This is a transfer of the question https://math.stackexchange.com/questions/4996853/an-lp-lp-inequality
Let $a\in (0,1)$ and $1<p<\infty$ and use $L^{p}$ to denote the space $L^{p}([0,\infty))$ and $\|\cdot\|_{p}$ to denote the norm $\|\cdot\|_{L^{p}([0,\infty))}$.
Assume that $f:[0,\infty)\rightarrow \mathbb{R}$ is continuously differentiable such that $\sup_{x\geq 0}|f(x)|=1$ and $f^{\prime}\in L^{p}$. Then, by Minkowski's integral inequality, one has $x\mapsto\int_{0}^{1}(1-t)^{-a}f^{\prime}(t x)dt$ in $ L^{p}$. This, in turn, together with the boundedness assumption on $f$ guarantees that $\int_{0}^{1}(1-t)^{-a}f(tx) f^{\prime}(t x)dt$ is in $ L^{p}$ as well.
I am studying the inequality $$\left\|\int_{0}^{1}\frac{f(t x) f^{\prime}(t x)}{(1-t)^{a}}dt\right\|_{p}\leq C \left\|\int_{0}^{1}\frac{f^{\prime}(t x)}{(1-t)^{a}}dt\right\|_{p}.\qquad (*)$$
My earliest attempts were focused on finding a counterexample for $(*)$. I discuss that in detail in the math.stackexchange version linked above. The inequality (*) is in fact false if $f$ is differentiable almost everywhere with a derivative continuous almost everywhere. Approximating the counterexamples found in a Sobolev space is not possible, and numerical tests show that mollifying $f$ or $f^{\prime}$ in the counterexample produces a smooth function for which $(*)$ holds.
Can we show that $(*)$ holds true ?
