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let $\Omega =\{(x,y)\in \mathbb{R}^2 : 0 < x < 1, 0 < y < x^2\}$ be a bounded and open domain and it is not $C^1$. And consider the function $u(x,y)=x^a.$

Use the function $u$ to prove that $H^1(\Omega)$ dose not inject in $L^p(\Omega)$ for $p > 6$.

Giovanni
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Math1995
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  • Please typeset your question appropriately, as it currently does not lend itself to an unambiguous and ready reading. (The community should not work harder on understanding your question than you do on asking it.) – avs May 08 '17 at 18:49
  • I did,but the symbols didn't appear so I edited it – Math1995 May 08 '17 at 18:53
  • @avs is it clear now? – Math1995 May 08 '17 at 18:59
  • The way your $\Omega$ is defined, it is closed, not open. Also, any restrictions on the exponent $a$? – avs May 08 '17 at 20:20
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    @avs: I edited the original question since it was written without using latex and I have mistakenly replace "less than" with "\le". I've fixed it now. – Giovanni May 08 '17 at 21:35

1 Answers1

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Notice that $$ \begin{align} \|u\|_{H^1}^2 = &\ \int_0^1\int_0^{x^2}x^{2a}\,dy\,dx + \int_0^1\int_0^{x^2}a^2x^{2a-2}\,dy\,dx\\ = &\ \int_0^1x^{2a + 2} + a^2x^{2a}\,dx, \end{align}$$ so that $u \in H^1(\Omega)$ if and only if $a > -\frac{1}{2}.$ On the other hand, $$\|u\|_{L^p}^p = \int_0^1\int_0^{x^2}x^{pa}\,dy\,dx = \int_0^1x^{pa + 2}\,dx$$ and hence $u \in L^p$ if and only if $pa + 2 > -1$, i.e. $a > -\frac{3}{p}$.

To conclude it is enough to notice that for any $p > 6$ we can find $a$ such that $$-\frac{1}{2} < a < -\frac{3}{p}.$$

Giovanni
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