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\begin{align} J:C[0,1]\rightarrow C[0,1], \quad f \rightarrow (J f)(x)=\int_0^1\frac{f(y)}{\lvert x-y\lvert^c} dy \end{align} For which $c>0$ is $(J f)_{f\in C[0,1]}$ equicontinous?

For $c\geq 1$ it's clear that $Jf$ is not even bounded. But for $c<1$ I don't even know how to prove that $Jf\in C[0,1]$. I'd like to use the dominated convergence theorem, but that will be difficult since $\frac{1}{\lvert x-y\lvert^c}$ is not bounded.

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    To see the continuity of $J(f)$, cut the integral in two parts ($0\leq y\leq x$ and the other). Then do a substitution to get $\frac 1{t^c}$. You will get something like $\int_0^x\frac{(x-t)}{t^c}dt+\int_0^{1-x}\frac{f(x+t)}{t^c}dt$. Take $x\in [0,1]$ and a sequence $x_n$ which converges to $x$. Then difference of the integrals is not two big (introduce $\int_0^x\frac{f(x_n-t)}{t^c}$) then use uniform continuity of $f$.2012-06-29

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