I have to prove this:


Let $\alpha\in(0,1)$ and $f\in L^q(a,b)$, $1\leq q<\frac 1\alpha$, and $\mathcal{I}\_{a+}^\alpha f=0$. Then $f(x)=0$ for almost all $x\in (a,b)$.


Where $(\mathcal{I}\_{a+}^\alpha f)(x):=\frac{1}{\Gamma(\alpha)} \int\_{a}^{x} f(y)(x-y)^{\alpha-1}$.


Can Somebody help me?
Thanks



If the Riemann-Liouville fractional integral of $f$ is zero then $f=0$ a.e
$\newcommand{\al}{\alpha}\newcommand{\Ga}{\Gamma}$Let $g:=\Ga(\al)\mathcal{I}\_{a+}^\al f$, so that $g=0$ on $(a,b)$. Then for any $z\in(a,b)$
\begin{equation}
\begin{aligned}
 0&=\int\_a^z dx\,(z-x)^{-\al}g(x) \\ 
 &=\int\_a^z dx\,(z-x)^{-\al}\int\_a^x dy\,f(y)(x-y)^{\al-1} \\ 
 &=\int\_a^z dy\,f(y)\int\_y^z dx\,(z-x)^{-\al}(x-y)^{\al-1} \\ 
 &=\int\_a^z dy\,f(y)\int\_0^1 du\,(1-u)^{-\al}u^{\al-1} \quad \Big[u:=\frac{x-y}{z-y}\Big] \\ 
 &=\int\_a^z dy\,f(y)\,\Ga(1-\al)\Ga(\al); 
\end{aligned}
\end{equation}
the interchange of the order integration is possible because $f\in L^q(a,b)$ for some $q\ge1$ and hence $f\in L^1(a,b)$.


So, $\int\_a^z f=0$ for all $z\in(a,b)$ and thus $f=0$ almost everywhere (a.e.) on $(a,b)$. $\quad\Box$




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*Details on the last sentence in the proof above:* Let $\mu$ be the measure over $(a,b)$ defined by the formula $\mu(A):=\int\_A f$ for all Lebesgue-measurable $A\subseteq(a,b)$. Then the condition that $\int\_a^z f=0$ for all $z\in(a,b)$ implies that $\mu=0$ on the algebra over $(a,b)$ generated by all intervals of the form $(a,z)$ for $z\in(a,b)$. By the uniqueness of the extension of measure, $\mu=0$ on all Borel subsets of $(a,b)$. Also, clearly $\mu=0$ on the sets of Lebesgue measure $0$. So, $\mu=0$ on all Lebesgue subsets of $(a,b)$. So, for each real $t>0$ and $A\_t:=f^{-1}((t,\infty))$ we have $0=\mu(A\_t)\ge t|A\_t|$, where $|\cdot|$ is the Lebesgue measure. So, $|A\_t|=0$ for all real $t>0$.
So, $f\le0$ a.e. on $(a,b)$. Similarly, $f\ge0$ a.e. on $(a,b)$. Thus, $f=0$ a.e. on $(a,b)$.




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The condition $q<1/\al$ was not needed or used in the proof above.


