Suppose that $f\in L^2$, $g\in \scrD'$, if $$\bex f=g,\mbox{ in }\scrD', \eex$$ then $f=g\in L^2$.

 

In fact, $\scrD\subset L^2 \ra L^2\subset\scrD'$. Thus $h=f-g=0\in \scrD'$, the zero element is the same in $L^2$ and $\scrD'$, and hence $h=f-g=0\in L^2$, $g=f-(f-g)\in L^2$.