# Sobolev imbedding theorem $H^{1,p}(\mathbb{R}^n)$ contained in $L^{{np}/(n-p)}(\mathbb{R}^n)$ (Taylor Michael)

i. I can not make sense of the following:

For (2.4) I imagine that it is the fundamental theorem of the calculation but I can not prove it formally.

Neither will it be understood how to arrive at equation 2.6. For this I had thought the following

$$(|u|_{L^{n/(n-1)}})^{(n-1)/n}=\int_{\mathbb{R}^n}|u(x)|^{n/(n-1)}dx\leq\int_{\mathbb{R}^n} |u_1\cdots u_n|dx$$ with $$u_j=(\int_{-\infty}^{\infty} |D_j u(x)|dx_{j})^{1/(n-1)}$$ but I can not get inequality 2.6

Observe by the fundamental theorem of calculus, you have \begin{align} u(x, \mathbf{x}_{n-1})=u(x, \mathbf{x}_{n-1})-u(x_0, \mathbf{x}_{n-1}) = \int^x_{x_0} D_1u(x_1, \mathbf{x}_{n-1})\ dx_1 \end{align} if $$(x_0, \mathbf{x}_{n-1})\notin \operatorname{supp} u$$ which means \begin{align} |u(\mathbf{x})| \leq \int^\infty_{-\infty}|D_1u|\ dx_1. \end{align} Likewise, we have \begin{align} |u(\mathbf{x})| \leq \int^\infty_{-\infty}|D_ju|\ dx_j. \end{align} which means \begin{align} |u(\mathbf{x})|^n \leq \prod^n_{j=1}\int^\infty_{-\infty}|D_ju|\ dx_j \ \ \implies \ \ |u(\mathbf{x})|^{\frac{n}{n-1}} \leq \prod^n_{j=1}\left\{\int^\infty_{-\infty}|D_ju|\ dx_j\right\}^{\frac{1}{n-1}}. \end{align} Finally, we see that \begin{align} \int_{\mathbb{R}^n} |u(\mathbf{x})|^{\frac{n}{n-1}}\ d\mathbf{x} \leq \int_{\mathbb{R}^n}\prod^n_{j=1}\left\{\int^\infty_{-\infty}|D_ju|\ dx_j\right\}^{\frac{1}{n-1}}\ d\mathbf{x}. \end{align}
Next, fix $$(x_2, \ldots, x_n)$$ and note that \begin{align} \int^\infty_{-\infty} dx_1\ \prod^n_{j=1}\left\{\int^\infty_{-\infty}|D_ju|\ dx_j\right\}^{\frac{1}{n-1}} =&\ \left\{\int^\infty_{-\infty} |D_1u(x_1, \mathbf{x}_{n-1})|\ dx_1 \right\}^{\frac{1}{n-1}}\int^\infty_{-\infty} dx_1\prod^n_{j=2}\left\{\int^\infty_{-\infty}|D_ju|\ dx_j\right\}^{\frac{1}{n-1}}\\ \leq&\ \left\{\int^\infty_{-\infty} |D_1u(x_1, \mathbf{x}_{n-1})|\ dx_1 \right\}^{\frac{1}{n-1}}\prod^n_{j=2}\left\{\int^\infty_{-\infty} \int^\infty_{-\infty}|D_ju|\ dx_1dx_j\right\}^{\frac{1}{n-1}}. \end{align} Next, fix $$(x_3, \ldots, x_n)$$, we see that \begin{align} \int^\infty_{-\infty}\int^\infty_{-\infty} \prod^n_{j=1}\left\{\int^\infty_{-\infty}|D_ju|\ dx_j\right\}^{\frac{1}{n-1}}dx_2dx_1\leq&\ \int^\infty_{-\infty} \ \left\{\int^\infty_{-\infty} |D_1u(x_1, \mathbf{x}_{n-1})|\ dx_1 \right\}^{\frac{1}{n-1}}\prod^n_{j=2}\left\{\int^\infty_{-\infty} \int^\infty_{-\infty}|D_ju|\ dx_1dx_j\right\}^{\frac{1}{n-1}}dx_2\\ =&\ \left\{\int^\infty_{-\infty}\int^\infty_{-\infty}|D_2u|\ dx_1dx_2 \right\}^{\frac{1}{n-1}}\int^\infty_{-\infty}\left\{\int^\infty_{-\infty} |D_1u(x_1,x_2, \mathbf{x}_{n-2})|\ dx_1 \right\}^{\frac{1}{n-1}}\prod^n_{j=3}\left\{\int^\infty_{-\infty} \int^\infty_{-\infty}|D_ju|\ dx_1dx_j\right\}^{\frac{1}{n-1}}dx_2\\ \leq&\ \left\{\int^\infty_{-\infty}\int^\infty_{-\infty}|D_2u|\ dx_1dx_2 \right\}^{\frac{1}{n-1}}\left\{\int^\infty_{-\infty}\int^\infty_{-\infty}|D_1u|\ dx_1dx_2 \right\}^{\frac{1}{n-1}}\prod^n_{j=3}\left\{\int^\infty_{-\infty}\int^\infty_{-\infty} \int^\infty_{-\infty}|D_ju|\ dx_1dx_2dx_j\right\}^{\frac{1}{n-1}}. \end{align} Continuing with this procedure yields \begin{align} \int_{\mathbb{R}^n} \prod^n_{j=1}\left\{\int^\infty_{-\infty}|D_ju|\ dx_j\right\}^{\frac{1}{n-1}}d\mathbf{x}\leq \prod^n_{j=1} \left\{\int_{\mathbb{R}^n}|D_ju(\mathbf{x})|\ d\mathbf{x}\right\}^{\frac{1}{n-1}} \end{align} which is the desired inequality.