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.