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The following assertion is true in the case $1<p<\infty$ but for $p=\infty$ in general not. So I want to find a counterexample. Can someone help me?

So find $f_n, g_n \in L^{\infty}(0,1)$ such that $f_n \rightharpoonup^* f,g_n \rightharpoonup^* g$ but $f_n g_n$ does not weak$^*$ converge to $fg$.

tubmaster
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  • Let $f_n(x)=g_n(x)=n$ for $x\in (0,1/n^2)$ and $ f_n(x)=g_n(x)=0$ for $x\in [1/n^2,1)$ . If $h(x)=1$ for all $x$ then we have $\int_0^1 f_ng_nh=1$ but I have not confirmed that $f_n$ converges weak$^*$ to $0$. – DanielWainfleet Mar 15 '17 at 14:13
  • This statement is also not true for $p \in (1,\infty)$! Btw. you can usually take $f_n = g_n$. – gerw Mar 15 '17 at 18:40
  • any other suggestions? :) – tubmaster Mar 23 '17 at 08:49

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