Consider $F\colon M\subseteq X\to [-\infty,\infty], M\neq\emptyset$. Then $\min\limits_{u\in M}F(u)=\alpha$ has a solution, if
1.) $X$ is reflexive. 2.) $F$ is coercive. 3.) $F$ is weak low semi continuous.
Now to the task:
Let $f$ be defined by $f(x,u,\xi)=u^2+(\xi^2-1)^2$.
The Bolza problem is defined by
$$ \inf\left\{F(u)=\int\limits_0^1 ((1-u'^2)^2+u^2)\, dx; u\in W^{1,4}(0,1); u(0)=0=u(1)\right\}. $$
a) Check, if $X$ is reflexive.
b) Check, if $F$ is coercive.
c) Check, if $F$ is weak low semi continuous.
Tip: You should use the Young equation $2ab\leq \epsilon a^2+\frac{b^2}{\epsilon}~\forall~a,b,\epsilon >0$.
Hello, i have some problems with this tasks.
a) I guess here is the answer YES, it is reflexive, because $W^{1,4}(0,1)$ is a Hilbert space.
b) and c) is difficult to me.
b) The professor gave the advice to find a sequence $(u_n)$with $F(u_n)\to 0$.
So I guess I have to find such a sequence with $\lVert u_n\rVert_{W^{1,4}}\to\infty$, but $F(u_n)\to 0$ what would show, that $F$ is NOT coercive.
But how to find such a sequence, can you help me? By the way: How is $\lVert \cdot \rVert_{W^{1,4}}$ defined? Is it $\lVert\cdot\rVert_{W^{1,4}}=\lVert\cdot\rVert_{L^2}+\lVert\cdot\rVert_{L^4}$?
a) $W^{1,4}_0(0,1)$ is not a Hilbert space. But it is reflexive. Probably the easiest way to see this is to use the map $F:W^{1,4}_0(0,1)\to L^4(0,1)$ defined by $F(u)=u'$. This map is an isomorphism onto its image (a closed subspace of $L^4$). Since the space $L^4$ is reflexive, so are its subspaces.
b) The advice you quoted pertains to c) not to b). The functional $F$ is coercive. You can prove coercitivity using an algebraic estimate of the form $u^2+(\xi-1)^2\ge \frac12 \xi^4-A$ where $A$ is some constant.
c) Was already answered: the sequence $u_n(x)=\int_0^x \operatorname{sign}\sin 2\pi n t\,dt$ converges to $0$ weakly, but $F(u_n)\to 0$ while $F(0)=1$. Hence, $F$ is not weakly lower semicontinuous.
I suggest to avoid posting multiple questions which address the same problem. It scatters the information around.