3 typo

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 $\del \partial U$ - 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.

2 Fixed some tex

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. $\forall u \in W^{1,p}_0 : \Vert u \Vert _ {W^{1,p}0} <= W^{1,p}_0} \le c \Vert \vert \nabla u \vert\Vert{L^p(U)}$. 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 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)$coincedes coincides with the form defined by the rhs. But then this$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. 1 # 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.$\forall u \in W^{1,p}_0 : \Vert u \Vert _ {W^{1,p}0} <= c \Vert \vert \nabla u \vert\Vert{L^p(U)}$. 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.