Riemann-Liouville integral of $f$ is zero implies $f =0$ a.e

Question

The Riemann-Liouville integral is defined by $$ I^\alpha f(x)=\frac{1}{\Gamma(\alpha)} \int_a^x f(t)(x-t)^{\alpha-1} d t $$ where $\Gamma$ is the gamma function and $a$ is an arbitrary but fixed base point. Take $a = 0$ and $\alpha = 1/2$ and $f \in L^1(0,\infty)$. Therefore we look at: $$ I^x_{\frac{1}{2}} := \frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} d t $$ Suppose $I^x_{\frac{1}{2}} = 0$ for all $x$. Can we then conclude $f=0$ a.e.?

My approach so far has been to take the Fourier transform and use the convolution theorem. I cannot conclude because I do not know if $f \in L^2(\mathbb{R})$. Otherwise, I could conclude just by using the fact that the Fourier transform is an isometry between $L^2$ spaces. See here for the same question on MathStack.

Answer 1

Let $H=\mathbf 1_{\mathbb R^+}$ be the Heaviside function and let us define $g$ by $ g(x)=H(x) x^{-1/2}. $ We note that $g$ is homogeneous with degree $-1/2$ and thus its Fourier transform is homogeneous with degree $\frac12-1=-\frac12$. Moreover the function $F=Hf$ belongs to $L^1(\mathbb R)$ and you have assumed that $$ F\ast g=0. $$ Thanks to the Riemann-Lebesgue Lemma, the Fourier transform of a function in $L^1(\mathbb R)$ is a (uniformly) continuous function (going to 0 at infinity). As a consequence we get that $ \hat F \hat g=0. $ Since we have \begin{multline}\hat g(\xi)=\int_0^{+\infty}e^{-2iπ s\text{ sign} \xi} s^{-1/2}ds\vert\xi\vert^{-1/2} \\=\vert\xi\vert^{-1/2}\Bigl\{H(\xi)\int_0^{+\infty}e^{-2iπ s} s^{-1/2}ds + H(-\xi)\int_0^{+\infty}e^{2iπ s} s^{-1/2}ds\Bigr\}, \end{multline} we obtain that the continuous function $\hat F$ is supported at the origin, thus is the zero function. As a consequence $F=0$.