Generalized Poincaré Inequality on H1 proof.

let’s see if someone can help me with this proof.

Let $\Omega\subset\mathbb{R}^n$ be a bounded domain. And let $L^2\left(\Omega\right)$ be the space of equivalence classes of square integrable functions in $\Omega$ given by the equivalence relation $u\sim v \iff u(x)=v(x)\, \text{a.e.}$ being a.e. almost everywhere, in other words, two functions belong to the same equivalence classes if they only are different in a zero meassure set.

This space is a Hilbert space and its norm is:

$$ \Vert v \Vert ^2_{L^2 \left( \Omega \right) }= \int_\Omega v(x)^2\,\mathrm{d}x $$

Let $H^1\left(\Omega\right)=\left\{ v\in L^2\left( \Omega \right);\,\vert\mathbf{grad}(v)\vert \in L^2\left( \Omega \right ) \right\}$ be also a Hilbert space and its norm :

$$ \Vert v \Vert ^2_{H^1 \left( \Omega \right) }= \Vert v \Vert ^2_{L^2 \left( \Omega \right) } + \Vert \vert\mathbf{grad}(v)\vert \Vert ^2_{L^2 \left( \Omega \right) } $$

I’ve been asked to proove that:

$$ \Vert v \Vert ^2_{H^1 \left( \Omega \right) } \le C\left( \Omega\right) \left( \Vert \vert\mathbf{grad}(v)\vert \Vert ^2_{L^2 \left( \Omega \right) } + \int_\Gamma \left ( \gamma v(x) \right) ^2\,\mathrm{d}\sigma \right) \quad \forall v\in H^1 \left( \Omega \right) $$

where $\Gamma = \partial \Omega $, $\Omega $ is smooth enough, $\mathrm{d}\sigma$ is a meassure on the boundary of $\Omega$, and the operator $\gamma: L^2\left( \Omega\right )\mapsto L^2\left( \Gamma\right )$ is the trace operator. ( if it is needed I can define it also )

Sincerely, I just don’t know how to begin, and maybe someone could give me some clue about it.(by the way the $C$ is a positive constant depending only on the domain $\Omega$).

Thank you very much!

Solutions Collecting From Web of "Generalized Poincaré Inequality on H1 proof."

One way is by contradiction: suppose that for all $n\in\mathbb N$, there exists $u_n\in H^1(\Omega)$ such that $$\|u_n\|_{L^2(\Omega)}> n\|\nabla u_n\|_{L^2(\Omega)}+n\|u_n\|_{L^2(\partial\Omega)}.$$ Then $\|u_n\|_{L^2(\Omega)}>0$ for all $n$, and if we set $v_n=u_n/\|u_n\|_{L^2(\Omega)}$, then $\|v_n\|_{L^2(\Omega)}=1$ for all $n$, and also $$n\|\nabla v_n\|_{L^2(\Omega)}+n\|v_n\|_{L^2(\partial\Omega)}<1,$$ which implies that $$\|\nabla v_n\|_{L^2(\Omega)}<\frac{1}{n}.$$ Therefore $\nabla v_n\to 0$ in $L^2(\Omega)$.

Now, $(v_n)$ is bounded in $H^1(\Omega)$, so there exists a subsequence, denoted by $(v_n)$, that converges weakly in $H^1(\Omega)$ to some $v$. Then $\nabla v_n\to \nabla v$, weakly in $L^2$, therefore $\nabla v=0$, which shows that $v$ is a constant $c$ in $\Omega$ (if $\Omega$ is also connected). In addition, from the Rellich–Kondrachov theorem, a further subsequence, still denoted by $v_n$, converges to $c$, strongly in $L^2(\Omega)$, if $\Omega$ is nice enough. Hence, $$1=\|v_n\|_{L^2(\Omega)}\xrightarrow[n\to\infty]{}\|c\|_{L^2(\Omega)}.$$ This shows that $c\neq 0$. Since the trace operator is continuous, we also obtain that $$\|v_n\|_{L^2(\partial\Omega)}\xrightarrow[n\to\infty]{}\|c\|_{L^2(\partial\Omega)}.$$ But, $$\|v_n\|_{L^2(\partial\Omega)}<1/n,$$ therefore $c=0$, which is a contradiction.

This shows that, for a constant $C$, $$\|u\|_{L^2(\Omega)}\leq C\|\nabla u\|_{L^2(\Omega)}+C\|u\|_{L^2(\partial\Omega)}$$ for all $u\in H^1(\Omega)$. Adding $\|\nabla u\|_{L^2(\Omega)}$ to both sides shows the inequality.