Problem: I'm trying to prove that the space $H=\{v\in H^{1}(I): v(0)=v(1)\}$ is a Hilbert's space equpped with the $H^{1}$ scalar product.
We know that and $$\forall u,v \in H^{1}(I):\quad \langle u,v \rangle_{H^{1}(I)}=\int_{I}uv+\int_{I}\nabla u \cdot \nabla v$$ and the induced norm is $$||v||_{H^{1}(I)}^{2}=||v||_{L^{2}(I)}^{2}+||\nabla v||^{2}_{[L^{2}(I)]^{d}}$$
Let $\{v_{n}\}$ a Cauchy sequence in $H$, so I need to prove that $v_{n}\to v\in H$. Now, we pick a convergent sequence $\{v_n\} \subset H$ such that $v_n \to v$, so I need to show that $v \in H$. Now, how $\{v_n\}$ converges in the norm $||\cdot ||_{H^1(I)}$ then $v_n \to v$ uniformly,
But how can I continue form here? Perhaps there is an easier way to approach the solution of the problem? or maybe I need to use a more strong result as the Sobolev-embedding?
Apologize me if my solution-approach is not right, I'm a bit confused with this space and your approach with the solution.
Since you already know that $H^1((0,1))$ is complete, it suffices to show that $H$ is a closed subspace of $H^1$. For that purpose note that $H$ is the kernel of the continuous map $$ H^1((0,1))\hookrightarrow C([0,1])\overset{\Phi}{\to}\mathbb{K}, $$ where the first arrow is the continuous embedding guaranteed by the Sobolev embedding theorem and $\Phi(v)=v(1)-v(0)$, which is obviously continuous (as discussed in the comments). The kernel of a continuous linear map is a closed subspace, so we are done.