Show that a sequence $(f_n)_n$, that is uniformly bounded in $L^2(\mathbb{R})$ and converges in $\mathcal{D}'(\mathbb{R})$, converges weakly in $L^2(\mathbb{R})$.
Consider $M > 0$ such that $\| f_n\|_{L^2} \leq M$ for all $n$. Let $g$ be in $L^2$ and $\mathcal{C}^{\infty}_c$. Then
\begin{align} | \langle f_n , g \rangle - \langle f , g \rangle | &\leq |\langle f_n, g \rangle - \langle f_n, h \rangle + \langle f_n, h \rangle - \langle f, h \rangle + \langle f, h \rangle - \langle f, g \rangle | \\ &\leq | \langle f_n, g-h \rangle + \langle f_N - f, h \rangle + \langle f, h-g \rangle| \\ &\leq \|f_n \|_{L^2} \| g-h\|_{L^2} + \epsilon + \|f \|_{L^2}\|h-g \|_{L^2}\\ &\leq (M + \|f \|_{L^2}) \| g-h\|_{L^2} + \epsilon \\ & \leq \epsilon + \epsilon \\ & \leq 2 \epsilon \end{align}
Thus $\langle f_n, g \rangle \to \langle f, g \rangle$. So $f_n$ converges weakly to $f$ in $L^2$. Is this correct or is there missing something? Do I need to show that $f$ is in $L^2$?
Here is a simpler argument: Since $L^2(\mathbb{R})$ is reflexive, bounded sequences have a weakly convergent subsequence. Denote any such subsequence by $f_{n_k}$ and its limit by $f_* \in L^2(\mathbb{R})$. Since $f_n$ converges to $f$ in $D'(\mathbb{R})$ we must have $f_*=f$. Now since the limit is unique the whole sequence $f_n$ must converge to $f$ weakly in $L^2(\mathbb{R})$.
Edit (slightly more detailed explanation): We know $f_{n_k}$ converges weakly to $f_* \in L^2(\mathbb{R})$ and $f_n$ converges to $f$ in $\mathcal{D}'(\mathbb{R})$. Furthermore, we also know $f_{n_k}$ converges to $f$ in $\mathcal D'(\mathbb{R})$, since it is just a subsequence of the original sequence.
Additionally, since weak convergence in $L^2(\mathbb{R})$ implies convergence in the sense of distributions, we also know that $f_{n_k}$ converges to $f_*$ in $\mathcal D'(\mathbb{R})$. Since $\mathcal D'(\mathbb{R})$ is Hausdorff, limits of sequences are unique and so we must have $f=f_*$.