Assume $\Omega$ is a bounded and regular domain of $\mathbb{R} ^n$. Show that
$$\inf_{u\in C^1_0(\Omega) } \frac{\int_\Omega|\nabla u|^2}{\int_\Omega| u|^2} >0. $$
My attempt was that since those $u$ are continuous on the bounded domain it is okay to consider closed subsets of $\Omega$, say $K$ so that $u$ are defined over a compact set. By implication, those functions $u\ne0$ are bounded over $K$ and the norm over $K$ becomes
$$\inf_{u\in C^1_0(\Omega) } \frac{\int_\Omega|\nabla u|^2}{\int_\Omega| u|^2} \ge \inf_{u\in C^1_0(K) } \frac{\int_K|\nabla u|^2}{\int_K| u|^2}>0. $$
That's how I see it. Please any help how to do this the right way?
Set $$ K := \inf_{u\in C^1_0(\Omega) } \frac{\int_\Omega|\nabla u|^2}{\int_\Omega| u|^2} \in [0,\infty), $$ noting this quantity is necessarily finite by choosing a non-trivial element $u \in C^1_0(\Omega) \setminus \{0\}$. By definition of $K$ we have $$ \int_{\Omega} \lvert u \rvert^2 \,\mathrm{d}x \leq K^{-1} \int_{\Omega} \lvert \nabla u \rvert^2 \,\mathrm{d}x $$ for all $u \in C^1_0(\Omega)$, understanding the inequality to be vacuous when $K = 0$. Morever if there exists another constant $C>0$ such that $$ \int_{\Omega} \lvert u \rvert^2 \,\mathrm{d}x \leq C \int_{\Omega} \lvert\nabla u \rvert^2 \,\mathrm{d} x \tag{$\dagger$}$$ for all $u \in C^1_0(\Omega)$, then we have $K \geq C^{-1}>0$ necessarily.
Thus $K^{-1}$ is the best constant in the Poincaré inequality, and to show that $K>0$ one has to show that $(\dagger)$ is valid for some $C \in (0,\infty)$. But this is a classical result that can be proven by writing $$ u(x) = \int_{-\infty}^{x_n} \partial_{x_n}u(x_1,\cdots,x_{n-1},t) \,\mathrm{d}t,$$ extending $u$ by zero outside $\Omega$. Taking $L^2$-norms on both sides and using e.g. the Minkowski integral inequality leads to the desired estimate (with a constant depending on the volume of $\Omega$). Many texts will also deduce this by chaining the Sobolev inequality with Hölder's inequality, but the direct proof I outline above is shorter.