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Let $\Omega$ be a bounded set of $\mathbb{R}^n$ and $(f_n)_n\subset L^2(\Omega)$ such that $f_n\rightharpoonup f\in L^2(\Omega)$ weakly in $L^2(\Omega)$. Then for any given test function $\phi\in C^\infty_c(\Omega)$, do we have the following convergent property: $$ \int_\Omega |f_n|\phi\,dx\to \int_\Omega |f|\phi \,dx,\quad \textrm{as $n\to \infty$.} $$

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  • $\begingroup$ @RhysSteele Thanks for pointing out the typos. $\endgroup$
    – John
    Aug 4, 2019 at 15:50

1 Answer 1

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Consider $n = 1$, $\Omega = (0,1)$ and let $$u_n(x) = \begin{cases} 1 \qquad k2^{-n} \leq x < (k+1)2^{-n} \text{ for an even } k\\ -1 \qquad \text{otherwise} \end{cases}$$

Then you can check that $u_n \rightharpoonup 0$ weakly in $L^2(\Omega)$ but $|u_n(x)| = 1$ for every $x$ so $\int |u_n| \phi \not \to 0$ for $\phi$ such that $\int \phi \neq 0$.

To check that $u_n \rightharpoonup 0$, first note that $\|u_n\|_{L^2} = 1$ for each $n$ and so it is enough to check that $\langle u_n, \phi \rangle \to 0$ for each $\phi \in C^\infty(0,1)$.

To do this, for any Lipschitz continuous $\phi$ write $$|\langle u_n, \phi \rangle| \leq \sum_{0 \leq k < 2^n, k \text{ even}} \int_{k2^{-n}}^{(k+1)2^{-n}} |\phi(x) - \phi(x+ 2^{-n})| dx \lesssim \sum_{0 \leq k < 2^n, k \text{ even}} 2^{-2n} = 2^{-n}.$$

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  • $\begingroup$ Could you explain more on why we have the first $\le$ in the last line? $\endgroup$
    – John
    Aug 4, 2019 at 16:21
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    $\begingroup$ I did do a few things at once there, sorry! You have that $\langle u_n, \phi \rangle = \sum_{0 \leq k < 2^n} \int_{k2^{-n}}^{(k+1)2^{-n}} (-1)^k \phi(x) dx = \sum_{0 \leq k < 2^n, k \text{ even}} \int_{k2^{-n}}^{(k+1)2^{-n}} \phi(x) - \phi(x+ 2^{-n}) dx$. Now you get the inequality by moving the modulus through the sum and then the integral using the triangle inequality. $\endgroup$ Aug 4, 2019 at 16:23
  • $\begingroup$ A similar counterexample, which may be a little easier to work with algebraically, is $u_n(x) = e^{i \pi n x}$. $\endgroup$ Aug 4, 2019 at 16:28

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