Oh, stupid me! Thank you! I was already so confused that the positive definitness of the derivative (which I had used in my proof) and the boundedness of the functional from below (which you had used) seemed unrelated to me at a first glance... Of course, the former must imply the latter (and actual convexity of the functional).

I haven't calculated carefully, but I think that if you have $G(v)=G(|v|)$ the the derivative for e.g. strictly negative $v$ has the wrong sign. My reason to assume this without calculation is that in case $\zeta(0)\ne0$ the derivative at $0$ does not exist while I think that this should be the case for the "correct" definition of $G$.

Why is $G$ non-negative? $v$ can be negative.

I edited the reply to require an a-priori bound. This is usual technique needed to prove the existence of a solution in such cases. Without any such a-priori bound, I doubt that it is possible to apply Brouwer's fixed point theorem directly for the problem.

If $u\mapsto\zeta(u)u$ is (globally) bounded, then $F$ (and thus $G$) is globally bounded, that is, there is some $R$ such that $\lVert G(v)\rVert\le R$ for every $v$. As mentioned in the previous comment, instead of the boundedness, sublinear growth near $\infty$ is suffiicient.

I know, but I guess that some assumption was forgotten, because it was used in the question e.g. that $\zeta(v_k)v_k$ is bounded. Moreover, I am quite sure that the assertion is not provable from the finite-dimensional reduction if practically nothing more than the continuity of $u\mapsto z(u)u$ is assumed - local boundedness of the derivative is practically an empty hypothesis concerning Brouwer. It is sufficient that $\frac{\lVert F(u)\rVert}{\lVert u\rVert}\to0$ as $\rVert u\rVert\to\infty$, though. This is a bit weaker than global boundedness.

Is $\varphi=\phi$ and does $\phi(v_k)$ mean the composition then? And what is the product $v_k\cdot b_k$ of a function with a vector?

@András Bátkai, the spectrum is upper semicontinuous with respect to the operator norm.

This is essentially the question about (uniform) smoothness with respect to the function on the right-hand side of a differential equation, isn't it? So the answer is $C^k$. Just consider the usual proof of differentiability with respect to a parameter from an infinite-dimensional space.

Thanks. Should be fixed now.