Prove an inequality in the space of $L^{2}(\Omega)$. EDIT: First of all, I thank Calvin Khor's exhaustive comments on the inequality that I proposed regarding the semi-norm in $L^{2}(\Omega)$, which after his teachings and his final comments allowed me to realize that this inequality did not make sense for any semi-norm of $L^{2}(\Omega)$. For that reason I have modified my problem, changing from the semi-norm $|\cdot|_{L^{2}(\Omega)}$ in $L^{2}(\Omega)$ to the norm $||\cdot||_{L^{2}(\Omega)}$ in $L^{2}(\Omega)$. Again, thank you very much Calvin Khor.

Fixed issue: My goals is to prove that there exists a positive-strictly constant $C_{0}>0$ such that for all $v\in H^{1}(\Omega)$, we have that $$||v||^{2}_{L^{2}(\Omega)} \leq C_{0} \left[ ||v||^{2}_{L^{2}(\Phi)}+||\nabla v||^{2}_{L^{2}(\Omega)}\right]$$
where $||\cdot||_{L^{2}}$ be norms in $L^{2}(\Omega)$ and $\Omega=(0,1)\times (0,1)$ and $\Phi=\{0\}\times (0,1)$.

My approach:  Maybe I should use the Poincaré's inequality:

Poncaré's inequality: Let $\Omega$ a bounded open set of $\mathbb{R}^{n}$. Then, there exists a positive $C_{p}(\Omega)$ such that: $$\color{blue}{\boxed{||v||_{L^{2}(\Omega)}\leq C_{p}(\Omega)|v|_{H^{1}(\Omega)}=C_{p}(\Omega)||\mathbf{\nabla v}||_{L^{2}(\Omega)}}}$$

We have that $$ ||v||_{L^{2}(\Omega)}^{2}\leq C_{p}(\Omega)^{2}||\mathbf{\nabla v}||_{L^{2}(\Omega)}^{2} $$
how can I continue from here? maybe the Rellich's compactness theorem help here?
 A: You are right that your inequality is reminiscent of Poincaré. In fact, you can just follow the proof by contradiction of the normal Poincaré inequality. By usual density arguments we can assume $v\in C^\infty(\overline \Omega)$. Suppose such a $C_0>0$  did not exist. Thus for each $k\in\mathbb N$ we can find $v_k\in C^\infty(\overline\Omega)$ such that $\|v_k\|_{L^2(\Omega)} = 1$ and
$$\|v_k\|_{L^2(\Phi)}^2 + \|\nabla v_k \|_{L^2(\Omega)} \le \frac1k.$$
Then $v_k$ are bounded in $H^1$ and therefore after passing to a subsequence using Rellich compactness theorem, there is some $v\in L^2$ such that
$$ v_k|_{\Phi} \to 0 \text{ in } L^2(\Phi), \ \nabla v_k \to 0 \text{ in } L^2(\Omega),\ v_k\to v \text{ in } L^2(\Omega). $$  In
particular $v_k\to v$ in $H^1$ with $\nabla v \equiv 0$, and hence $v\equiv c$ for some $c\in\mathbb R$ ($\Omega$ is connected). Therefore, $v|_{\Phi} \equiv c$ in the sense of trace. By continuity of the trace map, $v_k|_\Phi\to v|_{\Phi}$ in $L^2(\Phi)$. But also $\|v_k\|_{L^2(\Phi)} \to 0$, so $c=0$. This is in contradiction with the assumption that $\|v\|_{L^2}=1$.
