# 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 $f(y)$?

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?: $$$$\int_{\epsilon}^1 \Bigl\lvert \int_0^x \frac{f(y)}{\lvert x-y\rvert^{1/2}} dy\Bigr\rvert^2 dx$$$$

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..

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.

• Amazing.....thank you so much... Commented Jul 28, 2023 at 6:02
• Could you also please help me with the following question? mathoverflow.net/questions/451574/… Commented Jul 28, 2023 at 10:52

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 $$\|K\| \le \sqrt{2\cdot 2|1-\epsilon|^{1/2}} = 2|1-\epsilon|^{1/4} .$$

• Thank you for your answer as well. Commented Jul 28, 2023 at 7:49
• Perhaps, could you also help me with the following question as well? mathoverflow.net/questions/451574/… Commented Jul 28, 2023 at 8:03

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 $$h\notin L^2([0, 1])$$.