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 $1<p\leq \infty$ 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 ?

  • 1
    $\begingroup$ Yes indeed, no idea. $\endgroup$
    – user130023
    Oct 4, 2022 at 19:31

2 Answers 2


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 $1<p<\infty$ (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
    $\begingroup$ +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. $\endgroup$ Oct 5, 2022 at 13:58
  • $\begingroup$ @Giorgio Metafune It is easy to see that $\int_{|y|<r<1}\frac{dy}{|y|(|\log|y||)^\alpha}=\infty$, but why is $\liminf_{|x|\rightarrow 0} \int_{|x-y|<t,\,|y|<r<1}\frac{dy}{{|x-y|^{\frac{n}{p^{\prime}}}|y|^{\frac np}} (|\log|y||)^\alpha}=\infty$ ? $\endgroup$
    – Medo
    Oct 5, 2022 at 16:57
  • $\begingroup$ Because it is bigger of the integral with liminf inside. $\endgroup$ Oct 5, 2022 at 17:02
  • $\begingroup$ So we are using Fatou's lemma twice actually ? $\endgroup$
    – Medo
    Oct 5, 2022 at 17:07
  • 1
    $\begingroup$ Yes, one for the integral in $y$, the other in $t$. $\endgroup$ 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 $n > 2p'$.


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