Is the functional $I(u) = \int_{\Bbb{R}^N}h(x) |u|^q \ dx $ weakly lower semicontinuous?

Question

Is the functional $$ I(u) = \int_{\Bbb{R}^N}h(x) |u|^q \ dx $$ weakly lower semicontinuous for $N \geq 3$ and $1 < q < 2$?

This is part of an exercise that asks to solve the problem $$ \begin{cases} - \Delta u + u = h(x)|u|^{q - 2}u \quad \text{ in } \Bbb{R}^N, \\ u \in H^1 (\Bbb{R}^N) \end{cases} $$ where $1 < q < 2$ and $h \in L^{\frac{2^*}{2^* - q}} \cap L^\infty$ is nonnegative, using the method of minimization.

I showed that weak solutions are critical points of the functional $$ J(u) = ||u||^2 - \int_{\Bbb{R}^N}h(x)|u|^q \ dx. $$

Now I am trying to show that $J$ is weakly lower semicontinuous and coercive. Coerciveness is easy, as is the weak lower semicontinuity of the norm. Hence, it remains to show that $I$ is weakly lower semicontinuous.

Thanks in advance and kind regards.

score 3 · Accepted Answer · edited Jun 12 '20 at 10:38

3

I believe the following works. Because strongly lower semicontinuous convex functionals are weakly lower semicontinuous, assume that $u_n\rightarrow u\in H^1(\mathbb{R}^N)$. Using the fact that $|a|^q\leq |b|^{q}+q|a|^{q-2}a(a-b) $, which follows from the convexity of $x\mapsto |x|^q$ we see that $$h(x)|u(x)|^q\leq h(x)|u_n(x)|^q+qh(x)|u(x)|^{q-1}|u_n(x)-u(x)|$$ which implies that

$$I(u)\leq I(u_n)+q\int_{\mathbb{R}^n}h(x)|u(x)|^{q-1}|u(x)-u_n(x)|\,dx$$

If we can show that $\lim_{n\rightarrow\infty}\int_{\mathbb{R}^n}h(x)|u|^{q-1}|u-u_n|\,dx=0$ then we are done by taking the liminf of both sides of the above. To see this, note that $|u-u_n|$ goes to $0$ in $L^{2^*}$ by Sobolev embedding, so we use Hölder's inequality $$\int_{\mathbb{R}^n}h(x)|u(x)|^{q-1}|u(x)-u_n(x)|\,dx\leq ||h|u|^{q-1}||_{L^{\frac{2^*}{2^*-1}}(\mathbb{R}^N)}||u-u_n||_{L^{2^*}(\mathbb{R}^N)}$$ which goes to $0$ as long as $h|u|^{q-1}\in L^{\frac{2^*}{2^*-1}}(\mathbb{R}^N)$. But this is true because $|u|^{q-1}\in L^{\frac{2^*}{q-1}}(\mathbb{R}^N)$ and $$\frac{2^* -q }{2^*}+\frac{q-1}{2^*}=\frac{2^*-1}{2^*}$$

edited Jun 12 '20 at 10:38

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answered Apr 27 '20 at 16:39

user293794

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Thank you very much. I understood almost everything. Could you only develop on the first inequality you claimed, please? – Danilo Gregorin Afonso Apr 28 '20 at 18:19
I edited to clarify this. Does it make sense now? – user293794 Apr 29 '20 at 16:13
Yes, it is very clear, but I instead referred to the fact $|a|^q \leq |b|^q + q|a|^{q - 1}(a - b)$. Could you show it as well, please? I can't see how to get to it from the convexity of $|x|^q$. – Danilo Gregorin Afonso Apr 29 '20 at 17:31
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Oh I see. This is simply the fact that any convex function lies above its tangent line or in symbols $f(x)\geq f(y)+f'(y)(x-y)$. See here (https://en.wikipedia.org/wiki/Convex_function#Functions_of_one_variable), for instance. – user293794 Apr 29 '20 at 17:53
Now I understand. Thank you very much for your help :) – Danilo Gregorin Afonso Apr 29 '20 at 17:55
My pleasure - this is an interesting problem. I hope what I wrote now is correct. – user293794 Apr 29 '20 at 17:55
Dear @user293794, please, shouldn't it be $|a|^q \leq |b|^q + q |a|^{q - 2}a(a - b)$? – Danilo Gregorin Afonso May 06 '20 at 13:30
Also, in the last equation, it should be $\frac{2^* - q}{2^*}$, right? – Danilo Gregorin Afonso May 06 '20 at 13:36
Yes you are right on both counts. I will edit the post to reflect this. – user293794 May 06 '20 at 17:01
The solution works, though. Thanks again :) – Danilo Gregorin Afonso May 06 '20 at 17:04

Is the functional $I(u) = \int_{\Bbb{R}^N}h(x) |u|^q \ dx $ weakly lower semicontinuous?

1 Answers1