1
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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...)

$\endgroup$
  • $\begingroup$ (Also thanks for your answer. I will have a look at it tomorrow.) $\endgroup$ – Keba Jul 29 '16 at 17:36
  • $\begingroup$ 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? $\endgroup$ – Keba Jul 31 '16 at 12:34
  • $\begingroup$ @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 $$\left|\int_{j/n}^{j+1/n} g\left(f_n-\frac{s+t}{2}\right)\right|<\epsilon/n.$$ $\endgroup$ – David C. Ullrich Jul 31 '16 at 13:25
  • $\begingroup$ That makes sense, thanks. :) $\endgroup$ – Keba Jul 31 '16 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.