Does convergence in $D'(\Omega)$ imply convergence in $L^2(\Omega)$?

Question

Consider $\Omega \subset \mathbb R^n$ open.

Let $\{f_n\}_{n\geq 1}$ and $f$ belong to $L^2(\Omega)$. Suppose that $f_n \xrightarrow{n\to\infty} f$ in $D'(\Omega)$, i.e., $\int_\Omega f_n \varphi \xrightarrow{n\to\infty} \int_\Omega f\varphi$, for all $\varphi \in C_0^\infty(\Omega)$. Are we able to conclude that $f_n \xrightarrow{n\to\infty} f$ in $L^2(\Omega)$?

I think that, since $D(\Omega)$ is dense in $L^2(\Omega)$, we can conclude that $\int_\Omega f_n v \xrightarrow{n\to\infty} \int_\Omega f v $, for all $v \in L^2(\Omega)$ (am I right?). Then, maybe we can use Riesz representation theorem to write $$ ||f_n-f||_{0,\Omega}= \sup_{\substack{v \in L^2(\Omega)\\v\neq 0}} \frac{\langle f_n-f,v\rangle_{0,\Omega}}{||v||_{0,\Omega}}, $$ and somehow take the limit $n\to\infty$ inside the supremum.

EDIT:

Kavi Rama Murthy has shown that this is false if $\Omega$ is unbounded. Is it different if $\Omega$ is bounded?

Answer 1

This is false. Let $f_n =I_{(n,n+1)}$, $f=0$. Then $\int f_n \phi \to 0$ for all $\phi \in C_0^{\infty}$ but $\int |f_n-f|^{2}=1$ for all $n$.

Answer 2

Suppose $\Omega = (0,2\pi).$ Let $f_n(x) = \sin (nx).$ Then for any $\varphi \in C_c^\infty(0,2\pi),$ $\int_0^{2\pi} f_n \varphi \to 0$ by Riemann-Lebesgue. Thus $f_n\to 0$ in $D'(0,2\pi),$ but certainly not in $L^2(02\pi).$

Answer 3

Another way to see this is the following:

It is known that $D(\Omega)$ is dense in $D'(\Omega)$ and $L^2(\Omega)$ with respect to their natural topologies. By considering delta functions, evidently $L^2(\Omega) \neq D'(\Omega)$ if $\Omega \subset \mathbb R^n$ is open.

So if we take $u \in D'(\Omega) \setminus L^2(\Omega)$ and choose $u_n \in D(\Omega)$ such that $u_n \rightarrow u$ in $D'(\Omega),$ I claim $(u_n)$ does not converge in $L^2(\Omega).$ Indeed if $u_n \rightarrow v$ in $L^2(\Omega),$ it converges weakly in particular and hence $u_n \rightarrow v \in D'(\Omega).$ So $u = v \in L^2(\Omega),$ which is a contradiction.