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 ?