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: $$ I^{\frac{1}{2}} f(x) := \frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} d t $$ Suppose $I^{\frac{1}{2}}f(x) = 0$ 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 MathOverflow.

  • 1
    $\begingroup$ You should be able to use any existing proof that $\int_a^x f(t) dt \equiv 0 \implies f(t) = 0 a.e.$, e.g. the one in this post, as long as $f(x)$ is a measurable function. $\endgroup$
    – ConMan
    Commented Apr 18, 2023 at 1:17

1 Answer 1


Assume $f \in L^{1}(\lambda)$, where $\lambda$ is the Lebesgue measure. First, use that (property of Riemann-Liouville integral)

$$I^{\alpha} (I^{\beta} f(x)) = I^{\alpha + \beta} f(x)$$

Therefore: $I^{1/2} (I^{1/2} f(x)) = I^1 f(x) = \int_{0}^{x} f(t) dt =0$. Then use (an adaptation of) Theorem 2.1 from these notes or this under the assumption that $f$ is integrable wrt to the Lebesgue measure. From this conclude that $f =0$ a.e.

I don't think it is possible to circumvent the assumption $f \in L^{1}(\lambda)$.

Edit: I found an even easier solution. The claim follows from a direct application of Titchmarsh convolution theorem, that you can find here or, for an even more useful extension of this result, see here.

  • $\begingroup$ Theorem 2.1 is not applicable here since $x \in (0,\infty)$. $\endgroup$
    – Sam
    Commented Apr 18, 2023 at 14:24
  • $\begingroup$ You can adapt Theorem 2.1 also to the case $x \in (0,\infty)$. See math.stackexchange.com/questions/274702/… for an alternative proof that works without that assumption. $\endgroup$ Commented Apr 18, 2023 at 15:11

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