A few lessons ago, my professor proved Poincaré inequality in the following form:
Let $\Omega$ be a domain contained in $\mathbb{R}^{N-1}\times(0,a)$ for some $N\in\mathbb{N},a>0$. Then for all $u\in W_0^{1,p}(\Omega)$ we have:
$$\|u\|_{L^p(\Omega)}\leq\frac{a}{2}\|\nabla u\|_{L^p(\Omega)}.$$
I understood the whole proof, except for a passage at the beginning. The idea is to first prove such an inequality for test functions on $(0,a)$, then extend it to test functions on $\Omega$, and then of course everything passes to the limit for $W_0^{1,p}$ functions. The first passage for test functions on $(0,a)$ is that, if $v\in\mathcal{D}(0,a)$, then:
$$|v(x)|\leq\frac12\int\limits_0^a|v'(t)|\mathrm{d}t.$$
Now, by the fundamental theorem of calculus:
$$v(x)=\int\limits_0^xv'(t)\mathrm{d}t.$$
Take absolute values, major by taking the absolute values inside the integral, then you are integrating a positive functions so extending the domain to $(0,a)$ (since $x$ has to be in $(0,a)$) makes the integral bigger. But that $\frac12$? It looks as if he actually has an extra half in the fundamental theorem, and this half is actually responsible for the half in the theorem statement! So where is that half from? Did he just put it in as a mistake? Or is it actually supposed to be there?
Since $v(0) = v(a) = 0$, you have $$v(x) = \int_0^x v'(x) \, \mathrm{d} x = -\int_x^a v'(x) \, \mathrm{d} x.$$ Hence, $$|v(x)| \le \int_0^x |v'(x)| \, \mathrm{d} x$$ and $$|v(x)| \le \int_x^a |v'(x)| \, \mathrm{d} x.$$ Thus, $$|v(x)| \le \min\{\int_0^x |v'(x)| \, \mathrm{d} x, \int_x^a |v'(x)| \, \mathrm{d} x\} \le \frac12 \int_0^a |v'(x)|\,\mathrm{d}x.$$