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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 $\lVert\cdot\rVert_{W^{1,4}}=\lVert\cdot\rVert_{L^2}+\lVert\cdot\rVert_{L^4}$?

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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.

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a) Why is the image closed? c) Why does $u_n$ converge WEAKLY to 0? – math12 Jan 21 '13 at 21:23
@math12 a) because $\|F(u)\|_{L^4}\ge c \|u\|_{W^{1,4}}$ by the Poincaré inequality. c) clearly $u_n$ does not converge to zero strongly since $\int |u'_n|^4=1$ for all $n$. Weak convergence can be verified directly, using integration by parts and Riemann-Lebesgue lemma. An easier way is to argue that since $u_n$ is bounded, it has a weakly convergent subsequence, and this subsequence satisfies the requirements of c). – user53153 Jan 21 '13 at 21:37
I do not see how to show that $F$ is coercive. Suppose that $\lVert u_n\rVert_{W^{1,4}}=\lVert u_n\rVert_{L^4}+\lVert u_n'\rVert_{L^4}\to\infty$. Why then $F(u_n)=\int\limits_0^1 1\, dx-2\int\limits_0^1 u_n'(x)^2\, dx+\int\limits_0^1 u_n'(x)^4\, dx+\int\limits_0^1 u_n(x)^2\, dx\to\infty$? – math12 Jan 22 '13 at 11:11

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