poincaré inequality direct proof

Solution 1:

In the first inequality, integrate with respect to $x_1$ from $0$ to $L$. Since the right hand side is independent of $x_1$ you end up with $$\int_0^L \lvert u(x_1,x')\rvert^2 dx_1 \le L^2 \int^L_0 \lvert \nabla u(s,x') \rvert^2 ds.$$ This is the inequality you apply to derive the second one.

EDIT: The initial inequality that proved is $$\lvert u(x_1,x') \rvert^2 \le L \int^L_0 \lvert \nabla u(s, x') \rvert^2 ds.$$ In this inequality, the left hand side depends on $x_1$, but the right hand side does not. Thus we can integrate with respect to $x_1$ from $0$ to $L$ and switch the order of integration on the right (I suppose this requires Tonelli's theorem): \begin{equation} \begin{aligned} \int^L_0 \lvert u(x_1,x')\rvert^2 dx_1 \le L \int^L_0 \int^L_0 \lvert \nabla u(s,x') \rvert^2 dsdx_1 &= L \int^L_0 \lvert \nabla u(s,x') \rvert^2 \left( \int^L_0 dx_1 \right) ds \\ &=L^2\int^L_0 \lvert \nabla u(s,x') \rvert^2 ds. \end{aligned} \end{equation} Then we see \begin{align*} \int_\Omega |u|^2 \, dx &= \int_{\mathbb R^{n-1}} \left[\int_0^L |u(x_1,x')|^2 \, dx_1\right] \, dx' \\& \leq \int_{\mathbb R^{n-1}} \left[L^2 \int_0^L |\nabla u(s,x')|^2 \, ds\right] \, dx' \\ &= L^2 \int_\Omega |\nabla u|^2 \, dx.\end{align*} where the inequality of the two terms in the square braces is exactly the inequality we arrived at above.