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When $\Omega$ is a bounded open set of $\mathbb{R}^N$ with the help of Poincare inequality, we know that $H_0^1(\Omega)$ with $(u,v)_{H_0^1} = \int\nabla u \cdot \nabla v$ is a Hilbert space.

Clearly $(u,v)_{H_0^1}$ is not an inner product for $H^1(\Omega)$ since for any non zero constant function $u$, we have $(u,u)_{H_0^1} = 0 \not\Rightarrow u \equiv 0$.

When taking $\Omega = \mathbb{R}^N$, we have $$H_0^1(\mathbb{R}^N) = H^1(\mathbb{R}^N),$$ And I was wondering is $$(u,v)_{H_0^1} = \int\nabla u \cdot \nabla v$$ still an inner product? If yes, is $(H^1(\mathbb{R}^N), (u,v)_{H_0^1})$ a Hilbert space?

Second question, take $\Omega$ is bounded again, why do most books choose to define the inner product on $H_0^1(\Omega)$ to be $$(u,v) = \int uv + \int \nabla u\cdot \nabla v$$ instead of $$(u,v)_{H_0^1} = \int\nabla u \cdot \nabla v .$$

Is this just a preference? I feel like when working with energy functionals that only depends on $\nabla u$ such as $E(u) = \frac{1}{2}\int|\nabla u|^2$, it is more natural to work $(u,v)_{H_0^1} = \int\nabla u \cdot \nabla v $. For example, when calculating the Fenchel Legendre transform, we have $$E^*(u) = \sup_{v\in H_0^1(\Omega)} \Bigg\{\int \nabla u\cdot \nabla v - \frac{1}{2}\int|\nabla v|^2\Bigg\} = \frac{1}{2}\int|\nabla v|^2 = E(u) $$ rather than $$E^*(u) = \sup_{v\in H_0^1(\Omega)} \Bigg\{\int uv + \int \nabla u\cdot \nabla v - \frac{1}{2}\int|\nabla v|^2\Bigg\}$$ which I don't think there is a closed form.

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For your first question, no $H_0^1(\mathbb{R}^n)$ is not a Hilbert space with the inner product $(u,v)=\int \nabla u \cdot \nabla v$. Its completion however is, and is usually denoted by $\mathcal{D}^{1,2}(\mathbb{R}^n)$ or $L^{1,2}(\mathbb{R}^n)$ and consists of functions in $L^{2^*}(\mathbb{R}^n)$ with integrable gradient, where $2^*=2n/(n-2)$ is the usual Sobolev exponent (to see this think Gagliardo-Nirenberg-Sobolev inequality).

As for why we don't give this norm in bounded domains. Recall that the usual way of introducing $H^1_0$ is by defining it as the completion of test functions in $H^1$. This is because we want something to encode, in some sense, zero boundary values of an $H^1$ function (and indeed if the domain is smooth enough this is justified by the trace theorems). If what you have in mind is say minimization procedures like the Dirichlet principle it totally makes sense to introduce your norm directly, but most texts will want some familiarity with Sobolev sapces before tackling the applications and for this the norm induced by that of $H^1$ is more natural as I said.

I can't comment on the Frenchel-Legendre transform since I'm not at all familiar with it, sorry.

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  • $\begingroup$ Thank you for the answer. I am studying the $p$-Laplace operators on the space $H_0^1(\Omega)\cap W^{1,p}(\Omega)$ so the functions that I work with always have trace zero. I just want to make sure that $(u,v)_{H_0^1}$ is okay to work with because I rarely see it being introduced in the books. $\endgroup$ – Xiao Dec 2 '14 at 6:15

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