The equality of Sobolev norm and Lebesgue norm, $\|v\|_{-2} = \|v\|_{L^2}$

Question

We know that $L^2 \subset H^{-2}$ (where $H^{-2}$ is the dual space of $H^2_0$) so that $\|v\|_{-2} \leq \|v\|_{L^2}$.

Could we show that:

If $v \in L^2$, then $v \in H^{-2}$ and

$$\|v\|_{-2} = \|v\|_{L^2}$$

Answer 1

No, these are very different norms. There is no more reason for $L^2$ norm to agree with the $H^{-2}$ than for it to agree with the $H^2$ norm. (Indeed, $H^{-2}$ norm is dual to $H^2$ while $L^2$ is dual to itself.) Considering that the $H^2$ norm is basically $\int |f''|^2$ (plus some lower order terms), it's like asking for what functions $\int |f|^2 = \int |f''|^2$.

The answer is: equality sometimes happens, but by chance and not for any interesting reason; the equality depends on how the norms are defined and how the underlying domain is scaled. It's a bit like for what subsets of $\mathbb{R}^2$ the area is equal to the perimeter.