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

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.

• 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. Apr 18 at 1:17

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.

• Theorem 2.1 is not applicable here since $x \in (0,\infty)$.
– Sam
Apr 18 at 14:24
• 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. Apr 18 at 15:11