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Consider the function $$ f(x,y):=\lVert x\rVert^{1-n}\ln(\lVert x\rVert)(\arctan(\lVert x-y\rVert))^{-\alpha},~~0<\alpha<n,~~n>1,~~(x,y)\in\Omega\times\Omega,~~~\Omega\subset\mathbb{R}^n $$ with $x\neq y$.

I am searching for an estimation

$$ \lvert f(x,y)\rvert\leq\frac{\lvert a(x,y)\rvert}{\lVert x-y\rVert^{\alpha}} $$ with $a\in L^{\infty}(\Omega\times\Omega)$. The hint is, to use polar coordinates.

Here (Find a weakly singular kernel function for an estimation of a kernel) the same task was asked for the function $g(x,y):= (\arctan(\lVert x-y\rVert))^{-\alpha}$ and I found, using the main value theorem, that $$ \lvert g(x,y)\rvert\leq\frac{(\lVert x-y\rVert^2+1)^{\alpha}}{\lVert x-y\rVert^{\alpha}}. $$

So I think to estimate now f, I have to use this result, getting for now

$$ \lvert f(x,y)\rvert\leq\lvert\lVert x\rVert^{1-n}\ln(\lVert x\rVert)\rvert\frac{(\lVert x-y\rVert^2+1)^{\alpha}}{\lVert x-y\rVert^{\alpha}} $$

But I do not know how I can continue now, especially using polar coordinates.

Hope you can help me.

Edit: Correction of the task!

There was a mistake in the function f!

It has to be $$ f(x,y):=\lVert x\rVert\ln(\lVert x\rVert)(\arctan(\lVert x-y\rVert))^{-\alpha}. $$

Community
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    Is the origin in $\Omega$? If so, your desired estimate doesn't look true, since $|x|^{1-n}\ln|x|$ isn't essentially bounded (bounded a.e.) near the origin. Or: is the exponent in the denominator of your upper bound necessarily the same as in the definition of $f$...? – Andrew D. Hwang Dec 07 '13 at 13:13
  • In fact I do not know if my estimation with which I started is right, it was just an idea, because i showed the estimation of the term $(\arctan(\lVert x-y\rVert))^{-\alpha}$ in another task. Maybe one does not have to use that at all... I think it does not have to be the same exponent. The aim is only to find any estimation of the form $\ldots\leq \frac{\lvert a(x,y)\rvert}{\lVert x-y\rVert^{\alpha}}, 0<\alpha<n$. –  Dec 07 '13 at 13:16
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    Your $\arctan$ estimate looks correct, and is "(asymptotically) optimal" since $\arctan u = u + O(u^3)$. But if your estimate is allowed to contain a larger exponent, you're in fine shape. (Assuming that's the case: Did you really mean $|x|$, etc., or are these factors $|x - y|$, etc.?) – Andrew D. Hwang Dec 07 '13 at 13:21
  • I just cited the task.. so it is indeed $\lVert x\rVert^{1-n}$ and $\ln(\lVert x\rVert)$. But maybe this is a mistake and it is meant $\lVert x-y\rVert^{1-n}$ etc.? Our prof makes a lot of mistakes on worksheets... I am confused about the hint, to use polarcoordinates.. do you know what is meant? –  Dec 07 '13 at 13:24
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    Polar coordinates: Every non-zero vector in $\mathbf{R}^n$ can be written as $|x|\cdot\dfrac{x}{|x|}$, the product of a positive scalar and a unit vector. (This is a precise phrasing of the principle that "a vector is a quantity having magnitude and direction".) – Andrew D. Hwang Dec 07 '13 at 13:34
  • I see that one can write it that way, ok. But I do not see what thas has to do with polar coordinates and how I can use that here to find the estimation.. sorry, I am a little bit overcharged to be honest. Maybe you can write it down more explicitly in an answer? I would be very thankful because I already think about that task too(!) long. –  Dec 07 '13 at 13:39
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    Regarding $|x|$ versus $|x-y|$, the former is conceivably right (i.e., what your instructor meant). However, I'd guess they meant $|x - y|$, and they want you to "absorb" the logarithmic singularity into a small power. The relevant calculus factoid is that for every $\delta > 0$, $|x|^\delta \ln|x| \to 0$ as $x \to 0$. – Andrew D. Hwang Dec 07 '13 at 13:39
  • let us continue this discussion in chat –  Dec 07 '13 at 13:44
  • I got an answer: No, the task is allright, there are no mistakes. –  Dec 07 '13 at 14:49
  • By the way: The instructor said that he extended with $\lVert x-y\rVert^{\alpha}$ to find the estimation for the arctan-expression. And with polar coordinates he means to estimate an integral of $\lVert x\rVert^{1-n}\ln(\lVert x\rVert)$. But i do not understand what he means... –  Dec 07 '13 at 15:03
  • I asked the instructor what he means with the integral.. he only said... a has to be in $L^{\infty}(\Omega\times\Omega)$. Nevertheless, I do not understand what this has to do with an integral... –  Dec 07 '13 at 17:22
  • LATEST NEWS: The instructor means $\lVert x\rVert\ln(\lVert x\rVert)(\arctan(\lVert x-y\rVert))^{-\alpha}$. So the exponent of the norm of x is 1, not 1-n. –  Dec 08 '13 at 10:19
  • LATEST NEWS: 10+ comments to (not even) reach a definitive version of the question is MUCH TOO MUCH. – Did Dec 08 '13 at 11:02
  • I regret that. I would have prefered a shorter way to the solution, too. –  Dec 08 '13 at 11:10
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    Sorry but I fail to see what could prevent you to arrive AT THE ONSET with a well-formed question. – Did Dec 08 '13 at 11:12

2 Answers2

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One can guess that a missing hypothesis is that $\Omega$ is bounded, say $\|x\|\leqslant c$ for every $x$ in $\Omega$, for some $c\geqslant2$, then $|\|x\|\log\|x\||\leqslant c\log c$ for every $x$ in $\Omega$. Since $\|x-y\|\leqslant2c$ for every $x$ and $y$ in $\Omega$, the task is complete if one shows that $$ (\arctan t)^{-\alpha}\leqslant Ct^{-\alpha}, $$ for some finite $C$, for every $0\lt t\leqslant2c$. Equivalently, one wants $$ \arctan t\geqslant C't, $$ for some positive $C'$, for every $0\lt t\leqslant2c$. Since $\arctan$ is concave, $C'=\arctan(2c)/(2c)$ fits. Finally, since $2c\geqslant4$ and $\arctan4\geqslant1$, for each positive $\alpha$, $a(x,y)=A_\alpha$ works, with $$ A_\alpha=2^\alpha\cdot c^{\alpha+1}\log c. $$

Did
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  • Yes, $\Omega$ is indeed bounded. I have one question. Because in the link that I gave I already found an estimation for the term with $\arctan$, the task here reduces to show that $\lVert x\rVert\ln(\lVert x\rVert)\leq\lvert a(x,y)\rvert$ for a $a\in L^{\infty}(\Omega\times\Omega)$, right? The instructor said, this is equivalent to that the integral $\int_{\Omega\times\Omega}\lvert\lVert x\rVert\ln(\lVert x\rVert), dy, dy$ exists. Do you understand that? –  Dec 08 '13 at 11:04
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    Yes: for every bounded domain, bounded function implies integrable function (but I would object to "equivalent", if really the instructor said it). – Did Dec 08 '13 at 11:11
  • But I do not understand how I can find the function a when I show that this integral is existing. –  Dec 08 '13 at 11:12
  • If I show that $\int_{\Omega\times\Omega}\lvert f(x,y)\rvert, dy, dy<\infty$ how can I then say that there is a function a with $\lvert f(x,y\rvert\leq\lvert a(x,y)\rvert$ and $a\in L^{\infty}(\Omega\times\Omega)$? And why $dy dy$ and not $dxdy$? –  Dec 08 '13 at 11:34
  • You can't. As I already explained. Do you read the comments? – Did Dec 08 '13 at 11:37
  • Of course. You said that for a bounded domain bounded function implies integrable function. But that this is npt equivalent. But then I really wonder what the instructor meant. –  Dec 08 '13 at 11:41
  • I am really sorry for being the xaver here, but if I already found an estimation for the term with arctan, why then not choosing $a(x,y):=c\ln c\cdot b$, where b is the term estimating the term with arctan? –  Dec 08 '13 at 11:52
  • Are you going to engage in a new looooong string of comments, just after having been signaled this was not the way to go? Please STOP THIS. New question? Then new post. Forgot to mention important pieces of the setting, in the question? Too bad, too late, hence new post. – Did Dec 08 '13 at 12:26
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Easy Peasy. The new function ||x|| ln ||x|| is continuous (although negative near he orign), so it is bounded on $\Omega$ if $\Omega$ is bounded, so you can just use your old estimate times the bound on xlnx.

Brian Rushton
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