Let $2\leq p,q <\infty$ and fix $0<\alpha<1$ such that $\frac{1}{p}+\frac{1}{q}\leq 2-\alpha$.

Suppose that $f\in L^{p}([0,1])$ and $g\in L^{q}([0,1])$.

What is the optimal value of $t=t(\alpha,p,q)$ such that

\begin{equation*} \int_{\substack{0<x<2 \\\\ 1-\delta<y<1}} \frac{f(x)g(y)}{|x-y|^{\alpha}} dxdy\leq C \delta^t \|f\| _{p} \|g\| _{q}. \end{equation*}

The only thing I can think of is to apply the Hardy-Littlewood-Sobolev inequality then use H{o}lder's inequality. There must be some more sophisticated approach.

Let $2\leq p,q <\infty$ and fix $0<\alpha<1$ such that $\frac{1}{p}+\frac{1}{q}\leq 2-\alpha$.

Suppose that $f\in L^{p}([0,1])$ and $g\in L^{q}([0,1])$.

What is the optimal value of $t=t(\alpha,p,q)$ such that

\begin{equation*} \int_{\substack{0<x<2 \\\\ 1-\delta<y<1}} \frac{f(x)g(y)}{|x-y|^{\alpha}} dxdy\leq C \delta^t \|f\| _{p} \|g\| _{q}. \end{equation*}

The only thing I can think of is to apply the Hardy-Littlewood-Sobolev inequality then use Holder's inequality. There must be some more sophisticated approach.