelm-bk}. \section{Monotonicity and concavity properties of fractional derivatives}\label{s-concavity} The Caputo and R-L derivatives are nonlocal, that is, the fractional derivative at a point $t$ depends on the values of $u(s)$ for all $s \in [0,t]$. However, the Caputo derivative, in particular, has some similarities with derivatives of integer powers. For example if $u$ is non-decreasing on $[0,T]$ then for any one $\alpha \in (0,1)$, both $D^{\alpha}_{*}u(t)$ and $D^{\alpha}u(t)$ are nonnegative for a.e.\ $t$ but the converse is not true as we show below, though it is falsely claimed in the recent paper \cite{q-y}. The correct result implying monotonicity is given in \cite{diethelm-fcaa-16}, see Remark~\ref{rem-monotone} below. Similarly if $u \in C^2$ and $u''(t)\le 0$, that is $u$ is concave, then $D^{1+\alpha}_{*}u(t)$ is non-positive but the converse does not hold. We first note the following simple fact, we include the proof for completeness. \begin{lemma}\label{lemACinc} If $u \in AC[0,T]$ then $u$ is non-decreasing if and only if $u'(t) \ge 0$ for a.e.\ $t \in [0,T]$. \end{lemma} \begin{proof} Since $u \in AC$ the derivative $u'$ exists for a.e.\ $t$. Suppose that $u$ is non-decreasing and that $u'(\tau)$ exists at a point $\tau \in (0,T)$. For $h \ne 0$ sufficiently small we have $\dfrac{u(\tau+h)-u(\tau)}{h} \ge 0$. Taking the limit as $h \to 0$ shows that $u'(\tau) \ge 0$. Conversely, suppose that $u'(t) \ge 0$ for a.e.\ $t$. Since $u \in AC$ we have $u(t)-u(0)=\int_0^t u'(s)\,ds$. Then for $t>\tau$, $u(t)-u(\tau)=\int_{\tau}^t u'(s) \,ds \ge 0$, that is $u$ is non-decreasing. \end{proof} \begin{proposition}\label{prop-mon} Let $0<\alpha<1$ and let $0 \le \gamma<\alpha$. Suppose that $u,v$ are continuous and that $I^{1-\alpha}u \in AC$, $I^{1-\alpha}v \in AC$ so that $D^{\alpha}_{*}u$ and $D^{\alpha}_{*}v$ exist a.e. \begin{enumerate} \item\label{mon1} If $u$ is non-decreasing then $D^{\alpha}_{*}u(t) \ge 0$ and $D^{\alpha}u(t) \ge 0$ for a.e.\ $t \in [0,T]$. \item\label{mon2} If $D^{\alpha}_{*}u \in C_{-\gamma}$ and if $D^{\alpha}_{*}u(t) \ge 0$ for $t>0$ then $u(t) \ge u(0)$ for every $t \in [0,T]$. If $D^{\alpha}_{*}u(t) > 0$ for $t>0$ then $u(t)> u(0)$ for every $t \in (0,T]$. \item\label{mon3} If $D^{\alpha}_{*}u \in C_{-\gamma}$, $D^{\alpha}_{*}v \in C_{-\gamma}$ and $D^{\alpha}_{*}u(t) \ge D^{\alpha}_{*}v(t)$ for $t>0$ and $u(0) \ge v(0)$ then it follows that $u(t) \ge v(t)$ for $t \in [0,T]$. If $D^{\alpha}_{*}u(t)> D^{\alpha}_{*}v(t)$ for $t>0$ and $u(0) \ge v(0)$ then $u(t)>v(t)$ for all $t \in (0,T]$. \item\label{mon4} If, for some $\alpha \in (0,1)$, we have $D^{\alpha}_{*}u(t) \ge 0$ for $t \in [0,T]$ (or if $D^{\alpha}u(t) \ge 0$) it does not follow that $u$ is non-decreasing even if $D^{\alpha}_{*}u$ or $u$ is a $C^{\infty}$ function. \end{enumerate} \end{proposition} \begin{proof} \eqref{mon1} Let $u_0$ be the constant function taking the value $u(0)$. By definition, $D^{\alpha}_{*}u(t)=DI^{1-\alpha}(u-u_0)(t)$ and $ I^{1-\alpha}(u-u_0)(t)$ is a non-decreasing function of $t$ by Proposition~\ref{propfintprop}\,\eqref{rlint9}. Thus, by Lemma~\ref{lemACinc}, its derivative $D^{\alpha}_{*}u$ is $\ge 0$ a.e. \eqref{mon2} Let $g(t):=D^{\alpha}_{*}u(t)$, then, by assumption, $g(t):=t^{-\gamma}f(t)$ where $f$ is continuous. It follows from Theorem~\ref{ivporderm} part 3 that we have \[ u(t)=u(0)+\int_0^t (t-s)^{\alpha-1} g(s)\,ds \] therefore $u(t) \ge u(0)$ since $g(s) \ge 0$ for a.e.\ $s >0$. When the inequality is strict, we have $(t-s)^{\alpha-1} g(s)>0$ for a.e.\ $s \in (0,t)$, so the integral is positive for $t>0$, thus $u(t)>u(0)$ for $t>0$. \par \eqref{mon3} We have $D^{\alpha}_{*}(u-v) \ge 0$ so $u(t)-v(t) \ge u(0)-v(0) \ge 0$ by part (2). The strict inequality case is proved in the same way. \eqref{mon4} We give two simple counter-examples, it is easy to give many other similar ones. We note that for for a function $u \in AC$ with $u(0)=0$ the following equality holds: $D^{\alpha}u(t)=D^{\alpha}_{*}u(t)= D^{\alpha}_{C}u(t)$. For $p>0$, the function $h_{p}(t):=t^p$ satisfies $h_{p} \in AC$ and $h_{p}(0)=0$ and we will use the well-known and easily verified fact that for $0<\alpha<1$ \[ D^{\alpha}_{C}h_{p}(t) =\frac{t^{p-\alpha}\Gamma(p+1)}{\Gamma(p+1-\alpha)}. \] Let $\alpha=1/2$ and $T=1$ and firstly let $u(t)=5t^{3/2}-4t^{5/2}$. Then $u \in C^1$ and we obtain \[ D^{1/2}_{C}u(t)=5\Gamma({5}/{2})(t-t^2) =\dfrac{15\sqrt{\pi}}{4}(t-t^2)\ge 0, \quad\text{for } t \in [0,1], \] and $D^{1/2}_{C}u$ is a $C^{\infty}$ function, but $u(t)$ is increasing on $[0,3/4]$ and decreasing on $[3/4,1]$. Secondly, for $\alpha=1/2$ and $T=1$ take $v(t)=6t^{2}-5t^{3}$ so $v \in C^{\infty}$. By the above formulas we obtain $D^{1/2}_{C}v(t)=16(t^{3/2}-t^{5/2})/\sqrt{\pi} \ge 0$ for $t \in[0,1]$ but $v$ is increasing on $[0,4/5]$ and decreasing on $[4/5,1]$. \end{proof} \begin{remark}\label{rem-monotone} \rm The results with $\gamma \ne 0$ may be new. Diethelm~\cite{diethelm-fcaa-16} discusses monotonicity properties for $C^1[0,T]$ functions in terms of $D^{\alpha}_{C}u$. He proves part (1) for $C^1$ functions in his Theorem 2.1; the proof is also simple with the definition $D^{\alpha}_{C}u$ available. He further shows that monotonicity of $u \in C^1$ is equivalent to requiring that there is $\alpha_0 \in (0,1)$ such that the Caputo derivatives $D^{\alpha}_{C}u$ of orders $\alpha \in (\alpha_0,1)$ do not change sign, and this is equivalent to the Caputo derivatives $D^{\alpha}_{C}u$ of all orders $\alpha \in (0, 1)$ do not change sign. For part~\eqref{mon4} he gives a different example to ours where $D^{\alpha}_{*}u(t)$ does not change sign for some, but not all, $\alpha$ and $u$ is not monotone. Theorem 2.4 in \cite{d-b-s} partially proves part~\eqref{mon3} when $\gamma = 0$ for the fractional derivative $D^{\alpha}_{C}$ but omits to state that $f$ must be continuous and that $u$ should be in $AC$. The proof also claims the fractional differential and integral equations are \emph{equivalent} but the correct equivalence is given in Theorem~\ref{ivporderm}, however the implication of part 1 of Theorem~\ref{ivporderm} is sufficient there. \end{remark} We now discuss concavity properties, for simplicity we consider functions defined on $[0,1]$. If $0<\alpha<1$ is given and $D_{*}^{1+\alpha}u(t)=g(t)$ we might hope that $g \le 0$ implies that $u$ is concave but this is not valid. \begin{proposition} If $0<\alpha<1$, $u \in AC[0,1]$, $I^{1-\alpha}u' \in AC$, and $u$ is concave, that is $u'(t)$ is non-increasing, then $D_{*}^{1+\alpha}u(t) \le 0$. But $D_{*}^{1+\alpha}u(t) \le 0$ (for a fixed $\alpha$) does not imply that $u$ is concave even if $u \in C^2[0,1]$ or $D_{*}^{1+\alpha}u \in C^2[0,1]$ (or more regular). \end{proposition} \begin{proof} The assumption $I^{1-\alpha}u' \in AC$ ensures that the fractional derivative $D_{*}^{1+\alpha}u$ exists. Then by Lemma~\ref{lem-high-ord-star} we have \[ D_{*}^{1+\alpha}u(t)=D^{\alpha}(u'-u'(0))(t)=D(I^{1-\alpha}(u'-u'(0)))(t) \le 0 \] by Proposition~\ref{propfintprop} \eqref{rlint9} since $u'(t)-u'(0)$ is non-increasing. For the second part, we take $\alpha=1/2$ and note that for $h_{p}(t)=t^p$ we have \[ D^{3/2}_{C}h_{p}(t)= \begin{cases} 0, &\text{if } p=0 \;\text{or if}\;p=1,\\ \frac{\Gamma(p+1)}{\Gamma(p-1/2)}t^{p-3/2}, &\text{if } p>1. \end{cases} \] Firstly we take the following $C^{\infty}$ function $u(t)=a+bt+t^3-2t^2$ where $a,b$ can be arbitrary constants. Then by the above formula $D^{3/2}_{C}u(t)=\frac{8}{\sqrt{\pi}}(t^{3/2}-t^{1/2}) \le 0$ but $u''=6t-4$ changes sign at $t=2/3$ so $u$ is not concave, also not convex, on $[0,1]$. Secondly take, $u(t)=a+bt-\dfrac{8t^{5/2}}{15\sqrt{\pi}}+\dfrac{32t^{7/2}}{105\sqrt{\pi}}$ so that $D^{3/2}_{C}u(t)=-t+t^2 \le 0$. Then $u''(t)=-2t^{1/2}/\sqrt{\pi}+8t^{3/2}/(3\sqrt{\pi})$ which changes sign at $t=3/4$ so $u$ is neither convex nor concave. \end{proof} \begin{remark} \rm Using Diethelm's result on monotonicity \cite{diethelm-fcaa-16} mentioned above applied to $u'$, it follows that if $u \in C^2[0,1]$ then $u$ is concave if and only if for \textbf{every} $\alpha \in (0,1)$ the fractional derivative $D^{1+\alpha}_{C}u(t)$ is non-positive for $t \in [0,1]$. The application of this result could be limited by the fact that a solution of a problem $D^{1+\alpha}_{*}u(t)=f(t,u(t))$ with $f$ continuous is not a $C^2[0,1]$ function in general, see \cite[Theorem 6.25] {diethelm-bk} and Theorem~\ref{ivporder2} below. \end{remark} There have been some papers which make incorrect claims regarding concavity properties. Ntouyas and Pourhadi~\cite{ntouyas-pourhadi} discuss the following BVP with $0<\alpha<1$ (we have changed to the interval $[0,1]$): \begin{equation}\label{eq-n-p} D^{1+\alpha}_{C}u(t)+g(t)=0, \;\;u(0)=0, u(1)=\beta u(\eta), \end{equation} where $\beta\eta \ne 1$. They claim the following. \begin{quote} \cite[Lemma 2.6]{ntouyas-pourhadi} Suppose that $g \in C^2([0,1];\mathbb{R})$ and $g(0) \ge 0$.\\ (a) If $g$ is convex, then the unique solution of \eqref{eq-n-p} is concave.\\ (b) If $g$ is concave, then the unique solution of \eqref{eq-n-p} is convex. \end{quote} Unfortunately this is not correct. We can take the same example as above with $\alpha=1/2$, that is consider $D_{C}^{3/2}u$ for $u(t)=a+bt-\dfrac{8t^{5/2}}{15\sqrt{\pi}}+\dfrac{32t^{7/2}}{105\sqrt{\pi}}$, taking $a=0$ and $b$ chosen to make $u(1)=\beta u(\eta)$, then $u$ satisfies $D_{C}^{3/2}u(t)+t-t^2=0$, and we have $g(t)=t-t^2$ which is concave and non-negative but the solution is neither convex nor concave. For example, if $\eta=1/2$ and we also ask for $u(t) \ge 0$ for $t \in[0,1]$ (the paper \cite{ntouyas-pourhadi} is interested in existence of positive solutions and assumes $\beta\eta<1$ in some Lemmas) then we need $b \gtrapprox 0.129$ and then for any for any $\beta \in [0.0016, 1.944]$ such $b$ exists, for example when $\beta=0.01, \eta=1/2$ we can take $b \approx 0.1292$, for $\beta=1, \eta=1/2$ we can take $b \approx 0.182$, or for $\beta=1.944, \eta=1/2$ we can take $b \approx 1.968$. Eloe and Neugebauer \cite{eloe-neug} discuss concavity properties of the R-L fractional derivative. One of their results, \cite[Theorem 3.7]{eloe-neug}, claims that if $\alpha \in (0,1)$, and $D^{1+\alpha}u$ is continuous with $u(0)=0$ and $u(1) \ge 0$, then, if $u \in C^2(0,1]$ (not necessarily at $0$), $D^{1+\alpha}u(t) \le 0$ implies that $u''(t) \le 0$ for $t \in (0,1]$. Unfortunately this is not correct. We again take $\alpha=1/2$ and consider \[ u(t)=bt^{1/2}-\dfrac{8t^{5/2}}{15\sqrt{\pi}}+\dfrac{32t^{7/2}}{105\sqrt{\pi}} \] and we can choose, for example, $b \ge 0.129$ to make $u(1) \ge 0$; a simple choice is $b=1/6$ which we now make. Note that, by properties of the R-L derivative, $D^{3/2}t^{1/2}=0$. Then we obtain $D^{3/2}u(t)=-t+t^2 \le 0$ but by calculation $u''(t)$ changes sign when $t \approx 0.794$, thus $u$ is neither concave nor convex. An error is that the `if' part of their Lemma 2.1 which is used in the proof of their Theorem 3.7 is not correct, a counter example to that implication on $[0,1]$ is $u(t)=t/(1+4t^2)$ where $u''$ changes sign at $t=\sqrt{3}/2$. We note that for the `only if' part of their Lemma 2.1 (for which there is a simpler proof directly using the definition of concavity) the condition $u(0) \ge 0$ is necessary for its validity; in the paper \cite{eloe-neug} the main results have $u(0)=0$. \section{Existence of Caputo IVPs with nonlinearity depending on a fractional derivative}\label{s-IVP-Exist} We will now consider the Caputo IVP: \[ D^{1+\alpha}_{*}u(t)=t^{-\gamma}f(t, u(t),D^{\beta}_{C}u(t)) \quad\text{for } t>0, \;u(0)=u_0, u'(0)=u_1, \] where $0 \le \gamma<\alpha<1$ and $0<\beta \le 1$. We believe our results here are new even for the case $\beta=1$ since we include the singular term $t^{-\gamma}$. For this problem it is natural to work in the space $C^1[0,T]$ in which we will use the norm $\|u\|_1:= \|u\|_{\infty}+\|u'\|_{\infty}$. Since $u \in C^1$ we have $D^{\beta}_{*}u(t)=D^{\beta}_{C}u(t)$ by Proposition~\ref{propcap=rl1}, but we do not necessarily have $D^{1+\alpha}_{*}u(t)=D^{1+\alpha}_{C}u(t)$. We write $D^{\beta}_{C}u$ since this is the form we use in the proof below. First we have the following equivalence. \begin{theorem} \label{ivpCapderivorder2} Let $0 \le \gamma<\alpha<1$ and $0<\beta \le 1$. Let $f$ be continuous on $[0,T]\times \mathbb{R}\times \mathbb{R}$. If a function $u \in C^1[0,T]$ is such that $D^{1+\alpha}_{*}u$ exists a.e. and satisfies the Caputo fractional initial value problem \begin{equation}\label{eqCderiv2g} D^{1+\alpha}_{*}u(t)=t^{-\gamma}f(t, u(t),D^{\beta}_{C}u(t)), \quad \text{for a.e. }t>0, \quad \;u(0)=u_0, \quad u'(0)=u_1. \end{equation} then $u$ satisfies the Volterra integral equation \begin{equation} \label{eqVIEderivord2} u(t)=u_0+u_1\,t+I^{1+\alpha}(t^{-\gamma}f(t,u(t),D^{\beta}_{C}u(t))). \end{equation} Conversely, if $u \in C^1[0,T]$ satisfies \eqref{eqVIEderivord2} then $D^{1+\alpha}_{*}u$ exists a.e.\ and $u$ satisfies \eqref{eqCderiv2g}. \end{theorem} \begin{proof} Let $g(t):=f(t,u(t),D^{\beta}_{C}u(t))$. First we note that for $u \in C^1$ the Caputo fractional derivative $D^{\beta}_{C}u$ is continuous. In fact, \[ D^{\beta}_{C}u=I^{1-\beta}u' \in C_{1-\beta} \subset C, \quad \text{by Proposition~\ref{propfintprop} \eqref{rlint5}}. \] Hence for $u \in C^1$, $g$ is continuous. The proof is now almost identical with that of Theorem~\ref{ivporderm} so is omitted. \end{proof} We will prove a global existence theorem under the growth assumption \[ |f(t,u,p)| \le a(t)+M(|u|+|p|) \quad\text{for some $a \in L^{\infty}$ and constant $M>0$}. \] Under an extra Lipschitz condition we will also prove continuous dependence on initial data, and a uniqueness result. The problem \[ u(t)=u_0+\frac{1}{\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}s^{-\gamma}f(s, u(s))\,ds \] when $0 \le \gamma < \alpha<1$ and $f$ is continuous was recently studied by this author in \cite{jw-jmaa} by employing a new $L^{\infty}$ Gronwall inequality which established an a priori bound for solutions of a problem with weakly singular kernel in the space $C[0,T]$. For problem \eqref{eqCderiv2g} the appropriate space is $C^1[0,T]$ and we must obtain a priori bounds on both the function $u$ and its derivative $u'$. We will use the Gronwall inequality from \cite{jw-jmaa} which we now recall in a simplified form. \begin{theorem} \label{t-first-Gron} Let $a \ge 0$ and $b>0$ be constants and suppose that $\eta>0$, $\gamma \ge 0$ and $ \eta+\gamma < 1$. Suppose that $u \in L^{\infty}_{+}[0,T]$ satisfies the inequality \begin{equation} \label{eq4.1} u(t) \le a+b \int_0^t (t-s)^{-\eta}s^{-\gamma}u(s)\,ds, \quad \text{for a.e. } t \in [0,T]. \end{equation} Then there is an explicit constant $B=B(b, \eta, \gamma)$ (see \cite{jw-jmaa} for details) such that \begin{equation} \label{eq4.2a} u(t) \le \frac{a(1-\gamma)}{1-\eta-\gamma} \exp\,\Bigl(B t^{1-\gamma}\Bigr), \quad \text{for a.e. } t \in [0,T]. \end{equation} In particular, there is an explicit constant $C=C(b,\eta,\gamma,T)$ such that $u(t) \le aC$ for a.e.\ $t \in [0,T]$. \end{theorem} When $u$ is continuous the inequality holds for all $t$ not only a.e.\ $t$. We will now discuss solutions in the space $C^1[0,T]$ of the Volterra integral equation \begin{equation} \label{eqVIEderivord2bis} u(t)=u_0+u_1\,t+I^{1+\alpha}(t^{-\gamma}f(t,u(t),D^{\beta}_{C}u(t))), \end{equation} where $0<\alpha<1$ and $0<\beta\le1$. If $u\in C^1$ is a solution then its derivative satisfies \begin{equation} \label{eqVIEderivbis} u'(t)=u_1+I^{\alpha}(t^{-\gamma}f(t,u(t),D^{\beta}_{C}u(t))). \end{equation} Our result reads as follows. \begin{theorem} \label{thm-exC} Let $f:[0,T]\times \mathbb{R}^{2}\to \mathbb{R}$ be continuous, let $0<\alpha<1$, $0 \le \gamma < \alpha$, and $0 < \beta \le 1$, and let $u_0,u_1 \in \mathbb{R}$. Suppose there are $a \in L^{\infty}$ and a constant $M>0$ such that $|f(t,u,p| \le a(t)+M(|u|+|p|)$ for all $t \in [0,T]$ and $u,p \in \mathbb{R}$. Then the integral operator \[ Nu(t):=u_0+tu_1+\frac{1}{\Gamma(1+\alpha)}\int_0^t (t-s)^{\alpha}s^{-\gamma}f(s, u(s),D^{\beta}_{C}u(s))\,ds \] has a fixed point in $C^{1}[0,T]$. If, in addition, there exists $L>0$ such that $f$ satisfies the Lipschitz condition \[ |f(t,u,p)-f(t,v,q)| \le L(|u-v|+|p-q|), \quad\text{for all }t \in [0,T], \;u,v,p,q \in \mathbb{R}, \] then fixed points depend continuously on the initial data and, in particular, for a given $u_0,u_1$ there is a unique fixed point. \end{theorem} \begin{proof} The proof of existence consists of showing that $N:C^1 \to C^1$ is completely continuous, that is $N$ is continuous and maps bounded subsets of $C^1$ into relatively compact subsets of $C^1$, and that there is a bounded open set $U$ containing $0$ such that $Nu \ne \lambda u$ for all $u \in \partial{U}$ and all $\lambda \ge 1$. This will prove the Leray-Schauder degree $\deg(N, U,0)=1$ which proves that $N$ has a fixed point in $U$. To show that $N$ is completely continuous we will show $N$ is continuous and will apply the Arzel\`{a}-Ascoli criterion to show that $N(U)$ is bounded and that $(Nu)(t)$ and $(Nu)'(t)$ are equicontinuous for $u \in U$. We will use the simple fact that \begin{equation} \label{eqdbetssmgam} s^{-\gamma}|D^{\beta}_{C}u(s)| \le I^{1-\beta}(s^{-\gamma}|u'(s)|) \quad\text{for a.e. } s. \end{equation} In fact, since $u'$ is continuous $s \mapsto s^{-\gamma}|u'(s)|$ is an $L^1$ function and we have \begin{align*} s^{-\gamma}|D^{\beta}_{C}u(s)| &=s^{-\gamma}|I^{1-\beta}u'(s)|\\ &\le\frac{1}{\Gamma(1-\beta)}s^{-\gamma}\int_0^s (s-\sigma)^{-\beta}|u'(\sigma)|\,d{\sigma}\\ &\le \frac{1}{\Gamma(1-\beta)}\int_0^s {\sigma}^{-\gamma}(s-\sigma)^{-\beta}|u'(\sigma)|\,d{\sigma}\\ &=I^{1-\beta}(s^{-\gamma}|u'(s)|). \end{align*} We now show the existence of a suitable set $U$. In fact, if there exists $\lambda \ge 1$ and $u \ne 0$ such that ${\lambda}u = Nu$ then ${\lambda}u(t) = Nu(t)$ and ${\lambda}u'(t) = (Nu)'(t)$, and we have \begin{align*} |u(t)| &\le {\lambda}|u(t)| =|Nu(t)| \le |u_0|+t|u_1|+I^{1+\alpha}s^{-\gamma}|f(s, u(s),D^{\beta}_{C}u(s))|\\ & \le |u_0| +T|u_1|+I^{1+\alpha}(s^{-\gamma}(a(s)+M(|u(s)|+|D^{\beta}_{C}u(s)|)\\ & \le C_0(a,u_0,u_1,T)+MI^{1+\alpha}(s^{-\gamma}|u(s)|)+MI^{1+\alpha}I^{1-\beta}|s^{-\gamma}u'(s)|\\ &=C_0(a,u_0,u_1,T)+MI^{1+\alpha}s^{-\gamma}|u(s)|+MI^{2+\alpha-\beta}|s^{-\gamma}u'(s)|. \end{align*} The constant $C_0$ can be given explicitly but since it is not important to us we omit the explicit value; we also do this with other constants below. Similarly we obtain \begin{align*} |u'(t)| & \le |u_1|+I^{\alpha}|f(s, u(s),D^{\beta}_{C}u(s))|\\ & \le C_1(a,u_1,T)+MI^{\alpha}(s^{-\gamma}|u(s)|) +MI^{1+\alpha-\beta}|s^{-\gamma}u'(s)|. \end{align*} Now we note that \begin{equation} \label{eqcompfrint} I^{\alpha_1}v \le C_2(\alpha_1,\alpha_2,T)I^{\alpha_2}v \quad\text{whenever } \alpha_1 \ge \alpha_2. \end{equation} In fact, \begin{align*} I^{\alpha_1}v(t) &=\frac{1}{\Gamma(\alpha_1)}\int_0^t (t-s)^{\alpha_1-1} v(s)\,ds\\ &=\frac{1}{\Gamma(\alpha_1)}\int_0^t (t-s)^{\alpha_2-1}(t-s)^{\alpha_1-\alpha_2} v(s)\, ds\\ &\le \frac{1}{\Gamma(\alpha_1)}T^{\alpha_1-\alpha_2}\int_0^t (t-s)^{\alpha_2-1} v(s)\, ds\\ &\le \frac{\Gamma(\alpha_2)}{\Gamma(\alpha_1)}T^{\alpha_1-\alpha_2}I^{\alpha_2}v. \end{align*} Adding the above inequalities and using \eqref{eqcompfrint} several times (noting that $\alpha$ is the smallest exponent) gives \begin{align*} |u(t)|+|u'(t)| &\le C_3(a,u_0,u_1,T)+C_{4}(a,u_0,u_1,T)I^{\alpha}s^{-\gamma}(|u(s)|+|u'(s)|)\\ &=C_3+C_{4}(a,u_0,u_1,T)\int_0^t (t-s)^{\alpha-1}s^{-\gamma}(|u(s)|+|u'(s)|)\,ds. \end{align*} Since $1-\alpha+\gamma<1$, by Theorem~\ref{t-first-Gron} there is a constant $C_{5}$, independent of $u$, such that $|u(t)|+|u'(t)| \le C_{5}$ for all $t \in [0,T]$. Choose $R>C_{5}$, and let $U_{R}$, be the open ball in $C^1$ of radius $R$ centred at $0$. We have shown that $Nu \ne \lambda u$ for all $u \in \partial{U}_{R}$ and all $\lambda \ge 1$. Now we show that $N(\overline{U}_{R})$ is bounded. We have $\|u\|_1 \le R$ so that $|u(t)|+|u'(t)| \le R$ for all $t \in [0,T]$, and we obtain \[ |D^{\beta}u(t)|=|I^{1-\beta}u'|\le \frac{1}{\Gamma(1-\beta)} \int_0^t (t-s)^{-\beta}|u'|(s)\,ds \le R_1:=\frac{RT^{1-\beta}}{\Gamma(2-\beta)}. \] As $f$ is uniformly continuous on $[0,T] \times [0,R]\times [0,R_1]$, there exists $M=M(R,\beta,T)<\infty$ such that $|f(t,u,p)| \le M$ for all $t\in [0,T]$, $u\in [0,R]$, $p\in [0,R_1]$. Thus for $u \in \overline{U}_{R}$ we have \begin{align*} |(Nu)'(t)| &\le |u_1|+\frac{1}{\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}s^{-\gamma}|f(s,u(s),D^{\beta}u(s))| \,ds \\ &\le |u_1|+\frac{1}{\Gamma(\alpha)}M\int_0^t (t-s)^{\alpha-1}s^{-\gamma} \,ds\\ &=|u_1|+\frac{1}{\Gamma(\alpha)}Mt^{\alpha-\gamma}B(\alpha, 1-\gamma) \\ &\le |u_1|+\frac{1}{\Gamma(\alpha)}MT^{\alpha-\gamma}B(\alpha, 1-\gamma). \end{align*} A similar calculation is valid for $Nu(t)$. This proves that the set $N(\overline{U}_{R})$ is bounded. The proof of equicontinuity of $(Nu)'(t)$ is essentially identical with the proof in \cite{jw-jmaa} since the property of $f$ used is the uniform bound, hence we omit this. The proof of equicontinuity for $\{Nu(y):u \in\overline{U}_{R}\}$ is similar but easier. The equicontinuity shows that $N$ maps bounded sets into relatively compact sets by the Arzel\`{a}-Ascoli theorem. Also it shows that $Nu(t)$ and $(Nu)'(t)$ are continuous in $t$ so $N(\overline{U}_{R}) \subset C^1$. Finally we show that $N$ is continuous on $C^1$, that is, $u_n(s) \to u(s)$ and $u_{n}'(s) \to u'(s)$ uniformly in $s$ implies $Nu_n(t) \to Nu(t)$ and $(Nu)'_n(t) \to (Nu)'(t)$ uniformly in $t$. For $\varepsilon>0$ there exists $n_0(\varepsilon)\in \mathbb{N}$ such that $|u_{n}'(s)-u'(s)|<\varepsilon$ for $n>n_0$ and all $s \in[0,T]$. Then we have \begin{align*} |D^{\beta}u_n(t)- D^{\beta}u(t)| &\le \frac{1}{\Gamma(1-\beta)}\int_0^t (t-s)^{-\beta}|u_{n}'(s)-u'(s)|\,ds\\ &<\frac{1}{\Gamma(1-\beta)}\varepsilon\int_0^t (t-s)^{-\beta}\,ds \le \frac{{\varepsilon}T^{1-\beta}}{\Gamma(1-\beta)}. \end{align*} As $\varepsilon>0$ is arbitrary this shows that $D^{\beta}u_n(t) \to D^{\beta}u(t)$ uniformly in $t$. Hence, since $\{u_n\}$ is bounded, say $\|u_n\|_1 \le R$, and $f$ is uniformly continuous on $[0,T] \times [0,R]\times [0,R_1]$, for $\varepsilon>0$ there exists $n_1 =n_1(\varepsilon)\in \mathbb{N}$ such that \[ |f(s, u_n(s), D^{\beta}u_n(s)(s))-f(s,u(s),D^{\beta}u(s)|< \varepsilon, \] for all $n>n_1$ and all $s \in [0,T]$. Then we have, for every $t \in [0,T]$, \begin{align*} & | (Nu_{n})'(t)-(Nu)'(t)|\\ & \le \frac{1}{\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}s^{-\gamma}|f(s, u_n(s), D^{\beta}u_n(s))-f(s,u(s),D^{\beta}u(s))|\,ds\\ & < \frac{\varepsilon}{\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}s^{-\gamma}\,ds\\ &\le \frac{\varepsilon}{\Gamma(\alpha)}T^{\alpha-\gamma}B(\alpha,1-\gamma), \quad\text{for } n>n_1. \end{align*} This proves that $(Nu_n)'(t) \to (Nu)'(t)$ uniformly in $t$ and a similar calculation shows that $(Nu_n)(t) \to (Nu)(t)$ uniformly in $t$. This completes the proof for existence of a solution. Now suppose that $f$ satisfies the Lipschitz condition and let $u,v$ be fixed points with $u(0)=u_0, u'(0)=u_1$, $v(0)=v_0, v'(0)=v_1$. Then we have \begin{align*} u(t)-v(t) &=u_0-v_0+t(u_1-v_1)+\frac{1}{\Gamma(1+\alpha)} \int_0^t (t-s)^{\alpha}s^{-\gamma}\bigl(f(s, u(s),D^{\beta}u(s))\\ &\quad -f(s,v(s),D^{\beta}v(s)\bigr)\,ds, \end{align*} and \begin{align*} &u'(t)-v'(t) \\ &=u_1-v_1+\frac{1}{\Gamma(\alpha)} \int_0^t (t-s)^{\alpha-1}s^{-\gamma}\bigl(f(s, u(s),D^{\beta}u(s))-f(s,v(s),D^{\beta}v(s)\bigr)\,ds. \end{align*} By the arguments used above this gives \begin{align*} &|u(t)-v(t)|+|u'(t)-v'(t)|\\ &\le |u_0-v_0|+(1+T)|u_1-v_1| +C_{6}L\,I^{\alpha}\bigl(s^{-\gamma}(|u(s)-v(s)|+|u'(s)-v'(s)|\bigr). \end{align*} By Theorem~\ref{t-first-Gron} we obtain, for $t \in [0,T]$, \[ |u(t)-v(t)|+|u'(t)-v'(t)|\le ( |u_0-v_0|+(1+T)|u_1-v_1|)C_{7} \exp\bigl(Bt^{1-\gamma}\bigr). \] This proves the continuous dependence on initial data and taking $u_0=v_0,u_1=v_1$ proves uniqueness. \end{proof} \begin{remark} \rm We can obtain existence of nonnegative solutions if we suppose that $f(t,u,p)\ge 0$ and that $u_0\ge 0$ and $u_0+Tu_1 \ge 0$. The proof is almost the same using fixed point index theory in place of degree theory. \end{remark} \begin{remark} \rm Some previous work with nonlinearities involving fractional derivatives was done by Kosmatov \cite{kosmatov} and continued by Deng and Deng \cite{deng-deng}. They consider $D^{\alpha}_{C}u(t) =f(t, D^{\beta}_{C}u(t))$ (of any order $\alpha$) when the nonlinear term $f$ is assumed to be continuously differentiable. It is assumed in \cite{kosmatov} that $|f(t,p)| \le a_1(t)+a_2(t)|p|$ and that a certain integral involving $a_1$ is finite and another involving $a_2$ is smaller than one in order to prove appropriate a priori bounds. Deng and Deng \cite{deng-deng} use some of the results from Kosmatov \cite{kosmatov} and generalize the main existence theorem by imposing the condition $|f(t,p)| \le a|p|+b$ for $t \in[0,1]$ with $a,b$ positive constants but no restriction on the size. For $1<\alpha<2$ and $0<\beta<1$ (a particular case) Kosmatov~\cite[Lemma 3.1]{kosmatov} claims the equivalence of $D^{\alpha}_{C}u(t) =f(t, D^{\beta}_{C}u(t))$ and $u(0)=u_0, u'(0)=u_1$ with an integral equation for $C^2[0,T]$ functions. However, solutions of the integral equation do not have enough regularity for this in general, see Theorem~\ref{ivpudeporder2}, so this is not clear. Deng and Deng \cite{deng-deng} use the claimed result without comment. Our method uses Theorem~\ref{t-first-Gron}, the Gronwall inequality of \cite{jw-jmaa}, to obtain a priori bounds with fewer restrictions. We work in the space $C^1$ and we do not need $f \in C^1$ or a condition such as $f(0,0)=0$, also we allow $f=f(t,u, D^{\beta}u)$ to depend on $u$ too and we have the extra singular term $t^{-\gamma}$. \end{remark} The same methods as above apply to higher order problems. We only give the case for fractional derivatives of order between 2 and 3. We will consider the Caputo IVP: \begin{gather*} D^{2+\alpha}_{*}u(t)=t^{-\gamma}f(t, u(t),D^{\beta_1}_{C}u(t), D_{C}^{1+\beta_2}u(t)), \quad \text{for } t>0, \\ u(0)=u_0, \quad u'(0)=u_1, \quad u''(0)=u_2, \end{gather*} where $0 \le \gamma<\alpha<1$ and $0<\beta_1,\beta_2 \le 1$. For this problem we work in the space $C^2[0,T]$ in which we will use the norm $\|u\|_2:= \|u\|_{\infty}+\|u'\|_{\infty}+\|u''\|_{\infty}$. As previously in Theorem~\ref{ivpCapderivorder2} this is equivalent to the Volterra integral equation \begin{equation} \label{eqVIEderivord3} u(t)=u_0+u_1t+u_2t^2/2+I^{2+\alpha}(t^{-\gamma}f(t,u(t), D^{\beta_1}_{C}u(t),D_{C}^{1+\beta_2}u(t))). \end{equation} We note that $D_{C}^{1+\beta_2}u=I^{1-\beta_2}u'' \in C[0,T]$ and, as before, that \[ |s^{-\gamma}D_{C}^{1+\beta_2}u(s)| \le I^{1-\beta_2}|s^{-\gamma}u''(s)|. \] We now have the following result. \begin{theorem} \label{thm-exC3} Let $f:[0,T]\times \mathbb{R}^{3}$ be continuous, let $0<\alpha<1$, $0 \le \gamma < \alpha$, and $0 < \beta_{i} \le 1$ ($i=1,2$), and let $u_0,u_1,u_2 \in \mathbb{R}$. Suppose there are $a \in L^{\infty}$ and a constant $M>0$ such that $|f(t,u,p,q)| \le a(t)+M(|u|+|p|+|q|)$ for all $t \in [0,T]$ and $u,p,q \in \mathbb{R}$. Then the integral operator \begin{align*} Nu(t):&=u_0+tu_1+(t^2/2)u_2\\ &\quad +\frac{1}{\Gamma(2+\alpha)}\int_0^t (t-s)^{1+\alpha}s^{-\gamma} f(s, u(s),D^{\beta_1}_{C}u(s), D_{C}^{1+\beta_2}u(s)))\,ds \end{align*} has a fixed point in $C^{2}[0,T]$. If, in addition, there exists $L>0$ such that $f$ satisfies the Lipschitz condition \begin{gather*} |f(t,u_1,p_1,q_1)-f(t,u_2,p_2,q_2)| \le L(|u_1-u_2|+|p_1-p_2||+|q_1-q_2|), \\ \text{for all $t \in [0,T] $ and all $u_{i},p_{i},q_{i} \in \mathbb{R}$}, \end{gather*} then fixed points depend continuously on the initial data and in particular, for a given $u_0,u_1, u_2$ there is a unique fixed point. \end{theorem} \begin{proof} If $u=\lambda Nu$ for $\lambda \ge 1$ then we have: \begin{gather*} |u(t)| \le |u_0|+t|u_1|+(t^2/2)|u_2|+I^{2+\alpha} |s^{-\gamma} f(s, u(s),D^{\beta_1}_{C}u(s),D_{C}^{1+\beta_2}u(s)|, \\ |u'(t)| \le |u_1|+t|u_2|+I^{1+\alpha}|s^{-\gamma} f(s, u(s),D^{\beta_1}_{C}u(s),D_{C}^{1+\beta_2}u(s))|,\\ |u''(t)| \le |u_2|+I^{\alpha}|s^{-\gamma}f(s, u(s),D^{\beta_1}_{C}u(s), D_{C}^{1+\beta_2}u(s))|. \end{gather*} These are estimated exactly as in the proof of Theorem~\ref{thm-exC} and the Gronwall inequality Theorem~\ref{t-first-Gron} is used to get the a priori bound on $|u|+|u'|+|u''|$. The rest of the proof is exactly similar to that of Theorem~\ref{thm-exC} hence is omitted. \end{proof} A similar result can be given for higher order problems, since this requires no new ideas we do not state or prove this. \section{Regularity results for Caputo IVP}\label{s-reg} For the case of fractional derivatives $D^{\beta}_{*}$ with $1<\beta<2$ we can give some regularity results when $f$ has more regularity. \begin{theorem} \label{ivporder2} \begin{enumerate} \item\label{reg1} Let $f$ be continuous on $[0,T]$, let $0<\gamma \le \alpha<1$ and let $g(t):=t^{-\gamma}f(t)$. If $u \in C[0,T]$ is such that $D^{1+\alpha}_{*}u$ exists and satisfies the Caputo fractional initial value problem \begin{equation}\label{eqCmDE} D^{1+\alpha}_{*}u(t)=g(t) \quad\text{for a.e.\ $t$}, \quad u(0)=u_0, \quad u'(0)=u_1. \end{equation} then $u \in C^1[0,T]$ and satisfies the Volterra integral equation $u(t)=u_0+u_1\,t+I^{1+\alpha}g(t)$, that is, \begin{equation}\label{eqCapVIE2} u(t)=u_0+u_1\,t+\frac{1}{\Gamma(1+\alpha)}\int_0^t (t-s)^{\alpha}s^{-\gamma}f(s)\,ds, \quad t \in [0,T]. \end{equation} Conversely, if $u \in C[0,T]$ satisfies \eqref{eqCapVIE2} then $u \in C^1[0,T]$ and satisfies \eqref{eqCmDE}. \item\label{reg2} If $f \in AC$ and $u \in C[0,T]$ satisfies \begin{equation}\label{eqCmDEf} D^{1+\alpha}_{*}u(t)=f(t) \quad \text{for } t\in[0,T], \quad u(0)=u_0, \quad u'(0)=u_1. \end{equation} then $u(t)=u_0+tu_1+v(t)+\frac{f(0)t^{\alpha+1}}{\Gamma(\alpha+1)}$ where $v \in AC^1$, thus $u' \in AC$ and hence $D^{1+\alpha}_{*}u=D^{1+\alpha}_{C}u$. \item\label{reg3} If $f \in C^1$ and $u \in C[0,T]$ satisfies \eqref{eqCmDEf} then $u' \in AC$ and $u(t)=u_0+tu_1+v(t)+\frac{f(0)}{\Gamma(\alpha+1)}t^{1+\alpha}$ where $v \in C^2[0,T]$. In particular, $u \in C^2(0,T]$, moreover $u \in C^2[0,T]$ if and only if $f(0)=0$. \end{enumerate} \end{theorem} \begin{proof} \eqref{reg1} By definition $D^{1+\alpha}_{*}u(t)=g(t)$ means that $D I^{1-\alpha}(u-T_1(u)) \in AC$ and $T_1(u)$ exists, and $D^2 I^{1-\alpha}(u-T_1(u))=g \in L^1$. Integrating twice gives \[ I^{1-\alpha}(u-T_1(u))(t)=I^2g(t)+a+bt. \] By Proposition~\ref{propfintprop}\,\eqref{rlint5}, $I^{1-\alpha}(u-T_1(u))(0)=0$ so we must have $a=0$. Applying $I^{\alpha}$ gives \[ I(u-T_1(u))=I^{2+\alpha}g+bt^{\alpha+1}/\Gamma(\alpha+2), \] hence \[ (u-T_1(u))(t)=I^{1+\alpha}g(t)+bt^{\alpha}/\Gamma(\alpha+1). \] Since $u'(0)$ exists we must have $b=0$. Thus \eqref{eqCapVIE2} holds and $u-T_1(u)=I^{1+\alpha}g =I(I^{\alpha}g) \in C^1$ since $I^{\alpha}g \in C_{\alpha-\gamma} \subset C$ by Proposition~\ref{propfintprop} \eqref{rlint5}. Conversely, if $u \in C[0,T]$ satisfies \eqref{eqCapVIE2} then since $I^{1+\alpha}g=I(I^{\alpha}g) \in C^1$ we obtain $u \in C^1$ and $D^{1+\alpha}_{*}u=D^{1+\alpha}I^{1+\alpha}g=g$ a.e. \eqref{reg2} By part \eqref{reg1} $u-T_1u=I^{1+\alpha}f$ and $u \in C^1$. For $f \in AC$ we have $f' \in L^1$ and $f(t)-f(0)=If'(t)$. Therefore \[ u-T_1u=I^{1+\alpha}(If'+f(0))=v(t)+\frac{f(0)t^{\alpha+1}}{\Gamma(\alpha+1)}, \] where $v(t)=I\,I^{\alpha}(If') \in AC^1$ (using Proposition~\ref{propfintprop}) and $\frac{f(0)t^{\alpha+1}}{\Gamma(\alpha+1)} \in AC^1$ too. Thus $u \in AC^1$ and $D^{1+\alpha}_{*}u=D^{1+\alpha}_{C}u$ by Lemma~\ref{lemhigherorder}. \eqref{reg3} For $f \in C^1$, we have $f' \in C[0,T]$ and $f(t)-f(0)=If'(t)$ so \[ u-T_1u=I^{1+\alpha}(If'+f(0))=I^2I^{\alpha}f'+f(0)t^{1+\alpha}/\Gamma(1+\alpha), \] where $v(t):=I^2I^{\alpha}f' \in C^2[0,T]$. This equation shows that $u \in C^2[0,T]$ if and only if $f(0)=0$. \end{proof} \begin{remark} \rm Nothing much better than $u \in C^1$ is expected in the analogue to parts \eqref{reg2}, \eqref{reg3} when there is the singular term $t^{-\gamma}$ since we can only prove $I^{\alpha}(t^{-\gamma}f(t)) \in C_{\alpha-\gamma}$. \end{remark} For fractional differential equations of order $1+\alpha$ with $0<\alpha<1$ and when $f$ depends on $u$ but not on derivatives of $u$ we have the following regularity results. \begin{theorem} \label{ivpudeporder2} Let $0<\gamma \le \alpha<1$. \begin{enumerate} \item\label{regu1} Let $f$ be continuous on $[0,T]\times \mathbb{R}$. If a function $u \in C[0,T]$ is such that $D^{1+\alpha}_{*}u$ exists and satisfies the Caputo fractional initial value problem \begin{equation}\label{eqCmde2g} D^{1+\alpha}_{*}u(t)=t^{-\gamma}f(t, u(t)) \quad \text{for a.e. }t,\quad ;u(0)=u_0, \quad u'(0)=u_1. \end{equation} then $u \in C^1[0,T]$ and satisfies the Volterra integral equation \begin{equation} \label{eqVIEord2} u(t)=u_0+u_1\,t+I^{1+\alpha}(t^{-\gamma}f(t,u(t))). \end{equation} Conversely, if $u \in C[0,T]$ satisfies \eqref{eqVIEord2} then $u \in C^1[0,T]$ and satisfies \eqref{eqCmde2g}. \item\label{regu2} Let $f$ be continuous on $[0,T]\times \mathbb{R}$ and such that \begin{equation} \label{eqACfu} t \mapsto f(t,u(t)) \in AC \quad\text{for every}\;u \in C^1[0,T]. \end{equation} Then if $u \in C[0,T]$ satisfies \begin{equation}\label{eqCmde2f} D^{1+\alpha}_{*}u(t)=f(t, u(t)) \quad \text{for a.e. }t, \quad u(0)=u_0, \quad u'(0)=u_1. \end{equation} we have $u' \in AC[0,T]$ and $D^{1+\alpha}_{C}u(t)=f(t,u(t))$. \item\label{regu3} If $f \in C^1$, that is $\partial_{t}f$ and $\partial_{u}f$ are continuous functions, and $u \in C[0,T]$ satisfies \eqref{eqCmde2f} then $u(t)=v(t)+ct^{1+\alpha}$ where $v \in C^2[0,T]$ and $c$ is a constant, in particular, $u \in C^2(0,T] \cap AC^1[0,T]$ and satisfies $D^{1+\alpha}_{C}u(t)=f(t,u(t))$. Moreover $u\in C^2[0,T]$ if and only if $f(0,u_0)=0$. \end{enumerate} \end{theorem} \begin{proof} \eqref{regu1} Define $h(t):=f(t,u(t))$. If $u \in C[0,T]$ satisfies \eqref{eqCmde2g} then $D^{1+\alpha}_{*}u(t)=t^{-\gamma}h(t)$. By Theorem~\ref{ivporder2} \eqref{reg1} the result follows. \eqref{regu2} Let $h(t):=f(t,u(t))$. By part (1), $u \in C^1[0,T]$ hence by assumption \eqref{eqACfu} we obtain $h \in AC$ and Theorem~\ref{ivporder2} \eqref{reg2} applies. \eqref{regu3} In this case $h \in C^1$ so this follows from Theorem~\ref{ivporder2} \eqref{reg3}. \end{proof} \begin{remark} \rm The regularity in \eqref{regu3} on all of $[0,T]$ cannot be improved in general even if $f \in C^{\infty}$. For $u(t)=t^{1+\alpha}$ satisfies $D_{*}^{1+\alpha}u=\text{constant} \in C^{\infty}$ but we only have $u \in AC^1[0,T]$. In \cite[Theorem 6.25] {diethelm-bk} it is proved (also there are results for the higher order case) that $u \in C^1[0,T]$ as in \eqref{regu1}. In \cite[Theorem 6.26] {diethelm-bk} it is shown (also there is a higher order case result), as in \eqref{regu3}, that $u \in C^2(0,T]$ and that $u\in C^2[0,T]$ if and only if $f(0,u_0)=0$, by similar arguments but without the explicit functions. We have not seen the exact forms of our cases \eqref{regu2} and \eqref{regu3} in the literature so they may be novel. Although smooth solutions cannot exist in general, Diethelm \cite{diethelm-fcaa-07} give a full characterization of the situations where smooth solutions exist, even analytic. \end{remark} \begin{remark} \rm Of course it is important to have some explicit conditions that imply \eqref{eqACfu}. Clearly $f \in C^1$ is sufficient. The sum and pointwise product of functions in $AC[0,T]$ belong to $AC[0,T]$ and if $u \in AC$ then the composition $g \circ u \in AC$ if $g$ satisfies a Lipschitz condition. Hence we have the following examples: \begin{enumerate} \item\label{ac1} $f$ is Lipschitz in both variables, that is $|f(t,u)-f(s,v)| \le L(|t-s|+|u-v|)$ for all $t,s \in [0,T]$ and all $u,v \in \mathbb{R}$. Then for $u \in C^1$, $t \mapsto f(t,u(t)) \in AC$. \item\label{ac2} $f(t,u) =h_1(t)+h_2(t)g(u)$ where $h_{i}\in AC[0,T]$ and $g \in \mathrm{Lip}$. \item\label{ac3} Sums of terms of the above types. \end{enumerate} The case \eqref{ac1} where $f$ is Lipschitz in both variables is discussed in detail for several boundary value problems in \cite{lan-lin-caa} with somewhat different arguments. For the problem $D^{1+\alpha}_{*}u(t)=t^{-\gamma}f(t,u(t), D^{\beta}u(t))$ which was studied above in the space $C^1$ via the Volterra integral equation \[ u(t)=u_0+u_1\,t+I^{1+\alpha}(t^{-\gamma}f(t,u(t),D^{\beta}_{C}u(t))), \] even if $f\in C^1$ we only have $t \mapsto f(t,u(t),D^{\beta}_{C}u(t)) \in C^{0}$ so we can not prove more than $u \in C^1$. \end{remark} \section{Comments on Boundary Value problems} For the Caputo fractional differential equation $D^{\alpha}_{*}u(t)= f(t,u(t))$, $t \in [0,T]$, of order $\alpha \in (0,1)$ it is appropriate to impose exactly one boundary condition. A natural type of condition is $au(0)+bu(T) = c$. When $f$ is continuous it is shown in Lemma 6.40 of \cite{diethelm-bk} that the Caputo problem is equivalent to a Fredholm integral equation. Tisdell \cite{tisdell} studies systems of such equations with this type of boundary condition. We prove a new equivalence result exactly similar to \cite{diethelm-bk} for the single equation when the nonlinearity is allowed to be singular. \begin{theorem} \label{thmcbvp1} Let $0 \le \gamma<\alpha <1$, let $a,b,c \in \mathbb{R}$ with $a+b \ne 0$ and let $f :[0,T] \times \mathbb{R} \to \mathbb{R}$ be continuous. Then, the function $u \in C[0,T]$ with $I^{1-\alpha}u \in AC$ is a solution of the boundary value problem \begin{equation} \label{eqbvpcap0} D^{\alpha}_{*}u(t) = t^{-\gamma}f(t,u(t)), \quad t \in (0,T], \quad au(0)+bu(T) = c, \end{equation} if and only if $u \in C[0,T]$ is a solution of the integral equation \begin{equation} \label{eqbvpcint1} \begin{aligned} u(t)&=\frac{c}{a+b}+\frac{1}{\Gamma(\alpha)} \int_0^t (t-s)^{\alpha-1} s^{-\gamma}f(s,u(s))\,ds \\ &\quad -\frac{b}{a+b}\frac{1}{\Gamma(\alpha)} \int_0^T (T-s)^{\alpha-1} s^{-\gamma}f(s,u(s))\,ds. \end{aligned} \end{equation} \end{theorem} \begin{proof} Let $u \in C[0,T]$ satisfy \eqref{eqbvpcint1}, let $g(t):=t^{-\gamma}f(t,u(t))$ so $g \in L^1$. Then $u$ satisfies \[ u(t)=\frac{c}{a+b}+I^{\alpha}g(t)-\frac{b}{a+b}I^{\alpha}g(T). \] Thus $u(0)=\frac{c}{a+b}-\frac{b}{a+b}I^{\alpha}g(T)$ and $u(T)=\frac{c}{a+b}+\frac{a}{a+b}I^{\alpha}g(T)$, hence $au(0)+bu(t)=c$. Furthermore, $D^{\alpha}_{*}u=D^{\alpha}(u-u(0))=D^{\alpha}I^{\alpha}g=g$ a.e.\ so $D^{\alpha}_{*}u=t^{-\gamma}f(t,u(t)$ for $t \in (0,T]$. Moreover it follow that $I^{1-\alpha}u \in AC$ as in Theorem~\ref{ivporderm}. Conversely, let $u \in C[0,T]$ with $I^{1-\alpha}u \in AC$ satisfy \eqref{eqbvpcap0}. By Theorem~\ref{ivporderm}, $u(t)=u_0+I^{\alpha}g(t)$ and so $u_0=u(0), u(T)=u_0+I^{\alpha}g(T)$. The boundary conditions give $au_0+bu_0+bI^{\alpha}g(T)=c$ thus $u_0=\frac{c-bI^{\alpha}g(T)}{a+b}$, which proves that $u$ satisfies \eqref{eqbvpcint1}. \end{proof} \begin{remark} \rm When $\gamma=0$, $D^{\alpha}_{*}u=f$ means that $D(I^{1-\alpha}(u-u_0))=f$ so $I^{1-\alpha}(u-u_0)\in C^1$ and is automatically $AC$; this recovers the result of \cite[Lemma 6.40] {diethelm-bk}. \end{remark} For the Caputo fractional differential equation $D^{1+\alpha}_{*}u(t)= f(t,u(t))$, $t \in [0,T]$, where $\alpha \in (0,1)$ it is natural to impose two boundary conditions. The case with $D^{1+\alpha}_{C}u(t)= f(t,u(t))$ and general separated boundary conditions (also sometimes called Sturm-Liouville boundary conditions) was extensively studied in \cite{lan-lin-caa}. Lemma 6.43 of \cite {diethelm-bk} proves an equivalence when $f$ is continuous and $D^{1+\alpha}_{*}u(t)= f(t,u(t))$ together with boundary conditions of this type. We can allow a singularity as follows. \begin{theorem} \label{thmcbvp2} Let $0 \le \gamma<\alpha \in (0,1)$, let $a,b,c,d \in \mathbb{R}$ with $ad+bc+acT\ne 0$ and let $f :[0,T] \times \mathbb{R} \to \mathbb{R}$ be continuous. Then, the function $u \in C[0,T]$ with $I^{1-\alpha}u \in AC^1$ is a solution of the boundary value problem \begin{equation} \label{eqbvpcap1} \begin{gathered} D^{1+\alpha}_{*}u(t) = t^{-\gamma}f(t,u(t)), \quad t \in (0,T], \\ au(0)-bu'(0) = 0, \quad cu(T)+du'(T)=0, \end{gathered} \end{equation} if and only if $u \in C[0,T]$ is a solution of an integral equation equation \[ u(t)=\int_0^T G(t,s)s^{-\gamma}f(s,u(s))\,ds, \] where $G$ is called the Green's function. \end{theorem} \begin{proof} By Theorem~\ref{ivporderm} for $u \in C[0,T]$ with $I^{1-\alpha}u \in AC^1$ satisfying $D^{1+\alpha}_{*}u(t) = t^{-\gamma}f(t,u(t))$ we have $u(t)=u_0+tu_1+I^{1+\alpha}g(t)$ where $g(t)=t^{-\gamma}f(t,u(t))$. The boundary conditions are satisfied if \[ au_0-bu_1=0,\; cu_0+cTu_1+cI^{1+\alpha}g(T)+du_1+dI^{\alpha}g(T)=0. \] These equations have unique solution $u_0,u_1$ if and only if \[ \det\begin{bmatrix}a&-b\\c&cT+d \end{bmatrix} =ad+bc+acT \ne 0. \] These values are then substituted into $u_0+tu_1+I^{1+\alpha}g(t)$ to get the formula for the Green's function. The converse is similar to previous arguments. Since we are not pursuing any existence theory for boundary value problems here we omit the formulas for the Green's function which can be readily found and are given in \cite{lan-lin-caa}. \end{proof} \begin{remark} \rm In \cite{lan-lin-caa} the authors assume a Lipschitz condition on $f$ in order that the solutions of the integral equation satisfy $D^{1+\alpha}_{C}u(t)= f(t,u(t))$. \end{remark} For Riemann-Liouville fractional derivatives the type of boundary conditions that can be considered depend on the function space in which the solution is to be found. We will only discuss the case $D^{1+\alpha}$ for $0<\alpha<1$. The equation $D^{1+\alpha}u=f$ is equivalent to $D^2(I^{1-\alpha})u=f$ where we should suppose explicitly that $I^{1-\alpha}u \in AC^1$. By writing $v=I^{1-\alpha}u$ we see that $v' \in AC$ and $v$ satisfies the second order ordinary differential equation $v''=f$. Therefore natural BVPs would have boundary condition defined in terms of $I^{1-\alpha}u$ and its derivative $D(I^{1-\alpha})u=D^{\alpha}u$ evaluated at $t=0, t=T$. The physical interpretation of such boundary conditions is not obvious, but we recall that $I^{1-\alpha}u(0)$ is related to $\lim_{t \to 0+}t^{1-\alpha} u(t)$, see Lemma~\ref{limimplim}. If, as is often the case, solutions are to be found in the space $C[0,T]$ as fixed points of a Fredholm integral operator, with kernel the Green's function, then certain boundary conditions are not allowed. The Green's function would be found by using the equivalence from Theorem~\ref{rlequord2}. When $f$ is continuous the procedure would begin by claiming the solution is of the form \begin{equation} \label{eqRLint22} u(t)=c_1t^{\alpha-1}+c_2t^{\alpha}+I^{1+\alpha}f(t,u(t)) \end{equation} and then use the boundary conditions to determine unique values for $c_1,c_2$ as functions of $u,f$. We note that if $c_1 \ne 0$ then $u$ is not defined at 0 and if $c_2 \ne 0$ then $u'$ is not defined at 0. If we seek a solution $u \in C[0,T]$ of \eqref{eqRLint22} then it is necessary that $c_1=0$ irrespective of the value $u(0)$, but when $c_1=0$ in \eqref{eqRLint22} it is necessary that $u(0)=0$ so that is the only compatible boundary condition. If we seek a solution $u \in C[0,T]$ of \eqref{eqRLint22} and try to impose a boundary condition at $0$ involving $u'(0)$ then we must have $c_2=0$ in addition to $c_1=0$ so the only problem that can be considered is $u(t)=I^{1+\alpha}f(t,u(t))$ which is a Volterra equation, not a Fredholm equation, and any extra BC at $t=T$ would give an over-determined problem, in other words a BC at $0$ involving $u'(0)$ is not allowed with any other BC at $t=T$ for the R-L derivative case if solutions in $C[0,T]$ are sought. If one BC is $u(0)=0$ then one can impose many possible BCs at $t=T$ and solutions of the integral equation will give solutions of the BVP. This has been done in many papers, too large a number to be given here. However, it should be noted that one can never get periodic solutions (or anti-periodic ones) for fractional equations, it has been shown by Kaslik and Sivasundaram \cite{kaslik-Siv} in 2012 and by, Wang, Feckan, and Zhou \cite{wang-feckan-zhou} in 2013, and by Area, Losada, and Nieto \cite{A-L-N} in 2014 that the fractional derivative or integral of a periodic function $u$ defined on the whole real line cannot be periodic, except, in the Riemann-Liouville case for the trivial example of $u=0$, in the Caputo case for $u$ constant. Boundary value problems on a finite interval with boundary conditions of periodic type can be, and have been, considered. \begin{thebibliography}{99} \bibitem{Ag-OR-St} R. P. Agarwal, D. O'Regan, S. Stan\v{e}k; Positive solutions for mixed problems of singular fractional differential equations, \textit{Math. Nachr.}, 285 (2012), 27--41. \bibitem{A-L-N} I. Area, J. Losada, J. J. 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Proposition~\ref{propfintprop}~(10) is usually false when $u$ is increasing but is negative on some interval $(0, \delta)$. Then $I^{\alpha}f(t)$ is negative for $t \in (0,\delta)$. For example if $f\in L^p$ for $p>1/{\alpha}$ then $I^{\alpha}f(0)=0$ (see Proposition~\ref{propfintprop}~(3)), hence $I^{\alpha}f$ is not increasing for $t$ near 0. A simple example, for $t \ge 0$, is $u(t)=3kt-1$ and $\alpha=1/2$, where $k>0$ can be arbitrarily large. Then $u$ is increasing, negative for $t<1/(3k)$ and positive for $t>1/(3k)$. By a small calculation, $I^{1/2}u(t)=(4kt^{3/2}-2t^{1/2})/\sqrt{\pi}$ which is zero at $0$, is then negative and decreasing, has a negative minimum at $t=1/(6k)$, and then increases, crossing zero at $t=1/(2k)$. For $0<\gamma<1$ the $L^1[0,T]$ function $u(t)=t^{-\gamma}$ has \[ I^{\alpha}u(t)=\frac{\Gamma(1-\gamma)}{\Gamma(1+\alpha-\gamma)}t^{\alpha-\gamma}\] so gives an example of a positive decreasing $L^1$ function whose fractional derivative is increasing for $\alpha>\gamma$ and is decreasing for $\alpha<\gamma$. By applying the result to $-u$ there is the obvious \textbf{dual result} that\\ \emph{if $u(t) \le 0$ and non-increasing for $t\ge 0$ then $I^{\alpha}u(t)$ is non-increasing.} Proposition~\ref{propfintprop}~(10) is correctly used in two places, in Proposition~\ref{prop-mon} where it is applied to $v(t)=u(t)-u(0)$ with $u$ nondecreasing so $v$ is also non-negative, and in Proposition 7.4 where the dual is applied to $u'(t)-u'(0)$ with $u'$ non-increasing. \item Proposition~\ref{propfintprop}~(3) can be extended. Lemma A1. \emph{For $0<\alpha<1$ and $p>1/\alpha$, $I^{\alpha}$ is a compact operator from $L^p[0,T]$ to $C[0,T]$. Hence $I^{\alpha}$ is also compact from $C[0,T]$ to $C[0,T]$.} \smallskip This is because $I^{\alpha}$ maps into a H\"older space and the fact that for $0<\beta<1$ the H\"older space $C^{0,\beta}[0,T]$ is compactly embedded in $C[0,T]$. The proof of this compact embedding is simple, so we give it here. The norm in $C^{0,\beta}[0,T]$ is \[ \|u\|_{0,\beta}:=\sup_{x \in [0,T]}|u(x)|+\sup_{x \ne y}\frac{|u(x)-u(y)|}{|x-y|^{\beta}}. \] Lemma A2. \emph{If $\{u_n\}$ is a bounded sequence in $C^{0,\beta}[0,T]$, then $\{u_n\}$ is relatively compact in $C[0,T]$.} \begin{proof} Let $\|u_{n}\|_{0,\beta} \le M$. For $x \ne y$ we have $$|u_{n}(x)-u_{n}(y)| =\frac{|u_{n}(x)-u_{n}(y)|}{|x-y|^{\beta}}|x-y|^{\beta} \le M|x-y|^{\beta}$$ so $\{u_{n}\}$ is bounded and equicontinuous hence relatively compact in $C[0,T]$ by the Ascoli-Arzel\`a theorem. \end{proof} \item With a closer reading of the Hardy-Littlewood paper \cite{{Hardy-Lit}} I discovered that they have answered the following question which arises from Remark~\ref{remmodCequiv}: \begin{quote} Question. Does there exist a function $w \in C[0,T]$ such that $I^{\alpha}w \notin AC$? \end{quote} From the known results (see Proposition~\ref{propfintprop}~(6)) $w$ must not be $AC$. The answer is yes. In \S 5.5 of \cite{{Hardy-Lit}} they give the following example. For $a>1$ and any $k \in (0,1)$, let $w_{k}$ be the Weierstrass function \begin{equation*} w_{k}(t)=\sum_{n=0}^{\infty} a^{-nk}\cos(a^nt). \end{equation*} Then state (and have proved in other papers cited there) that $w_{k}$ is H\"older continuous with exponent $k$ but $I^{1-k}w_{k}$ is not differentiable at any point so is not $AC$. This implies that, for $0<\alpha<1$, if all we know is that $u$ and $f$ are continuous and $u(t)=u_0+I^{\alpha}f(t)$ then it cannot be inferred that $D^{\alpha}_{C}$ exists; $D^{\alpha}_{*}$ does exist since $I^{1-\alpha}(u-u_{0})=If \in C^1$. Thus, in Theorem~\ref{ivporderm} the often supposed equivalence between \eqref{eqCapDEm} and \eqref{eqCapVIEm} is false in general. Therefore to study a Caputo fractional differential equation via the corresponding fractional integral equation the definition $D^{\alpha}_{*}$ should be used, as is done in the well-known books \cite{diethelm-bk} and \cite{kst}. This supports the maxim `Read the masters', Hardy and Littlewood answered the question nearly a century before the answer was wanted by those studying Caputo fractional derivatives. \item Some typos: page 11, In the sentence `The result is a special case of the result given for fractional derivatives of all orders in \cite[Theorem 3.1]{diethelm-bk}, and which is proved below in Lemma 4.10.', Lemma 4.10 here should be replaced by Lemma~\ref{lemmthord}. page 18, In Theorem~\ref{rleqordmfu}, the sum should go from $k=1$ to $k=m+1$. page 19, $1/\Gamma(\alpha)$ is omitted in front of the integral sign in the proof of Proposition~\ref{prop-mon} step (2). \end{enumerate} \end{document}