Estimating the integral $\int_{\epsilon}^1 \Bigl\lvert \int_0^x \frac{f(y)}{\lvert x-y\rvert^{1/2}} dy\Bigr\rvert^2 dx$ for $L^2$ function $f(y)$?

Question

I guess the chances are slim but still curious about the integral in the title.

Let $f : [0, \infty) \to \mathbb{R}$ be a locally "square-integrable" function on $[0,\infty)$.

Then, for any $\epsilon \in (0,1)$, is it possible to estimate the following integral?: \begin{equation} \int_{\epsilon}^1 \Bigl\lvert \int_0^x \frac{f(y)}{\lvert x-y\rvert^{1/2}} dy\Bigr\rvert^2 dx \end{equation}

In particular, is this integral finite in general? Naive application of Jensen's inequality of course leads to divergent estimate, but I wonder if there is anything more precise..

Answer 1

First, we can observe that your integral depends solely on the behavior of $f$ on the interval $[0, 1]$. Its values outside that region do not affect the expression. So we may multiply by the cutoff function $\chi_{[0, 1]}$.

Thus we may consider this problem for elements of $L^2(\mathbb{R})$ with compact support.

We can look at the Hardy-Littlewood-Sobolev theorem on fractional integration (or more accurately, its proof). The argument found in Remark 2 here, partitioning into dyadic shells, shows that for $x > 0$, $$\left|\int_0^x \frac{f(y)}{|x - y|^{1/2}} \, dy \right| \leq \int_0^x \frac{|f(y)|}{|x - y|^{1/2}} \, dy \leq \int_{B(x, x)} \frac{|f(y)|}{|x - y|^{1/2}} \, dy \leq C x^{1/2} M f(x),$$ where $M f$ denotes the Hardy-Littlewood maximal function and $C$ is an absolute constant.

In your case, since we are integrating over $x \in [\epsilon, 1]$, we can bound this solely by a multiple of $M f$, and then the strong-type Hardy-Littlewood $L^p$ estimate gives (as a very rough upper bound), $$\left\|\int_0^x f(y) |x - y|^{-1/2} \, dy \right\|_{L^2_x[0, 1]} \leq C \|M f(x)\|_{L^2_x} \leq C' \|f\|_{L^2} = C' \|f\|_{L^2[0, 1]},$$ invoking the support restriction on $f$.

So your integral is bounded above by the quantity $K \int_{0}^{1} |f(x)|^2 \, dx$, for some dimensional constant $K$. (And this also indicates that you don't need to take $\epsilon > 0$; you can directly integrate over $[0, 1]$.) You specified $f$ is locally $L^2$, so this quantity is well-defined and non-infinite.

Thus, for any $f \in L^2_{\text{loc}}$, you can guarantee that your integral will be finite.

Answer 2

You are asking whether the operator $K\colon L^2((0,\infty)) \to L^2((\epsilon,1))$ given by $K(f)(x) = \int_0^\infty K(x,y) f(y) dy$ with the kernel $K(x,y) = \Theta(x-y) |x-y|^{-1/2}$ is bounded. Boundedness follows by applying the Schur test with the estimates \begin{align*} \int_0^\infty |K(x,y)| dy &\le \int_0^1 \frac{dy}{|1-y|^{1/2}} = 2 \quad \text{on } x\in (\epsilon,1) , \\ \int_\epsilon^1 |K(x,y)| dx &\le \int_\epsilon^1 \frac{dy}{|x-\epsilon|^{1/2}} = 2|1-\epsilon|^{1/2} \quad \text{on } y \in (0,\infty) \end{align*} on the kernel. The Schur test actually then estimates the operator norm $$ \|K\| \le \sqrt{2\cdot 2|1-\epsilon|^{1/2}} = 2|1-\epsilon|^{1/4} . $$

Answer 3

Isn't it just a convolution operator with $h(x) = \chi_{[0, 1]}(x)\frac{1}{\sqrt{x}}$? So, by the Young's convolution inequality, $f*h$ is in $L^p([0, 1])$ for all $p < \infty$ (in particular for $p = 2$) because $h\in L^q([0, 1])$ for all $q < 2$? But $f*h$ is not necessarily in $L^\infty$ because $h\notin L^2([0, 1])$.