$f_n \in L^{\infty}(\mathbb R)$ such that $f_n \to f$ and ${f_n}^2 \to g$ weakly$^*$ with $f^2 \neq g$

Question

The exercise metioned in the title was a part of my exams in Functional Analysis, which took place the previous week. Unfortunately I wasn't able to think of a sequence $f_n\in L^{\infty}(\mathbb R)$ such that

• $f_n \to f$ weakly$^*$
• ${f_n}^2 \to g$ weakly$^*$

but $f^2 \neq g$. However, I would like to know the answer.

I believe that such a sequence might be the product of a trigonometric sequence and the indicator sequence but I haven't managed to proceed.

Any help is much appreciated!

Thanks in advance

Answer 1

We can take, for example, $f_n(x) = \cos(nx) $. Then for every $h\in L^1(\Bbb R)$, we have $$ \int_{\Bbb R}h(x) f_n(x)\mathrm d x \xrightarrow{n\to\infty} 0 $$ by Riemann-Lebesgue lemma. On the other hand, we have $$ \left(f_n(x)\right)^2 =\cos^2(nx)=\frac{1+\cos(2nx)}{2}. $$ Also by Riemann-Lebesgue lemma, we have $f_n^2 \xrightarrow[\text{weak*}]{n\to\infty}\frac12$. This gives $f=0$ while $g=\frac12 \ne f^2$.