How to prove this change of variables?

Question

Consider the sets: $$Q_0=\{x\in\mathbf{R}^n:0<x_1<2, \ |x_\alpha|<1, \ for \ 1<\alpha\leq n\},$$ $$Q_l=\{x\in\mathbf{R}^n:0<x_1<2l, \ |x_\alpha|<l \ for \ 1<\alpha\leq n\}.$$ Suppose $0\leq u\leq K$ in $Q_l$, and $$|Mu|\leq A(|\nabla u|+u+k), \ \ in \ Q_0,$$ where $$M=\sum_{i,j=1}^na_{ij}(x)\frac{\partial^2}{\partial x_i\partial x_j}.$$ The, by scaling $x=ly$, $$|Mu|\leq A\left(\frac{|\nabla u|}{l}+\frac{u}{l^2}+k\right) \ \ in \ Q_l.$$

Answer 1

Take $y=\frac{x}{l}$. Then $y(Q_l)=Q_0$. Now, with $v=u \circ y$, by the chain rule:

$|Mv|=\frac{1}{l^2}|Mu| \leq A(\frac{|\nabla u|}{l^2} + \frac{u}{l^2} + \frac{k}{l^2})= A(\frac{|\nabla v|}{l} + \frac{v}{l^2} + \frac{k}{l^2})$.

If $l\geq1$, the result follows. For $l<1$, it is actually wrong:

Take $n=1$, $a_{ij}=1$ and $u(y)=y^2$. Then $u$ satisfies the first inequality (for $A=1, k=2$) but not the second since

$|v''|=\frac{2}{l^2} > \frac{2x}{l^3} + \frac{x^2}{l^4} + 2 = A(\frac{|v'|}{l} + \frac{v}{l^2} + k)$ for $x$ small enough.

So one has to scale $k$, too (or require a lower bound for $u$).