Can we conclude that a distribution is a $L^2$ function by testing with $L^2$?

Question

Let $T\colon \mathcal{D}\to\mathbb{R}$ be a distribution. Does $|T(f)|\leq\|f\|_2 \forall f\in\mathcal{D}$ imply $T=T_g$ for some $g\in L^2$? What if $T$ is tempered?

Answer 1

Your hypothesis leads to $$T:D\to\mathbb C,$$ linear continuous functional in the sense of $L^2(\mathbb R)$. $D$ is dense in $L^2$, therefore by continuity we can extend $T$ to the whole $L^2$. By Riesz representation theorem, there exists an $L^2$ function $g$ such that $T(f) = (g,f)_{L^2}$. Moreover, it implies that $T$ is given by $T_{\bar g}$ and therefore tempered.

Answer 2

If $T :\mathscr D(\mathbb R^n)\to \mathbb R$ satisfies $$ \lvert T(f)\rvert \le \|f\|_{L^2(\mathbb R^n)}, $$ then $T$ is a functional defined in a dense subset of $L^2(\mathbb R^n)$, and as it is bounded (equivalent, continuous) there, it extends continuously in the whole of $L^2(\mathbb R^n)$. And by Riesz representation, there exists a unique $g\in L^2(\mathbb R^n)$, such that $$ T(f)=\int_{\mathbb R^n} fg\,dx. $$ Clearly $\mathbb R^n$ can be replaced by any open subset it.

It $T$ is a tempered distribution nothing changes since $\mathscr S(\mathbb R^n)$ is still dense in $L^2(\mathbb R^n)$.