Vtyjkyuk

Consider $Omega subset mathbb R^n$ open.

Let $f_n_ngeq 1$ and $f$ belong to $L^2(Omega)$. Suppose that $f_n xrightarrowntoinfty f$ in $D'(Omega)$, i.e., $int_Omega f_n varphi xrightarrowntoinfty int_Omega fvarphi$, for all $varphi in C_0^infty(Omega)$. Are we able to conclude that $f_n xrightarrowntoinfty f$ 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 xrightarrowntoinfty int_Omega f v $, for all $v in L^2(Omega)$ (am I right?). Then, maybe we can use Riesz representation theorem to write
$$
||f_n-f||_0,Omega= sup_substackv in L^2(Omega)\vneq 0 fraclangle f_n-f,vrangle_0,Omega,
$$
and somehow take the limit $ntoinfty$ inside the supremum.

EDIT:

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

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 $n$.

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

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.