Does the norm of a normed linear space determine the form of its dual spaces elements?

Hello everybody,

As an introductory example, suppose $U \subset R^n$ is open and bounded, let $p = 2$. Then there is a constant $c>0$ s.t. $\forall u \in W^{1,p}_0 : \Vert u \Vert _ {W^{1,p}_0} \le c \Vert \vert \nabla u \vert\Vert_{L^p(U)}$. This implies that the latter expression defines an equivalent norm on ${W^{1,p}_0}$.

Let $f \in L^2(U), g \in C^1(\bar{U})$. Then there exists an unique solution $u \in W^{1,p}$ to the system $\triangle u = f$ over $U$, $u = g$ over $\partial U$ - or equivalently, there exists an unique solution $u \in W^{1,p}_0$ to the system $\triangle u = f - \triangle g$ over $U$, (in the distributional sense).

Proof: $W^{1,2}_0$ is a Hilbert space, hence self-dual. The rhs $f - \triangle g$, defines an element of $D'$, which by density can extended to $W^{1,2}_0$. On the other side, the equivalent norm as introduced above is defined the inner product $(u,v) = \int \nabla u \cdot \nabla v$, by riesz' representation theorem, there is an $u \in W^{1,2}_0$ s.t. the induced form $(u, \cdot)$ coincides with the form defined by the rhs. But then this $u$ is a weak solution to $\triangle u = f - \triangle g$.

So far, so good. I would like to ask some questions on this.

i) Can this be extended to other dual exponents $p$, $q$ ?

ii) The equivalent norm that regards first derivatives only is not only an equivalent norm for $p=2$, but also for $1 \leq p < \infty$. In the above case, it seems the norm imposes a form the dual vectors are subject to. I wonder whether in general - not only in the case of $L^p$ and its friends - there is some way how the form of linear functionals on some normed space $X$ are determined by the norm attached to the vector space $X$.

I hope this questions ain't too vacuous and there are interesting answers. In either case, thanks.

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I've cleaned up some of the TeX, but I don't know which of several symbols you intend by "del". Please make use of the preview feature when writing a question. –  Mark Meckes Jun 14 '10 at 13:47
Thanks. Unfortunately, I didn't see the preview feature, maybe due to js being disabled. –  shuhalo Jun 14 '10 at 14:38

The argument you have presented is an adaptation of the Lax-Milgram theorem which is essentially equivalent to the Riesz representation theorem (and generally speaking both of these results hold only in the Hilbert space framework). The Lax-Milgram theorem fails for the Laplace equation in $L^p$-spaces with $p\neq2$. Instead, some analogous results based on the ideas of coercivity, duality and monotonicity can be obtained in any reflexive Banach space.
The Dirichlet problem for the $p$-Laplace operator
$$-\nabla(|\nabla u|^{p-2}\nabla u)=f,\quad x\in\Omega,\qquad (*)$$ $$u=0,\qquad x\in\partial\Omega,$$ might be a "correct" $L^p$-analogue of the problem described in the question.
The right hand side of $(*)$ gives rise to the mapping $A: W_{0}^{1,p}\to(W_{0}^{1,p})^{*}$ defined by the identity $$\langle Au,v\rangle=\int_{\Omega}|\nabla u|^{p-2}\nabla u\cdot\nabla v\ dx\quad \mbox{for all } v\in W_{0}^{1,p}.$$ A straightforward check shows that $A$ satisfies the conditions of the following theorem (which might be viewed as an $L^p$-analogue of the Lax-Milgram theorem).
Theorem. Let $A$ be a strictly monotone, coercive operator from a reflexive Banach space $E$ to its dual $E^{* }$. If $A$ is continuous on finite-dimensional subspaces of $E$ then for every $f\in E^{*}$ there exists a unique solution to the problem $$Au=f.$$