Let H be a Hilbert space over $R$ , $r > 0$ and $F ∈ C^1(H, R)$ such that: 1)−F is weakly sequentially lower semicontinuous 2) $DF(u) = 0$ implies $u = 0$ (this is the Frechet derivative) 3) $F(0) = 0$ and there exists $v ∈ H \backslash 0$ with $F(v) > 0$ and $||v|| < r$ .
How do we show that there exists $u_0 ∈ H$ with $||u_0|| = r$ and $F(u_0) = max_{||u||=r} F(u)$.
So this problem is definitely from calculus of variations, however I did not get this problem from any calculus of variation text, but from a nonlinear functional analysis text so I don't have much theorems for calculus of variations that I can use except for the theorem about direct methods which state: given a weakly sequentially closed subset of a reflexive space, a mapping from the weakly sequentially closed subset to R that is sequentially lower semi continuous and coercive is bounded from below and attains a minimum.
So we are already given that $-F$ is weakly sequentially lower semicontinuous and all Hilbert spaces are reflexive. In order to apply the theorem, I want to ask whether the restriction of $-F$ on the set of all elements with norm $r$ is sequentially closed. The restriction is definitely still sequentially lower semicontinuous and it should be vacuously coercive since the norms of all elements are just r. So does this allow us to apply the theorem which gives us a minimum for $-F$ which would be a maximum for $F$?
Problem is that $\{ \|x\|: \ \|x\|=r\}$ is not weakly sequentially closed.
Here is a counterexample: Take $H=l^2$, $$ F(x) = -\sum_n \frac1n x_n^2. $$ It satisfies all the assumptions.
Then $F(x) \le 0$, and $F(re_n) = -r^2 / n$. Hence $\sup_{\|x\|=r} F(x) = 0$. But $F(0)=0$ if and only if $x=0$.