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Let ${f_n}$ a bounded sequence is $L^2(\mathbb{R})$ such that $\int f_n \phi dx \rightarrow0$ for each $\phi \in C_c^\infty(\mathbb{R})$. I want to prove that $f_n \rightharpoonup 0$.

The sequence is bounded so exists $f \in L^2$ and a subsequence $f_{n_k}$ of $f_n$ sucht that $f_{n_k} \rightharpoonup f$. If we pick $\phi \in C^\infty_c$ then $\int f_{n_k} \phi dx \rightarrow \int f \phi dx$ (and $\int f_{n_k} \phi dx \rightarrow 0$) so $\int f \phi dx=0$ for each $\phi \in C^\infty_c$ which implies $f=0$ a.e.

This proves that every convergent subsequence of $f_n$ converge to $0$. How can I conclude?

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    $\begingroup$ Do you mean $\int f_n\phi=0$ or $\int f_n\phi \to 0$? $\endgroup$ Commented Apr 14, 2020 at 12:37
  • $\begingroup$ I correct the question. I meant $\int f_n \phi \rightarrow 0$. $\endgroup$ Commented Apr 14, 2020 at 12:41

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Let $g \in L^{2}$ and $\epsilon >0$. There exists $\phi \in C_c^{\infty} (\mathbb R)$ such that $\|g-\phi\|_2 <\epsilon$. Hence $|\int f_n g|\leq |\int f_n \phi|+\|f_n\|\epsilon$ (by Hölder's/ C-S inequality). It is clear now that $\int f_n g \to 0$.

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  • $\begingroup$ Ok, I thought the same, but I got stuck due the fact $||f_n||\epsilon$ tends to $0$ if $\epsilon \to 0$ but for fixed $\phi$ is not true that $|\int f_n g| \to 0$ for $n \to \infty$ $\endgroup$ Commented Apr 14, 2020 at 12:46
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    $\begingroup$ @donovan The argument is $\lim \sup |\int f_n g| \leq 0+M\epsilon$ where $M$ is a bound for $\|f_n\|, n \geq 1$. Now $\epsilon$ is arbitrary. $\endgroup$ Commented Apr 14, 2020 at 12:50

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