Let $\Omega = (0, \frac{1}{2})$ and $f: \Omega \to \mathbb R: x \to (x(\log(x))^2)^{-\frac{1}{p}}$
In the lecture, we had this statement: $f \in L^p (\Omega)$ but $f \not \in L^{p+ \epsilon} (\Omega)$ for all $\epsilon > 0$.
Why does this hold true?
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Are you sure that $\Omega=(0,1/2)$? – Julián Aguirre Oct 03 '18 at 13:34
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Yes, I'm sure... – StMan Oct 03 '18 at 14:25
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It is just computation. \begin{align} \int_\Omega|f(x)|^p\,dx&=\int_0^{1/2}\frac{dx}{x\,(\log x)^2}\\ &=\lim_{t\to0^+}\int_t^{1/2}\frac{dx}{x\,(\log x)^2}\\ &=\frac{1}{\log 2}\\ &<\infty. \end{align} This shows $f\in L^p$. Now let $q=(p+\epsilon)/p>1$. $$ \int_\Omega|f(x)|^{p+\epsilon}\,dx=\int_0^{1/2}\frac{dx}{x^q\,(\log x)^{2q}}. $$ Comparison with $1/x$ shows that the integral diverges.
Julián Aguirre
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Thank you very much, @Julián! It seems quite clear to me now. But I have one last question: What if $\Omega = B(0, \frac{1}{2})$ (the open 2-dim disc centered at the origin with the radius 0.5)? An unbound function in $H^1(\Omega)$ could be, for example, the same function using our q, or not? – StMan Oct 04 '18 at 08:15
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A change to polar coordinates would lead to similar computations. – Julián Aguirre Oct 04 '18 at 09:50