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Let $\Omega\subset\mathbb{R}^n (n>1)$ be a bounded domain and $0<\alpha<n$. Show, that the kernel function $$ k(x,y):=(\arctan(\lVert x-y\rVert))^{-\alpha}\text{ for }x\neq y $$ can be estimated by a kernel function that is weakly singular, i.e. that $$ \lvert k(x,y)\rvert\leq\frac{\lvert a(x,y)\rvert}{\lVert x-y\rVert^{\alpha}}. $$ for a function $a\in L^{\infty}(\Omega\times\Omega)$.

My idea is to use the mean value theorem. $$ \arctan(\lVert x-y\rVert)=\frac{\lVert x-y\rVert}{1+\xi^2}>\frac{\lVert x-y\rVert}{1+\lVert x-y\rVert^2} $$ for a $\xi\in (0,\lVert x-y\rVert)$.

$$ \Rightarrow \lvert k(x,y)\rvert=\left\lvert\frac{1}{(\arctan(\lVert x-y\rVert))^{\alpha}}\right\rvert=\frac{1}{(\arctan(\lVert x-y\rVert))^{\alpha}}\leq\frac{(1+\lVert x-y\rVert^2)^{\alpha}}{\lVert x-y\rVert^{\alpha}} $$

So I found $$ a(x,y):=(1+\lVert x-y\rVert^2)^{\alpha}, $$ which is to my opinion in $L^{\infty}(\Omega\times\Omega)$, because $$ \lvert a(x,y)\rvert=(1+\lVert x-y\rVert^2)^{\alpha}\leq (1+(2M)^2)^{\alpha}<\infty $$ with $M>\lVert\omega\rVert~\forall~\omega\in\Omega$, because $\Omega$ is bounded.

So $a$ is bounded and then in $L^{\infty}(\Omega\times\Omega)$.

Could you please say me, if my proof resp. estimation is okay?

With greetings

math12

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    Your proof looks good, though you might want to be explicit that you're using the $1$-dimensional MVT (otherwise there's potential cause for skepticism, since $\Omega$ needn't be convex), and you should probably be explicit that your $a$ is bounded almost everywhere (i.e., not on the diagonal on $\Omega \times \Omega$). – Andrew D. Hwang Dec 07 '13 at 13:02
  • Hello, thanks for your answer. But I did mention that I used the mean value theorem - or waht do you mean? –  Dec 07 '13 at 13:06
  • There's a "multi-variable" version of the MVT that looks at $f\bigl((1-t)x + ty\bigr)$, but the proof requires convexity of $\Omega$ (since the argument of $f$ lies on the segment from $x$ to $y$). It took me a minute or two to realize this was not the basis of your argument. :) – Andrew D. Hwang Dec 07 '13 at 13:17
  • Ok, now I understand what you mean!... Thanks, then I will indeed add that I used the one-dim. MVT. –  Dec 07 '13 at 13:20

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