Coercive bilinear form satisfying $Az=\Phi'(z)$ Let $X$ be a Hilbert space, $A$ compact such that there exist $\alpha$ with
$<Au,u>  \ge \alpha||u||^2$ $\forall u $ (i.e coercive)
Let $\Phi \in C^1(X, \mathbb{R})$ such that $\Phi$ satisfies
$lim_{||u|| \to \infty}\frac{|\Phi(u)|}{||u||^2}=0$
Assume that $\Phi$ is weakly continuous. I want to show there exist $z \in X$ such that $Az=\Phi'(z)$
I'm not really sure where to start. I don't see how the assumptions lead to conclusion, any nudge in the right direction would be appreciated

If $A$ is self-adjoint then you should be able to show existence of solutions for
$$
\min_u \frac 12\langle Au,u\rangle - \Phi(u).
$$
Every solution $z$ of this problem then satisfies $Az=\Phi'(z)$.
(Compactness is not needed here)
