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I need to show that the weakly singular integral operator $T_\alpha$ defined by: $$(T_\alpha f)(x)=\int_0^1|x-y|^{-\alpha}f(y)dy$$ where $0<\alpha<1$, is well-defined and bounded on $L^p([0, 1])$ for all $1 ≤ p ≤ ∞$.

I am really stuck with this problem, can someone give me some hints on how to develop the proof? $\\$ I was also wondering, is this true for $\alpha=1$?

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Let's begin with $p=1$. $$\begin{align} \int_0^1|(T_\alpha f(x)|\,dx&=\int_0^1\Bigl|\int_0^1|x-y|^{-\alpha}f(y)\,dy\Bigr|\,dx\\ &\le\int_0^1|f(y)|\Bigl(\int_0^1|x-y|^{-\alpha}\,dx\Bigr)\,dy\\ &=\frac{1}{1-\alpha}\int_0^1\bigl(y^{1-\alpha}+(1-y)^{1-\alpha}\bigr)\,|f(y)|\,dy\\ &\le \frac{2^\alpha}{1-\alpha}\int_0^1|f(y)|\,dy. \end{align}$$ It is even simpler for $p=\infty$. For $1<p<\infty$ use interpolation.

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