$\int _a^{x_1} |\frac{\partial u}{\partial x_1}(s, x_2,...,x_n)|^p ds \le (x_1-a) |\frac{\partial u}{\partial x_1}(x_1, x_2,...,x_n)|^p$? How  to show $\int _a^{x_1} |\frac{\partial u}{\partial x_1}(s, x_2,...,x_n)|^p ds \le (x_1-a) |\frac{\partial u}{\partial x_1}(x_1, x_2,...,x_n)|^p$, where $\frac{\partial u}{\partial x_1}(a, x_2,...,x_n)=0$ and $u(a, x_2,...,x_n)=0$.
This question is a step of proof of Poincare inequality, I fail to calculate it, and I feel it is not right,  so ask here, thanks for any help.
I have add the book, but it is written by Chinese.



 A: AH, found it. There was no way the inequality you wanted was true : I had to see the context and get the answer. There's a hanging variable.
It is this : let us write $dx = dx_1dx_2...dx_n$ for the integral over $Q$. Then the line before the question mark is actually (by definition of $Q$):
$$
d^{p-1}\int_Q \int_{a_1}^{a_1+d} |D_1u(s,x_2,...,x_n)|^p dsdx_1dx_2...dx_n  \\ =d^{p-1} \underbrace{\int_{a_1}^{a_1+d}\int_{a_1}^{a_1+d}\cdots\int_{a_1}^{a_1+d}}_{n \text{ times}} \int_{a_1}^{a_1+d} |D_1u(s,x_2,...,x_n)|^p dsdx_1dx_2...dx_n \\ = d^{p-1}\int_{a_1}^{a_1+d} \left[ \underbrace{\int_{a_1}^{a_1+d}\int_{a_1}^{a_1+d}\cdots\int_{a_1}^{a_1+d}}_{n \text{ times}} |D_1u(s,x_2,...,x_n)|^p dsdx_2...dx_n\right]dx_1 \\ = d^{p-1} \times d \times \left[\underbrace{\int_{a_1}^{a_1+d}\int_{a_1}^{a_1+d}\cdots\int_{a_1}^{a_1+d}}_{n \text{ times}} |D_1u(s,x_2,...,x_n)|^p dsdx_2...dx_n\right] \\ = d^p \int_Q |D_1u(s,x_2,...,x_n)|^p dsdx_2...dx_n
$$
where the fourth line follows from the fact that the inside integral doesn't depend on $x_1$. So the $d$ term comes from this variable. Finally, if we just call $s = x_1$ now, we get the line you mark $?$ in the author's notes.
