Bijective transformation of $L^2$-weak convergence sequence again weak converging?

Question

Let $f_n$ converge weakly in $L^2(x)([0,1])$ to $f$, with $|f_n(x)|\leq C$ for almost all $x\in]0,1]$ and all $n$.

Let $H:R\rightarrow R$ be strong monotone increasing and continuous with $H(0)=0$. (e.g. $H(x)=x^3$)

By the boundedness of $f_n$: $H(f_n(.)) \in L^2([0,1])$.

The Question:

Is then $H(f_n(.))$ weakly converging to $H(f(.))$ in $L^2$?

(In the example is $f_n(.)^3$ weakly converging to $f(.)^3$?)

Edit: Or at least does a subsequence converges to $H(f(.))$?

Answer 1

No. Consider $H(x) = x^3$ and $$ f_n(x) = \begin{cases} 1 & \frac in \le x < \frac{i+1/2}n \quad i = 0,\ldots, (n-1)\\ 0 & \text{otherwise} \end{cases} $$ Then $H(f_n) = f_n$, $f_n \rightharpoonup \frac 12$, $H(f_n) = f_n \rightharpoonup \frac 12 \ne H(\frac 12) = \frac 18$. Note this trick works for any $H$ which has $H(\frac 12) \ne \frac{H(0) + H(1)}2$.