# Solution of a sublinear elliptic problem.

In a lecture notes, the author showed the problem

$\tag{$P$}$ $\begin{cases} -\Delta u = |u|^{q-2}u \textrm{ in } \Omega, \\ u(x)= 0 \textrm{ in } \partial\Omega, \end{cases}$

where $\Omega \subset \mathbb{R}^{N}$ is open, bounded and has smooth boundary. And $1<q<2$.

The aim was use the:

Theorem A: Let $X$ a Bannach reflexive space and $\phi: X \rightarrow \mathbb{R}$ a mapping weakly l.s.c such that $\phi(x) \rightarrow +\infty$ whenever $||x||_{X} \rightarrow +\infty$. Then exists $u_{o}$ such that

$\displaystyle\inf_{x \in X} \phi(x) = \phi(u_{0})$

Associated to the elliptic equation $(P)$, we have the functional $I: H^{1}_{0}(\Omega) \rightarrow \mathbb{R}$ given by

$I(u)= \dfrac{1}{2}\displaystyle\int_{\Omega}|\nabla u|^{2} - \dfrac{1}{q}\displaystyle\int_{\Omega}|u|^{q}$

By the Sobolev embedding theorem, we can show that if $||u||_{H^{1}_{0}} \rightarrow +\infty$ then $I(u) \rightarrow +\infty$.

Futhermore, by the Rellich–Kondrachov theorem, we can show that

$I(u) \leq \displaystyle\liminf_{n\rightarrow \infty}I(u_{n})$ whenever $u_{n} \rightharpoonup u$ (convergence in the weak topology) in $H^{1}_{0}(\Omega)$.

So, the author used Theorem A to show the existence of a minimun for $I$ and consequently the existence of a solution for $(P)$.

My comment is:

1. The fact that $I(u) \leq \liminf I(u_{n})$ whenever $u\rightharpoonup u$ in $H^{1}_{0}(\Omega)$ doesn't implies that $I$ is weakly l.s.c. (See Functional weakly lower-semicontinuous).

In my search for answers about it, I found that if a functional $\phi: X \rightarrow \mathbb{R}$ is convex then $\phi$ is l.s.c if, only if, $\phi$ is weakly l.s.c.

So, some doubts arise naturally:

(i) How to show that $I$, the functional defined above, is convex?

(ii) It's clear that $I$ is continuos , then l.s.c. If $I$ is convex, why use the Rellich–Kondrachov theorem?

(iii) In the notes, there is a comment saying that if $\Omega = \mathbb{R}^{N}$, the main difficult in the study of solution for $(P)$ by variational methods is the "lack of compactness". But, by (ii), why I need compactness?

Any help is welcome, these doubts really bothers me.

Have you checked if the proof to theorem A. uses weak lower-semicontinuity or sequential weak lower-semicontinuity? I would guess it is the latter, especially after reading @user127096 response in the link you provided.

(i) Note that for $t\in\mathbb R$, $$I(tu)=\frac{|t|^2}{2}\int|\nabla u|^2-\frac{|t|^q}{q}\int|u|^q\sim_{|t|\to0} -\frac{|t|^q}{q}\int|u|^q$$ which means that if $u\neq 0$, $I$ takes negative values on both side of $0$ on the line $\{tu,t\in\mathbb R\}$. Since $I(0)=0$, $I$ is not convex.
(ii) $I$ being continuous does not mean that $(u_n\rightharpoonup u \implies f(u_n)\to f(u))$. We would need strong convergence instead.
For example, take $f=\Vert .\Vert$ on any infinite dimensional normed vector space $(E,\Vert.\Vert)$, then $f$ is obviously continuous, but the unit sphere $S:=\{x\in E, \Vert x\Vert =1\}$ is both closed for the strong topology and not closed for the weak topology. More precisely, $\overline S^{\,\sigma(E,E')}=\{x\in E, \Vert x\Vert \leq1\}$, where $\bar{\vphantom{\S}.}^{\sigma(E,E')}$ is the closure with respect to the weak topology. You can very well have $u_n\rightharpoonup 0$ and $f(u_n)\not\to f(0)$.
(iii) The compact embedding theorem is not true anymore in $\mathbb R^N$, so you need to work a little bit more on the variational problem. What is meant with the 'lack of compactness', is that the problem is now invariant by translation: if $u$ is a solution, then so is $x\mapsto u(x-y)$ for any $y\in\mathbb R^N$. This must be taken care of, because if you were to take a weakly convergent subsequence $u_n\rightharpoonup u$, it might very well not have any strongly convergent subsequence. As a matter of fact, given any minimizing sequence for this problem, it is always possible to construct, using translations, another minimizing sequence that does not have a strongly convergent subsequence in $L^q(\mathbb R^N)$.