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

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. 
