Norm on the Space $H^{-1}$

Question

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?

Answer 1

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 $H^1_0$.

Answer 2

This is one of those instances in which going through abstract theory actually clarifies things.

Proposition. Let $(X, \lVert\cdot\rVert)$ be a Banach space and suppose that $\lVert\cdot \rVert_1$ is an equivalent norm on it. Then the norms on $X^\star$ defined by $$ \lVert f \rVert_\star:=\sup_{x\ne 0} \frac{\lvert f(x)\rvert}{\lVert x \rVert}, \qquad \lVert f \rVert_{1,\star}:=\sup_{x\ne 0} \frac{\lvert f(x)\rvert}{\lVert x \rVert_1}$$ are equivalent.

I omit the proof which follows immediately from the definition of equivalent norms, which I recall here: $$ c\lVert x\rVert_1\le \lVert x\rVert \le C \lVert x\rVert_1, $$ for some constants $c>0$ and $C>0$ independent on $x$.

This abstract proposition applies to the present question by letting $(X, \lVert \cdot\rVert)$ denote $(H^1_0, \lVert \cdot \rVert_{H^1_0})$, and letting $\lVert \cdot \rVert_1$ denote $\lvert \cdot\rvert_{1,2}$. The fact that the latter is a norm and it is equivalent to $\lVert \cdot\rVert_{H^1_0}$ is the content of the Poincaré inequality.

(I am here assuming that $H^1_0=H^1_0(\Omega)$ where $\Omega$ denotes a bounded open set in $\mathbb R^n$. If $\Omega$ is unbounded, then the Poincaré inequality needs not hold and $\lvert\cdot\rvert_{1,2}$ needs not be a norm).