MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $$Au=f.$$

Have a look at the textbook by Chipot or the free monograph by Showalter where the approach is explained in detail.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.