While reading the well known book *Minimax Methods in Critical Point Theory with Applications to Differential Equations* by Paul Rabinowitz, in the proof of a generalisation of the Mountain Pass Theorem (Theorem 5.29 in the book), I encountered the following abstract result:



> 
> Let $E$ be a real Hilbert space and consider $b\in C^1(E, \mathbb{R}$) such that $b'$ is compact. Then, $b$ is weakly continuous, i.e. if $(u\_n)\_n\subseteq E$ converges weakly to $u\in E$, then $b(u\_n)\to b(u)$ as $n\to \infty$.
> 
> 
> 


The reference for this result given in the book is



> 
> M. A. Krasnoselski, *Topological methods in the theory of nonlinear integral
> equations*, Macmillan, New York, 1964.
> 
> 
> 


However, even if this is a well renowned book in the field of nonlinear analysis, I do not have access to it. Does anyone know a modern reference for this result? The proof looks nontrivial to me, at least I do not know how to approach it.


EDIT: The fact that $b'$ is compact means that if $A\subseteq E$ is bounded, then the closure of $b'(A)$ is compact. Of course, $b'$ denotes the mapping $x\mapsto b'(x)$, i.e. $b'$ associates with each $x\in E$ the Frechet derivative of $b$ at $x$, which we denote by $b'(x)\in E^{\*}$.



If $b\in C^1(E, \mathbb{R})$ and $b'$ is compact, then $b$ is weakly continuous — a reference request
To obtain a contradiction, suppose that $b(u\_n)\not\to b(u)$. Passing to a subsequence, without loss of generality (wlog) we have $|b(u\_n)-b(u)|\ge c$ for some real $c>0$ and all $n$.


By the mean value theorem, $b(u\_n)-b(u)=b'(v\_n)(u\_n-u)$ for all $n$ and some $v\_n$ on the straight line segment from $u$ to $u\_n$.


Since the sequence $(u\_n)$ is weakly convergent, it is [bounded](https://en.wikipedia.org/wiki/Weak_convergence_(Hilbert_space)#Properties). So, recalling that $b'$ is compact and passing to a subsequence, wlog we have $b'(v\_n)\to B$ for some $B\in E^\*$. Also, $B(u\_n-u)\to0$, since $u\_n\to u$ weakly. Also, $\|b'(v\_n)-B\|\,\|u\_n-u\|\to0$, since $b'(v\_n)\to B$ and the sequence $(u\_n)$ is bounded.
So,
$$|b(u\_n)-b(u)|=|b'(v\_n)(u\_n-u)|\le\|b'(v\_n)-B\|\,\|u\_n-u\|+|B(u\_n-u)|\to0+0=0,$$
which contradicts the condition that $|b(u\_n)-b(u)|\ge c$ for some real $c>0$ and all $n$. $\quad\Box$


