When is the $L^{2}$ norm smaller than the $H^{-1}$ norm?

Question

If $u\in L^{2}$ then we can define the functional: $$u(\phi)=\int \phi u $$

for all $\phi \in H^{1}_{o} $.

which means that $u$ is a linear functional in $H^{-1}$.

Now for any $f\in H^{-1}$

$$\begin{multline}\|f\|_{H^{-1}} = \inf \Big\{ \left( \int_U \sum_{i=0}^n |f^i|^2 dx \right): \langle f, v \rangle = \int_U f^0 v + \sum_{i=1}^n f^i v_{x_i} dx,\\ \;f^0,\ldots,f^n \in L^2(U), \; \forall v \in H^1(U)\Big\}. \end{multline}$$

When is it true that if $u\in L^{2}$ then $\|u\|_{H^{-1}}>\|u\|_{L^{2}}$?

Answer 1

From your definitions, you directly obtain $\|u\|_{H^{-1}}\le \|u\|_{L^2}$.

To see this, take $u\in L^2(\Omega)$, and define $$ f(\phi) = \int u \phi. $$ Now setting $f_0=u$, $f_i=0$ for $i=1\dots n$, we immediately find $$ \|u\|_{H^{-1}}\le \|u\|_{L^2}. $$