# Understanding the proof of the Poincaré's inequality

This is from the text, finite method elements and their applications: I am trying to understand how we integrate the right hand side over $$\Omega$$. In the one dimensional case, the right hand side is constant, but in this case, it depends on $$x$$ since as $$x$$ varies, $$x_2,\ldots,x_d$$ will vary. So integrating the right hand side over $$\Omega$$ would mean :

$$\int_\Omega\left(\int_{0}^{l}\left|\frac{\partial v}{\partial x_{1}}\left(y, x_{2}, \ldots, x_{d}\right)\right|^{2} d y\right)dx$$

How would this give us the final inequality?

The right side of the inequality above (1.35) has the variables $$x_2,\dotsc,x_d$$ still free, while the right side has all of the variables $$x_1,\dotsc,x_d$$ still free. In this sense you can think of the inequality as being of the form $$F(x) = F(x_1,\dotsc,x_d) \le G(x_2,\dotsc,x_d)$$ for all $$x \in \Omega$$, where $$\Omega$$ is the cube $$(0,\ell)^d$$. The nice thing about a cube is that if you forget about its first variable (or, really, any one of them) it still gives you a cube for the remaining variables. So let's say that $$\Omega = (0,\ell)^d = (0,\ell) \times \Omega'$$, where $$\Omega' = (0,\ell)^{d-1}$$. Then we first integrate the inequality over $$\Omega'$$ and use Fubini-Tonelli: $$\int_{\Omega'} |v(x_1,x_2,\dotsc,x_d)|^2 dx_2 \cdots dx_d \le \ell \int_{\Omega'} \int_0^\ell |\partial_1 v(y,x_2,\dotsc,x_d)|^2 dy dx_2 \cdots dx_d \\ =\ell \int_0^\ell \int_{\Omega'} |\partial_1 v(y,x_2,\dotsc,x_d)|^2 dy dx_2 \cdots dx_d = \ell \int_\Omega |\partial_1 v(w)|^2 dw$$ where in the last integral $$dw$$ indicates Lebesgue measure in $$\mathbb{R}^d$$. Now we're back in a situation similar to before in that the left side still has $$x_1$$ as a free variable but the right has no variable and is in fact a constant. So we simply integrate once more with respect to $$x_1$$ and use Fubini-Tonelli again: $$\int_\Omega |v(w)|^2 dw = \int_0^\ell \int_{\Omega'} |v(x_1,x_2,\dotsc,x_d)|^2 dx_2 \cdots dx_d dx_1 \le \int_0^\ell \left( \ell \int_\Omega |\partial_1 v(w)|^2 dw\right) \\ = \ell^2 \int_\Omega |\partial_1 v(w)|^2 dw.$$ As the final step we forget about the special role of $$\partial_1$$ and replace by the less-efficient inequality that uses the entire $$H^1$$ seminorm: $$\Vert v \Vert_{L^2(\Omega)}^2 = \int_\Omega |v(w)|^2 dw \le \ell^2 \int_\Omega |\partial_1 v(w)|^2 dw \le \ell^2 \sum_{j=1}^d \int_\Omega |\partial_j v(w)|^2 dw = \ell^2 |v|_{H^1}^2.$$ Take a square root, and we're done.