I have a question about the norm on the space $H^{-1}$, the dual space of the Sobolev space $H_0^1$. The canonical choice for the norm on $H^{-1}$ would be
\begin{equation}
\|T\|_{-1} = \sup_{v\in H_0^1, \|v\|_{H_0^1}=1} \frac{\lvert Tv \rvert}{\|v\|_{H_0^1}},
\end{equation}
where $\|v\|_{H_0^1} = \left ( \|v\|_{L^2}^2 + \|v^{\prime}\|_{L^2}^2 \right )^{1/2}$
But in class we defined the norm to be
\begin{equation}
\|T\|_{-1} = \sup_{v\in H_0^1, \lvert v \rvert_{1,2}=1} \frac{\lvert Tv \rvert}{\lvert v \rvert_{1,2}},
\end{equation}
where for $v\in H_0^1$:
\begin{equation}
\lvert v \rvert_{1,2} = \|v^{\prime}\|_{L^2}.
\end{equation}
Now my question: Is there a specific reason, why we devide by the $\lvert \cdot \rvert_{1,2}$-norm (it is indeed a norm on $H_0^1$) instead of the $\|\cdot\|_{H_0^1}$-norm?
Best Answer
Generally speaking, I think that there is no "deep"/useful reason behind the choice of $| \cdot |_{1,2}$ in the definition of the dual norm $\| \cdot \|_{H^{-1}}$, while in some other cases there could be (for example, if you want to characterize functionals in $H^{-1}$ or just to simplify notations and calculations). By the way, it's worth to say that being just a norm in $H^1_0$ it's not enough to justify the equivalence of definitions: everything keeps consistent if we choose an equivalent norm (since equivalent norms preserve continuity, convergence...), which is the case for $| \cdot |_{1,2}$ in $H^1_0$.