# Gagliardo-Nirenberg-Sobolev inequality

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.

It is in fact easiest to integrate the third line with respect to $$x_2$$ (the line the OP indicated by (12) ) because the Hölder inequality has already been applied. The term in the product $$\bigg(\int_{-\infty}^\infty\int_{-\infty}^\infty |Du|dx_1dy_2\bigg)^{\frac{1}{n-1}}$$ comes out of the new integral because it is constant in $$x_2$$, as a result of integrating out $$y_2$$. What's left over inside the integral with respect to $$x_2$$ is then
$$\left( \int_{-\infty}^{\infty} |Du| \, dy_1 \right)^{\frac 1{n-1}} \prod_{i=3}^n \left(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |Du| \, dx_1dy_i \right)^{\frac 1{n-1}}$$ which is precisely $$\prod_{i=1 \atop i \not= 2}^n I_i^{\frac 1{n-1}}$$ where the $$I_i's$$ are as Evans and the OP have defined. Hence we end up with $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |u|^{\frac n{n-1}} \, dx_1 dx_2 \le \left(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |Du| \, dx_1 dy_2 \right)^{\frac 1{n-1}} \int_{-\infty}^{\infty} \prod_{i=1 \atop i \not= 2}^n I_i^{\frac 1{n-1}} \, dx_2$$ which is the desired result.