permuting two integrations

Question

Can someone explain to me what he used after he said 'and permuting the two integrations with respect t and x we conclude ..'

Lemma Let $\Omega$ be an open set of $\mathbb{R}^{N}$ bounded in at least one space direction. There exists a constant $C>0$ such that, for every function $v \in C^{1}(\bar{\Omega})$ which is zero on the boundary $\partial \Omega$

$$ \int_{\Omega}|v(x)|^{2} d x \leq C \int_{\Omega}|\nabla v(x)|^{2} d x $$ Proof

.....

$$ \int_{\Omega}|v(x)|^{2} d x \leq(b-a) \int_{\Omega} \int_{a}^{b}\left|\frac{\partial v}{\partial x_{1}}\left(t, x_{2}, \ldots, x_{N}\right)\right|^{2} d t d x $$ and permuting the two integrations with respect to $t$ and $x,$ we conclude $$ \int_{\Omega}|v(x)|^{2} d x \leq(b-a)^{2} \int_{\Omega}\left|\frac{\partial v}{\partial x_{1}}(x)\right|^{2} d x \leq(b-a)^{2} \int_{\Omega}|\nabla v(x)|^{2} d x $$

Answer 1

Denoting $f(t,x_{2},...x_{N})=\frac{\partial}{\partial{x_{1}}}v(t,x_{2},...x_{N})$ and using Fubini (i.e. "permuting the integrations") in the first step we get $$\int_{\Omega}\int_{a}^{b}f(t,x_{2},...x_{N})dtdx = \int_{a}^{b}\int_{\Omega}f(t,x_{2},...x_{N})dxdt = \int_{a}^{b} ||f||_{L^{2}(\Omega)}^{2}dt$$ So we´re only integrating the constant expression $||f||_{L^{2}(\Omega)}^{2}$ over $[a,b]$, which yields the inequality.