# sobolev space reflexivity

I am having problem with the following 1)Are $H^{1}$ nad $H^{1}_{0}$ a reflexive spaces?

2)If $u_{n} \rightarrow u$ weakly in $H^{1}_{0}$, can I say that it is same as $(\nabla u_{n} , \nabla w) \rightarrow (\nabla u, \nabla w)$ for any $w \in H^{1}_{0}$? Thanks a lot

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ad 1) Hilbert spaces are reflexive. –  martini Feb 10 '12 at 8:18

1) Assume you're in an open set $U\subset\mathbb{R}^n$ then the mapping $u\mapsto (u,\nabla u)$ from either $H^1$ or $H_0^1$ to $L^2\times(L^2)^n$ equipped with the appropiate product norm gives an isometry. Since $H^1$ and $H_0^1$ are Banach spaces, they're also closed subspaces of a reflexive space and so are reflexive themselves.
2) By the Poincaré inequality (the one that says $\| u\|_{H_0^1}\leq c\| \nabla u \|_{L^2}$ for every $u\in C_c^\infty$) the inner product $(\nabla u , \nabla v)_{L^2}$ is equivalent to the usual inner product in $H_0^1$ so the answer is yes.
It means there exists $c_1,c_2>0$ such that $c_1(u,v)_{H_0^1} \leq (\nabla u, \nabla v)_{L^2}\leq c_2 (u,v)_{H_0^1}$ for all $u,v\in H_0^1$. By the polarization identity this is the same as asking the norms to be equivalent. –  Jose27 Feb 10 '12 at 14:51