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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 $f$. – Davide Giraudo Jun 29 '12 at 12:06

I guess you need the equicontinuity for $f$ in the unit ball of $\mathscr{C}^0([0,1])$, and not the whole space (which cannot be, check it !). For this, first notice that given $f$ in the unit ball of $\mathscr{C}^0([0,1])$, and $x,z\in[0,1]$ \begin{align*} |Jf(x)-Jf(z)| \leq \int_0^1 \left|\frac{1}{|x-y|^c}-\frac{1}{|z-y|^c}\right|\mathrm{d}y, \end{align*} so that we just have to check the continuity of $Jf$, where $f=1$. Now in this case, for $(x_n)_n\in[0,1]^\mathbb{N}$ such as $(x_n)_n\rightarrow a\in[0,1]$, we have by (two) change(s) of variable \begin{align*} Jf(x_n) = \int_0^1 \frac{1}{|x_n-y|^c} \mathrm{d} y = \int_{x_n-1}^{x_n} \frac{1}{|u|^c}\mathrm{d}u \operatorname*{\longrightarrow}_{n\rightarrow+\infty} \int_{a-1}^{a} \frac{1}{|u|^c}=Jf(a),\end{align*} by dominated convergence, since \begin{align*} \mathbf{1}_{[x_n-1,x_n]}(t)\frac{1}{|t|^c} \leq \mathbf{1}_{[-1,1]}(t)\frac{1}{|t|^c} \in L^1(\mathbb{R}), \end{align*} since $c<1$.

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