# Is this an $L^p-L^{\infty}$ operator?

Let $$1\leq p <\infty$$ and let $$p^{\prime}$$ denote its conjugate exponent. Consider the following operator on Schwartz functions:

$$Tf(x)=\int_{0}^{\infty}t^{\frac{n}{2 p^{\prime}}-1}e^{-t} \int_{|x-y|^2\leq t}\frac{f(y)}{|x-y|^{\frac{n}{p^{\prime}}}}dy dt,\qquad x\in \mathbb{R}^{n}.$$

I have tried to prove that $$T$$ is bounded from $$L^{p}$$ to $$L^{\infty}$$ but failed so far.

Young's inequality for convolution is not useful with the $$y$$-integral as $$|\cdot|^{\frac{n}{p^{\prime}}}$$ is not in $$L^{p^{\prime}}(B(t))$$ with $$B(t)$$ the standard ball centered at the origin with radius $$t>0$$.

Hardy-Little-wood-Sobolev inequality is not useful for obtaining $$L^{\infty}$$ boundedness.

One could look at the Hardy-Littlewood maximal operator $$\displaystyle Mf(x)=\sup_{r>0} \frac{1}{B(x,r)} \int_{B(x,r)}\frac{f(y)}{|x-y|^{\frac{n}{p^{\prime}}}}dy$$ since
$$\frac{1}{t^{\frac{n}{2}}}\int_{|x-y|^2\leq t}\frac{f(y)}{|x-y|^{\frac{n}{p^{\prime}}}}dy\leq Mf(x).$$

The maximal operator is known to be bounded from $$L^{p}$$ to $$L^{p}$$ for all $$1 and from $$L^1$$ to weak $$L^{1}$$. I have no idea about the boundedness of $$M$$ from $$L^p$$ to $$L^{\infty}$$ when $$p<\infty$$.

Is it true that $$\|Tf\|_{L^{\infty}}\leq C \|f\|_{L^{p}}$$ for any $$1\leq p<\infty$$ or is there a counterexample ?

• Yes indeed, no idea. Oct 4, 2022 at 19:31

No this is not true. Take $$n=1$$, $$p=2$$ and $$f(y)=\frac{1}{\sqrt {|y|} (|\log|y||)^\alpha}\chi_{(-1,1)}(y)$$ with $$\frac 12 < \alpha <1$$. Then $$f \in L^2(\mathbb R)$$ but the innermost integral diverges at $$x=0$$ for every $$t>0$$.

EDIT The same counterexample can be done for general $$n$$ and $$1 (the case $$p=1$$ is in the answer by @Willie Wong). Take $$f(y)=\frac{1}{{|y|^{\frac np}} (|\log|y||)^\alpha}\chi_{B_r}(y)$$ with $$r<1$$ and $$\frac 1p <\alpha <1$$. Then $$Tf(0)=\infty$$ and, by Fatou, $$\lim_{x \to 0}Tf(x)=\infty$$.

• +1 for the general case! I got confused by the question last night, because the OP mentioned maximal functions. But because the inner integration is with $|x-y|^2 \leq t$ (and not the other way around) the outer integral cannot regularize away any singularity at zero. Oct 5, 2022 at 13:58
• @Giorgio Metafune It is easy to see that $\int_{|y|<r<1}\frac{dy}{|y|(|\log|y||)^\alpha}=\infty$, but why is $\liminf_{|x|\rightarrow 0} \int_{|x-y|<t,\,|y|<r<1}\frac{dy}{{|x-y|^{\frac{n}{p^{\prime}}}|y|^{\frac np}} (|\log|y||)^\alpha}=\infty$ ?
– Medo
Oct 5, 2022 at 16:57
• Because it is bigger of the integral with liminf inside. Oct 5, 2022 at 17:02
• So we are using Fatou's lemma twice actually ?
– Medo
Oct 5, 2022 at 17:07
• Yes, one for the integral in $y$, the other in $t$. Oct 5, 2022 at 17:10

The case $$p = 1$$ asks if $$\int_0^\infty t^{-1} e^{-t} \int_{|x-y|^2\leq t} f(y) dy dt \leq 1$$ for every $$f$$ with $$\int |f| = 1$$. Just take $$|f_n|$$ a family approximating the identity with support on $$B(0,1/n)$$, then you see that the corresponding integral blows up.

For $$n < 2p'$$ (this includes the case that Giorgio Metafune considered), let $$f$$ be a smooth function that is concentrated on the dyadic annulus of size $$2^{-k}$$ with height 1. Then its $$L^p$$ norm is of size $$2^{-kn/p}$$. For $$\sqrt{t} \approx 2^{-k}$$, the inner integral evaluates to the same value. So the full integral is bounded below by $$2^{-kn/p} \cdot 2^{-k(n/p' - 2)}$$ for $$k > 0$$. So we see that when $$n < 2p'$$ taking $$k\nearrow \infty$$ you also get counterexamples to the boundedness.

I think with some work a similar counterexample can be made for $$n = 2p'$$. But I am not sure about what happens when $$n > 2p'$$.