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.

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.

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 $\alpha = 1/2$. 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.

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.