# weak convergence in $L^2$ and convergence of integral involving test-functions

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: $$\int_\Omega |f_n|\phi\,dx\to \int_\Omega |f|\phi \,dx,\quad \textrm{as n\to \infty.}$$

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

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

• Could you explain more on why we have the first $\le$ in the last line?
– John
Aug 4, 2019 at 16:21
• 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. Aug 4, 2019 at 16:23
• A similar counterexample, which may be a little easier to work with algebraically, is $u_n(x) = e^{i \pi n x}$. Aug 4, 2019 at 16:28