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Let $f : M \rightarrow \mathbb{R}$ be a continuous function on $M \subset \mathbb{R}^2$, where $M$ is compact (so $f$ is uniformly cont.). I have to show that $T : C(M) \rightarrow C(M) : Tf := \int_M \frac{1}{|t-\tau|} f(\tau) d\tau$ is bounded in the usual norm on $C(M)$, i.e. $\vert \vert f \vert \vert = \sup_{\tau \in M} f$.

If the kernel function was continuous, this would be child's play, since it would also be bounded and the rest follows easily enough. However, since $1/|x-y|$ has a singularity, I am unsure of what the proper angle of attack is. Can anyone help?

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    Do you mean $\int_M\frac1{|t-\tau|}f(t)dt$, or perhaps $\int_M\frac1{|t-\tau|}f(\tau)d\tau$? – 2017-02-27
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    Yes, the last, sorry. My bad. I edited it now. – 2017-02-27

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The point is that $1/|t|$ locally integrable on $\mathbb R^2$: using polar coordinates, for suitable $R$

$$ \int_M \frac{1}{|t-\tau|} \; dt \le \int_{0}^{2\pi} d\theta \int_0^R dr $$

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    Hey, thanks for the answer, but could you elaborate a bit on that? – 2017-02-27
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    $$\|Tf\| = \sup_{\tau \in M} \int_M \frac{f(t)}{|t-\tau|}\; dt \le \|f\| \sup_{\tau \in M} \int_{M} \dfrac{dt}{|t-\tau|} $$ – 2017-02-27
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    Thanks a lot. Appreciate it. – 2017-02-27