Please check proof of convergence in $L^2(0,T;L^2)$ of a composition (uses Nemytskii operator) Suppose we have a continuous function $g:\mathbb{R} \to \mathbb{R}$ that satisfies $|g(x)| \leq C|x|$. Let $u_{n} \to u$ in $L^2(0,T;L^2)$. I want to show that $g(u_{n'}) \to g(u)$ in $L^2(0,T;L^2)$. for a subsequence $n'$. 
I was thinking since we only care about a subsequence, we can extract $u_{n'} \to u$ pointwise a.e. But the Dominated Convergence Theorem does not apply because we can't find a uniform bound (there is a bound because $g$ is bounded linearly but obviously it is not uniform).
However we can use this theorem from here 

where $N_f(u)(z) = f(z,u(z))$. 
Define $f\colon [0,T] \times L^2(\Omega) \to L^2(\Omega)$ by $f(t, u) = g(u)$. $f$ is Caratheodory since $g$ is continuous. Define $N_f(u)(t) = f(t,u(t)) = g(u(t))$. If $u \in L^2(0,T;L^2)$ then
$$|N_f(u)(t)|^2 = |g(u(t))|^2 \leq C|u(t)|^2$$
which is in $L^2(\Omega)$. So $N_f(u) \in L^2(0,T;L^2)$. Therefore the theorem applies and $N_f$ is continuous. So if $u_n \to u$ in $L^2(0,T;L^2)$ then $g(u_n) \to g(u)$ in $L^2(0,T;L^2)$.
Question:


*

*Is this a correct proof? 

*I feel weird about this because this removes the need for the dominated convergence theorem when we don't have a uniform bound. Surely this must be more well-known if what I wrote is correct.
 A: Yes, applying the textbook theorem about the continuity of a more general Nemytskii operator solves the problem stated in the first paragraph. 
That said, the proof of continuity in 3.4.4. is much like the proof you could give directly for your problem. The idea of picking a subsequence that converges a.e. is good. And you are right: we don't have a dominating function for the Dominated Convergence Theorem. So, use the Generalized DCT instead. 
To simplify writing, I assume   $u_n$ already converges a.e. Since $\int u_n^2 \to\int u^2$ and $C^2u_n^2$ dominates $g(u_n)^2$, it follows that $\int g(u_n)^2 \to\int g(u)^2$. That is, $$\|g(u_n)\|_2\to \|g(u)\|_2 \tag1$$ Since $g(u_n)$ is a bounded sequence in $L^2$, it contains a weakly convergent subsequence $g(u_{n_k})$. Recall that when   weak limit and a.e. limit both exists, they are equal. Therefore, the weak limit is $g(u)$.  The relation (1) upgrades the  weak limit to strong (with respect to the $L^2$ norm). 
Let's recap: every subsequence of $g(u_n)$ contains a subsubsequence that converges to $g(u)$ in the norm. By a general topological fact, $g(u_n)\to g(u)$ in the norm. 
