2

Given $\Omega\in R^N$ open bounded with smooth boundary. We define $H^{-1}$ to be the dual space of $H_0^1(\Omega)$ and from Evan's PDE book, chapter $5$, we know that for any $f\in H^{-1}$, there exists functions $f_0$, $f_1$,...$f_N$ in $L^2(\Omega)$ such that for any $u\in H_0^1(\Omega)$ \begin{equation} (1)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left<f,u\right>_{<H^{-1}, H_0^1>} = \int_\Omega f_0 \,u\, dx+\sum_{n=1}^N\int_\Omega f_n \partial_n u\,dx \end{equation} I understand this theorem. However, today I came across this theorem and find out a small corollary which troubles me a bit.

I wrote in my notes that if $f\in L^2\subset H^{-1}$, then I have $$(2)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left<f,u\right>_{<H^{-1}, H_0^1>} =\int_\Omega f\,u\,dx = \left(f,u\right)_{L^2}$$ I have no explanations in my notes...(Ok I wrote from $(1)$ we clearly have $(2)$...) But now I can not understand how I got equation $(2)$ from equation $(1)$.

Now I have a problem regrading to evaluate the dual... Let's take $I=(0,1)$ and define $$V:=\{u\in H^1(I),\,\,u(0)=0\} $$ Then it is a fact that if $u\in V$ then $$\frac{u(x)}{x}\in L^2(I)$$ Now I define my norm $\|\cdot\|_V$ on $V$ by using the following inner product $$ (u,v)_V:=\int_I u'v'dx+\int_I \frac{u}{x}\frac{v}{x}dx$$ It can be verified this is a inner product and hence the norm $\|\cdot\|_V$ is well-defined.

Now, given any $f\in L^2(I)$ such that $\frac{1}{x}f(x)\in L^2(I)$ as well. My question is, what is $$\left<f,v\right>_{<V^*,V>}$$ Should I have $$\left<f,v\right>_{<V^*,V>} = \int_I \frac{f}{x}\frac{v}{x}dx$$ or $$\left<f,v\right>_{<V^*,V>} = \int_I fv\,dx$$ Which one is correct? why? Thanks a lot for help!

spatially
  • 5,720
  • The answer has been provided in

    http://math.stackexchange.com/questions/183644/acting-of-a-dual-pairing-in-sobolev-spaces

     – Milly Oct 21 '14 at 03:01
  • Can you also help me the part I just added? Thx! @Milly – spatially Oct 22 '14 at 23:40
  • Generally, if nothing other is explicitly specified, the "standard" symbol most often denotes the induced scalar product. If different operator is used, then a "symbol" normally comes in (e.g. $\langle Au,v\rangle_{V^,V} :=\int \frac{uv}{x^2} dx$, $\langle Bu,v\rangle_{V^,V} :=\int u_x v dx$, etc.) – Milly Oct 23 '14 at 00:14

1 Answers1

3

The answer to your final question depends on what inner product you give $L^2(I)$. We have $\langle f, v \rangle_{V', V} = (f,v)_{L^2(I)}$.

To understand this properly, you should read up on Gelfand triples. This is the situation where you have $V \subset H \subset V^*$, where $V$ is a Banach space continuously and densely embedded in a Hilbert space $H$. By identification $H$ with its dual space, the inclusion $H \subset V^*$ follows. Then we use the notation $$\langle f, v \rangle_{V^*, V} = (f,v)_H$$ whenever $f \in H$ and $v \in V$.

This is often encountered in PDE with $V=H^1$ and $H=L^2$.

riem
  • 781