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. $\forall u \in W^{1,p}_0 : \Vert u \Vert _ {W^{1,p}_0} <= c \Vert \vert \nabla u \vert\Vert_{L^p(U)}$. 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 $\del U$ - 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 $(u,v) = \int \nable u \cdot nabla v$, by riesz' representation theorem, there is an $u \in W^{1,2}_0$ s.t. the induced form $(u, \cdot)$ coincedes 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.