Bound on convolution: $ | (h * f^2) (x)| \leq \| f\|^2_2 g(h)$

Question

I am trying to find bounds for the following quantity. Take two functions $f,h \in L^{1} \cap L^2$ but $\|h \|_{\infty} = \infty$. Is there a way to obtain a bound of the following type:

$$ | (h * f^2) (x)| \leq \| f\|^2_2 g(h)$$

where $g$ is some function of $h$? This would be the case if we could take the sup norm of $h$ out of the integral by using Holder's inequality, but in this case is not allowed as $\|h \|_{\infty} = \infty$.

PS: you can also assume that $(h * f^2) (x)$ is everywhere well-defined.

Answer 1

The given conditions do not imply $h * (f^2)\in L^\infty$, so no bound of the type can hold.

1. Let $\phi_a: \mathbb R\to [0,\infty]$ be defined by $\phi_a(x) = 1/|x|^a$. Pick any $a_{1},a_2>0$ such that $a_1 < 1/2$ and $1-a_1\le a_2<1$, and put $h = \phi_{a_1} \mathbf 1_{[-1,1]}$, $f=\sqrt{\phi_{a_2}} \mathbf 1_{[-100,100]}$. By construction $h,f\in L^1\cap L^2$. Now for say $|x|<1$, $$ h * (f^2)(x)= \int_{-1}^1 h(y) f^2(x-y)dy = \int_{-1}^1 \frac{dy}{|y|^{a_1}|y-x|^{a_2}}, $$ and $h * (f^2)(0)=\infty$ since $a_1+a_2\ge 1$. So $h * (f^2)$ is not bounded.
2. In fact, by triangle inequality and monotone convergence, $h * (f^2)(x)\to \infty$ as $x\to 0$. Hence, $h * (f^2)\notin L^\infty$.
3. The point of blowup can be easily translated to any point; if $\tau_c f (x)= f(x-c)$ is translation by $c$, then note $$ h* (\tau_cf)^2 = \tau_c(h*(f^2))$$ and now $h* (\tau_cf)^2$ blows up at $x=c$. Since $c$ can be chosen arbitrarily and independently from $h$, it is not even possible to have the bound $|h* (f^2)(x)| \le \|f\|_2^2 g(h)(x)$ where $g(h)$ is not identically $\infty$.
4. Asking for $|h * (f^2)(x)| < \infty$ for all $x$ does not help. We can simply replace $f$ from 1 with $f_N=\min(N,f)$ and take $N$ large. If $g(h)(c) < \infty$ at $c=0$ (wlog by point 3) then $\text{RHS}\le \|f\|^2_2 g(h)(0) < \infty$ uniformly in $N$, but $\text{LHS} =|h* (f_N)^2(x)|\to\infty$ as $N\to\infty$.