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 ?



Is this an $L^p-L^{\infty}$ operator?
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, $\lim\_{x \to 0}Tf(x)=\infty$.


