A simple bilinear estimate

Question

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.

Answer 1

Since $f$ is only defined for $x\in [0,1]$ and I can perform a change of variables to get $y \mapsto 1-y$, I think you are asking about the inequality

$$ \int_{[0,1]\times [0,\delta]} \frac{f(x)g(y)}{|x-y|^\alpha} \leq C \delta^t \|f\|_p \|g\|_q $$

The right hand side is equivalent to

$$ \int_{[0,1]\times [0,1]} \frac{f(x) g(y) \chi_\delta(y)}{|x-y|^\alpha} $$

where $\chi_\delta$ is the characteristic function of $[0,\delta]$.

Standard HLS tells you that this integral is bounded by

$$ \|f\|_p \|g\chi_\delta\|_{q'} $$

when $\frac{1}{p} + \frac{1}{q'} = 2 - \alpha$. If $q > q'$, we can further estimate

$$ \|g \chi_\delta\|_{q'} \leq \|g\|_q \delta^{\frac{1}{q'} - \frac{1}{q}} $$

which shows that the original inequality holds with $t = 2 - \alpha - \frac{1}{p} - \frac{1}{q}$. (This is probably what you said you did in the question statement.)

I claim this is the best you can do.

Indeed, as long as $g$ vanishes outside of $[0,\delta]$, the truncation by $\chi_\delta$ is a null operation. So let

$$ f_\delta(x) = f(x/\delta), \quad g_\delta(x) = g(x/\delta) $$ both extended by $0$ outside $[0,\delta]$.

Then we find that $f_\delta$ and $g_\delta$ are both still in $L^p$ and $L^q$ respectively. Change of variables tells us

$$ \int_{[0,1]\times[0,\delta]} \frac{f_\delta(x) g_\delta(y)}{|x-y|^\alpha} = \delta^{2-\alpha} \int_{[0,1]\times[0,1]} \frac{f(x) g(y)}{|x-y|^{\alpha}} $$

while

$$ \|f_\delta\|_p \|g_\delta\|_q = \delta^{\frac1p + \frac1q}\|f \|_p \|g\|_q $$

In order of the desired inequality to hold, as $\delta \searrow 0$ you must have

$$ \delta^{2-\alpha} \leq C \delta^{t + \frac1p + \frac1q} $$

uniformly, which requires $t \leq 2-\alpha - \frac1p - \frac1q$. So the previously found value is the optimal one.