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..



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 $f(y)$?
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](https://math.stackexchange.com/q/3639820), 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](https://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function#Proof)
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.


