# Function $a$ with $f_n \rightharpoonup f$ implies $a(f_n) \rightharpoonup a(f)$

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?

## 1 Answer

Suppose $a$ is not affine. There exist $s<t$ so that $$a\left(\frac{s+t}{2}\right)\ne\frac{a(s)+a(t)}{2}.$$ Define $$f_n(x)=\begin{cases} s,&(x\in[2j/2n,(2j+1)/2n), j=0,\dots,n-1), \\ t,&(x\in[(2j+1)/2n,(2j+2)/2n),j=0,\dots,n-1).\end{cases}$$

Then $f_n\to (s+t)/2$ weakly and $a(f_n)\to (a(s)+a(t))/2$ weakly. (Since $||f_n||_2$ is bounded it's enough to check $\int gf_n$ for $g$ in some dense subset...)

• (Also thanks for your answer. I will have a look at it tomorrow.) – Keba Jul 29 '16 at 17:36
• Did you mean “$j$” even and “$j$ odd” in the first and two lines defining $f_n$ respectively? Or do I miss something? Intuitively I see why $f_n \rightharpoonup (s+t)/2$ weakly, but I am unable to prove this. Any further hints? – Keba Jul 31 '16 at 12:34
• @Keba Typo, sorry. If $g$ is continuous on $[0,1]$ then $g$ is uniformly continuous; if $n$ is large enough then $|g(t)-g(t+1/2n)|<\epsilon$ for every $t$. It follows that $$\left|\int_{j/n}^{j+1/n} g\left(f_n-\frac{s+t}{2}\right)\right|<\epsilon/n.$$ – David C. Ullrich Jul 31 '16 at 13:25
• That makes sense, thanks. :) – Keba Jul 31 '16 at 14:10