There is one step of the textbook's proof that I wish could be clarified.
Pages 277-278 of PDE Evans, 2nd edition, says:
Integrate this inequality with respect to $x_1$: \begin{align} \int_{-\infty}^{\infty} |u|^{\frac{n}{n-1}} \, dx_1 &\le \int_{-\infty}^{\infty} \prod_{i=1}^n \left( \int_{-\infty}^{\infty} |Du| \, dy_i \right)^{\frac 1{n-1}} \, dx_1 \\ &= \left( \int_{-\infty}^{\infty} |Du| \, dy_i \right)^{\frac 1{n-1}} \int_{-\infty}^{\infty} \, \prod_{i=2}^n \left( \int_{-\infty}^{\infty} |Du| \, dy_i \right)^{\frac 1{n-1}} \, dx_1 \tag{12} \\ &\le \left( \int_{-\infty}^{\infty} |Du| \, dy_i \right)^{\frac 1{n-1}} \prod_{i=2}^n \left(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |Du| \, dx_1dy_i \right)^{\frac 1{n-1}} \end{align} $\quad$Now integrate $\text{(12)}$ with respect to $x_2$: $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |u|^{\frac n{n-1}} \, dx_1 dx_2 \le \left(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |Du| \, dx_1 dy_2 \right)^{\frac 1{n-1}} \int_{-\infty}^{\infty} \prod_{i=1 \atop i \not= 2}^n I_i^{\frac 1{n-1}} \, dx_2, $$ for $$I_1 := \int_{-\infty}^{\infty} |Du| \, dy_1, \quad I_i := \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |Du| \, dx_1 dy_i \quad (i=3,\ldots,n)$$
What steps are employed to derive that last expression (the one that results from integrating $\text{(12)}$ with respect to $x_2$)?
I suspect that the general Hölder inequality was used. The proof continues in the book explicitly saying this, and those derivations--for integrating with respect to $x_3$, $x_4$, ... all the way up to $x_n$--are similar to the one I'm posting here.