So the problem I have at hand should be rather elementary, but I can't figure out. It's in a proof I'm trying to understand completely. Here's what I need to do exactly:
Show that $|x-y| \approx \ell(Q) + |x - c_Q|$, for all $x \notin Q^*$ and $y \in Q$, where $Q \subsetneq Q^*$ with $\ell(Q^*) = C\ell(Q)$. Here $Q$ and $Q^*$ are cubes in $\mathbf{R}^n$, $c_Q$ is the center of $Q$, $\ell(Q)$ is the side length of $Q$, $Q^*$ also has center $c_Q$, and $C$ is a constant depending only on dimension.
First off, here's what the picture might look like in $\mathbf{R}^2$:
Here's what I have shown. I know by the triangle inequality and the fact that the largest $|y - c_Q|$ can be is $C_0 = \max(\sqrt{n}/{2},1) \ell(Q)$ that $$ |x - y| \leq |y - c_Q| + |x - c_Q| \leq C_0\big(\ell(Q) + |x - c_Q|\big) $$ proving that $|x - y| \lesssim \ell(Q) + |x - c_Q|$. Now the other direction is what I'm having issues with. Mainly, I can't show that $|x - y| \gtrsim|x - c_Q|$. I have shown that $|x - y| \gtrsim \ell(Q)$. In particular, $$ |x - y| \geq \mbox{dist}(x,Q) \gtrsim \ell(Q) $$ since $$ \mbox{dist}(x,Q) \geq \dfrac{\sqrt{n}}{2} \ell(Q^*) - \dfrac{\sqrt{n}}{2} \ell(Q) \approx \ell(Q). $$ Now if I can show that $|x-y| \gtrsim |x-c_Q|$ I'll be done since I can just average the two approximations. Any ideas? Thanks!
Edit: Please see my solution below and let me know what you think. Thanks!
Notice that $|x - y | \geq |x - c_Q| - |y - c_Q|$, so if one can show that $|y - c_Q| \leq A|x - c_Q|$ for some constant $A < 1$, we're done. We have $$|x - c_Q| \geq \dfrac{\sqrt{n}}{2} \ell(Q^*) = C\dfrac{\sqrt{n}}{2} \ell(Q)$$ and $$|y - c_Q| \leq \dfrac{\sqrt{n}}{2} \ell(Q).$$ Therefore, $$ |y-c_Q| \leq \dfrac{1}{C} \cdot C \dfrac{\sqrt{n}}{2} \ell(Q) \leq \dfrac{1}{C} |x - c_Q|. $$ Hence, setting $A = 1/C$ gives $|x - y| \geq (1-A) |x-c_Q| > 0$ since $C > 1$.