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Define $u : (-1,1) \to \Bbb R$ by $$ u(x) = \left( \log \frac{1}{1-x} \right)^{\alpha} ( 1/2 \leqslant x < 1 ), \;\;\;\;\;u(x) = (\log 2 )^{\alpha} \;( -1 < x \leqslant 1/2 ) $$ If $0 < \alpha < 1/2$ then I want to prove that $$ \int_{-1}^1 (1-x^2 ) | u'(x) |^2 dx + \int_{-1}^1 |u(x)|^2 dx < \infty . $$

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What did you try? What are the possible causes of divergence of these integrals? – Did Jul 30 '12 at 11:03
Ann: you realize that the convergence of the second integral is trivial and that the problem is with the FIRST integral, don't you? – Did Jul 30 '12 at 17:11
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Split your integrals on the two intervals $(-1,1/2)$ and $(1/2,1)$. For the function on the interval $(-1,1/2)$, it is easy to see that they are finite. On the interval $(1/2,1)$ we have
$\int_{1/2}^{1} |u(x)|^2\, dx $, and you can do the same technique with the other one,

$$\int_{1/2}^{1} |u(x)|^2\, dx = \int_{1/2}^{1} \left({\ln\left(\frac{1}{1-x}\right)}\right)^{2\alpha}\, dx = \int _{\ln \left( 2 \right) }^{\infty }\!{t}^{2\,\alpha}{{\rm e}^{-t} }{dt} < \int _{ 0 }^{\infty }\!{t}^{2\,\alpha}{{\rm e}^{-t} }{dt} = \Gamma(2\alpha + 1) < \infty \,,$$

where $\Gamma(2\alpha +1)$ is the gamma function, $0<\alpha<1/2$. We used the change of variables, $ x=1-{\rm e}^{-t} $. You can use the same change of variable and the same technique with the other integral.

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For $x$ in $[\frac12,1)$, $u'(x)=\alpha\,\frac1{1-x}\left(\log\frac1{1-x}\right)^{\alpha-1}$ and $1-x^2\leqslant2(1-x)$, hence the change of variable $\log(1-x)=-z$ yields $$ \int^1(1-x^2)(u'(x))^2\,\mathrm dx\leqslant\int^1\frac{2\alpha^2\,\mathrm dx}{(1-x)(\log(1/(1-x))^{2-2\alpha}}=\int^{+\infty}\frac{2\alpha^2\,\mathrm dz}{z^{2-2\alpha}}, $$ which converges since $2-2\alpha\gt1$, that is, $\alpha\lt\frac12$. The integral with $(u(x))^2$ causes no problem.

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