How can we prove $\int_{B(0,r)}|\log |x||dx\le cr^2|\log r|,r\to 0$? Any hint is appreciated.
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3Assuming you mean $\log|x|$ and you're working in ${\mathbb R}^2$, switch to polar coordinates and just do the integral. – Zarrax Mar 04 '13 at 16:38
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1How can this be true? If $r=1$, the upper bound is zero, but the integral is clearly not. – copper.hat Mar 04 '13 at 16:56
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It still cannot be true. The limit as $r \to 1$ of the right hand side is still zero, but the left hand side is increasing as a function of $r$ and is non-zero. – copper.hat Mar 04 '13 at 17:01
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Is this the final question? – copper.hat Mar 04 '13 at 17:05
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Yes, thank @copper.hat. – Sam Mar 04 '13 at 17:09
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Thank @Zarrax for your hint. – Sam Mar 04 '13 at 17:09
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For $r<1$: $$\int |\log |x||dx = r(1 - \log r) = r(|\log r|+1)$$ Now to find $c$, use the function: $$f(r) = c r |\log r|-|\log r|$$ Now for suitable $c$, $f(r) \ge 1$, so: $$\int |\log |x||dx = r(|\log r|+1) \le cr^2 |\log r|$$ Now for all $r$ in the open interval $(0,1)$ there exists such a $c$, but this is of course a function of $r$.
Nathaniel Bubis
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How can you conclude $r(|\log r|+1) \le cr^2 |\log r|$? Divide across by $r$ and let $r \uparrow 1$, then you get $1 \le 0$. – copper.hat Mar 04 '13 at 17:05
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@copper.hat - for large enough $c$, it works out, but as I stated only in the open interval. – Nathaniel Bubis Mar 04 '13 at 17:15
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If $c$ is a function of $r$, then the bound is rather meaningless. All it states is that $r^2 |\log r| \neq 0$. – copper.hat Mar 04 '13 at 17:20
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@copper.hat - Never claimed it meant anything, I just tried to show that $c$ cannot be constant. – Nathaniel Bubis Mar 04 '13 at 17:22
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Thanks a lot, @nbubis. I like your answer. Also thank copper.hat comments. – Sam Mar 04 '13 at 20:29