Let $f\in L^{p}([1,2])$ with $p>1$, and define $$Tf(x):= \int_{1<y<2}\frac{f(y)}{|x-y|}dy,\quad 0<x<1.$$
I am trying to prove/disporove that
$$\|Tf\|_{L^p([0,1])}\leq C \|f\|_{L^p([1,2])}\quad (1)$$
I did not go far enough by Minkowski's integral inequality which implies
$$\left( \int_{0}^{1}\left(\int_{1<y<2}\frac{f(y)}{|x-y|}dy\right)^{p} dx \right)^{1/p}\leq \int_{1<y<2}f(y)\left(\int_{0}^{1}\frac{dx}{(y-x)^{p}}\right)^{1/p}dy$$
One can estimate
$$\left(\int_{0}^{1}\frac{dx}{(y-x)^{p}}\right)^{1/p}=\frac{1}{p-1}\left(\frac{1}{(y-1)^{p-1}}-\frac{1}{y^{p-1}}\right)^{1/p}\leq \frac{C}{(y-1)^{1-\frac{1}{p}}}$$ for $y$ close enough to $1$.
But this is not very helpful since $\int_{0}^{1}\frac{f(y)}{(y-1)^{1-\frac{1}{p}}} dx$ is not bounded by $\|f\|_{L^{p}([1,2])}$ considering the counterexample $g_{p}(x):=\frac{\chi_{[1,3/2]}(x)}{|\log{(x-1)}|(x-1)^{\frac{1}{p}}}$.
Let $(Sf)(x)=f(x+1)$ and $(Rf)(x)=f(1-x).$ Then $S$ and $R$ are isometries from $L^p(1,2)$ to $L^p(0,1)$ and $L^p(0,1)$ to $L^p(0,1),$ respectively. Observe that $T=RPHS, $ where $$(Hf)(x)=\int\limits_0^\infty {f(y)\over y+x}\,dy$$ and $P$ is the restriction operator from $L^p(0,\infty)$ to $L^p(0,1).$
It suffices to show that the operator $H$ is bounded. The operator $H$ is a continuous version of the Hilbert matrix. By substitution $y={tx}$ we get $$(Hf)(x)=\int\limits_0^\infty {f(tx)\over 1+t}\,dt$$ Let $(H_tf)(x)=f(tx).$ Then $\|H_tf\|_p=t^{-1/p}\|f\|_p.$ Hence $$\|Hf\|_p\le \int\limits_0^\infty {1\over (1+t)t^{1/p}}\,dt\ \|f\|_p$$ Therefore $$\|T\|\le \int\limits_0^\infty {1\over (1+t)t^{1/p}}\,dt$$