Prove that the space $H=\{u\in H^1(\Omega):\int_{\Omega}u\,dx=0\},$ where $\Omega$ is bounded domain in $\mathbb{R}^N,\,N\geq 1$ is Hilbert with the inner product $<u,v>=\int_{\Omega}\nabla u\cdot\nabla v\,dx.$
The fact that $<,>$ is an inner product follows from the definition of $H.$ Now to prove Hilbert, we need to prove $H$ is complete, to rove which I am facing difficuly.
Indeed, let $u_n$ be Cauchy in $H.$ Then $\nabla u_n$ is Cauchy in $L^2(\Omega)$ which converges to some $v\in L^2(\Omega).$ Now $\int_{\Omega}u_n\,dx=0,$ as $u_n\in H.$After this step, I got stuck. Please help me how to proceed from here to prove the limit exists in $H.$ Thanks.
You need some version of the Poincare inequality. The well-known version states that, as long as $\partial \Omega$ isn't unreasonable, a function $u \in H^1_0(\Omega)$ will satisfy $$\|u\|_{L^2(\Omega)} \le C_{N,\Omega} \|\nabla u\|_{L^2(\Omega)}.$$ You should try to prove this result replacing $u \in H^1_0(\Omega)$ with $u \in H^1(\Omega)$ and $\displaystyle \int_\Omega u \, dx = 0$.
Once the above is established the solution is simple. If $\{u_n\}$ is Cauchy in $H$ then $\{\nabla u_n\}$ is Cauchy in $L^2(\Omega)$ and the Poincare inequality implies that $\{u_n\}$ is also Cauchy in $L^2(\Omega)$. Consequently $\{u_n\}$ is Cauchy in $H^1(\Omega)$ and thus converges to a function $u \in H^1(\Omega)$. But $$ \left| \int_\Omega u \, dx \right| = \left| \int_\Omega u \, dx - \int_\Omega u_n \, dx\right| \le |\Omega|^{1/2} \|u - u_n\|_{L^2(\Omega)} \to 0$$ so that $u \in H$.