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Let $u,v$ be arbitrary elements of a function space $X$ defined on $\Omega \subset \mathbb{R}^n$. Define

$$ (u,v)_2 = \int_\Omega \partial_x u \, \partial_x v + \partial_y u \, \partial_y v \:dx $$

Now, my question is what should $X$ be so that $(\cdot,\cdot)_2$ defines an inner product?

It is clear that $(\cdot,\cdot)_2$ is symmetric and linear but the problem seems to be positive-definitiveness.

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In particular, $(u,u)_2 = 0$ for every constant function $u=c$. – chango Apr 29 '12 at 9:25
Hint: $(u,u)_2=\int_\Omega |\nabla u|^2$. Hence the only constant function in $X$ must be $0$. Anyway, the most used choice is $X=W_0^{1,2}(\Omega)$. But you can also think of a space of functions $u$ with $\int_\Omega u=0$. – Siminore Apr 29 '12 at 9:43
Ok, so $X=W_0^{1,2}(\Omega)$ is the space of functions in $H^1(\Omega)$ which vanish on $\partial \Omega$. And I have that $(u,u)_2=0$ implies that $u$ is constant a.e. But how do I conclude from that that $c=0$? – chango Apr 29 '12 at 10:50
Since $u$ vanishes on $\partial \Omega$. Actually, by Poincaré's inequality, $\int_\Omega |\nabla u|^2 =0$ implies $\int_\Omega |u|^2=0$. This is a more refined viewpoint. – Siminore Apr 29 '12 at 11:26

The norm involving the inner product of the first derivative shall be associate with the Sobolev norm for the space $W^{1,2}(\Omega)$, in which the elements and their weak derivative are both $L^2$-integrable. And as you mentioned in the comment, the fact that having constant plugging in would make the inner product be zero makes the induced "norm" only a semi-norm, we denote it as $|\cdot|_{W^{1,2}(\Omega)}$

So Here what we wanna do is to find $X$ such that this semi-norm behaves the same as the full $W^{1,2}(\Omega)$-norm on $X$, ie, we want to quotient the kernel of $\nabla$ out to get an equivalence class $W^{1,2}(\Omega)/ \mathbb{R}$, there are normally two ways to do this:

  • For every element $u\in W^{1,2}(\Omega)$, consider the new space for $\displaystyle \bar{u} = u - \frac{1}{|\Omega|}\int_{\Omega} u $, and we have $\displaystyle \int_{\Omega} \bar{u} = 0$, naming this equivalence class as space $\mathring{H}^1(\Omega) = X$, then by Poincaré inequality, for any $w\in \mathring{H^1}(\Omega)$: $$ \|w \|_{L^2(\Omega)}\leq C\|\nabla w \|_{L^2(\Omega)} $$ hence $$ |w |_{W^{1,2}(\Omega)} \leq \|w \|_{W^{1,2}(\Omega)}\leq(1+ C)|w |_{W^{1,2}(\Omega)} $$ we have the norm equivalence, and this construction is normally associated with the Pure Neumann problem for the second order elliptic PDEs.

  • Another construction is we set the boundary value to be zero like $H^1_0 = X$ like you said, and use Friedrichs' inequality(Poincaré inequality's counterpart for zero-trace functions), we could get the same norm equivalence relation above whereas the geometry constant $C$ might be different. This space relates to the Dirichlet problem for Poisson equation.

  • We also could have a mixed version of above two which associates to the following problem $$ \left\{ \begin{aligned} -\Delta u &= f &\text{ in } \Omega \\ u &= g &\text{ on } \Gamma_D\subset \partial \Omega \\ \nabla u\cdot \boldsymbol{n} &= g_N &\text{ on } \Gamma_N\subset \partial \Omega \end{aligned} \right. $$ Define the space $X = H^1_{g}(\Omega) = \{u\in H^1(\Omega): u = g \text{ on } \Gamma_D\}$, if the Dirichlet boundary is not empty, $|\cdot|_{W^{1,2}(\Omega)}$ is a norm too.

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up vote 0 down vote accepted

Ok, so from what I gather on the basis of Siminore's comments:

$X=H_0^1$ and using Poincare's inequality we have that

$$ \|u\|_{L^2}\leq C \| \nabla u \|_{L^2}. $$

Hence $(u,u)_2=\| \nabla u \|_{L^2}=0$ implies $u=0$ a.e.

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