This is from PDE Evans, 2nd edition: Chapter 9, Exercise 2:
Assume $a: \mathbb R \rightarrow \mathbb R$ is continuous and $a(f_n) \rightharpoonup a(f)$ weakly in $L^2(0,1)$ whenever $f_n \rightharpoonup f$ weakly in $L^2(0,1)$. Show $a$ is an affine function.
My “work” so far:
Wlog. it can be assumed that $a(0) = 0$.
If $a(x) = \alpha x^2 + \beta x$ with $\alpha \neq 0$, then taking $f_n(x) = \sin(\pi n x)$ and $f(x) = 0$ leads to a contradiction, i.e. $a$ is not a polynomial of degree $2$ (alternatively, chose an arbitrary orthonormal sequence in $L^2(0,1)$). That is because if $a(f_n) \rightharpoonup a(f)$, then $$ \alpha \int_0^1 f_n^2 + \beta \int_0^1 f_n = \int_0^1 a(f_n) \cdot 1 \rightarrow \int_0^1 a(f) \cdot 1 = 0. $$ Since $f_n \rightharpoonup 0$ one has $\beta \int_0^1 f_n \cdot 1 \rightarrow 0$, ie. also $\alpha \int_0^1 f_n^2 \rightarrow 0$, which is a contradiction.
- Can one generalize this for all polynomials and then use the density of the polynomials in $C(\mathbb R)$?
Instead of trying to “create” a contradiction, one could also show directly that $a$ is linear, but I have no idea how to do that.
Any hints how to solve this problem?