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.


1 Answer 1


I too became confused by that exact step this morning as I tried to make my way through Evans, and was very disappointed to see that this had no answer. I've since figured it out, and while 4 years too late to help the OP here, I at least hope to be of use to future Evans users seeking clarification.

In my copy of Evans (also 2nd edition), the (12) is on the 2nd line of the calculation, and not the 3rd like the OP has indicated, so that caused my main confusion and perhaps confusion for others with the same Evans book as me.

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.


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