An estimate for $x/|x|$

Question

I want to show that $$\left|\frac{x-y}{|x-y|}-\frac{x}{|x|}\right|\leq 4\frac{|y|}{|x|}$$ whenever $|x|>2|y|$ and $x,y\in\mathbb R^n$. This is at many places in the literature and related to singular integrals. Can anyone tell me how to prove this? I tried to prove this using mean value theorem but the derivative of the function $x/|x|$ does not seem to have the required bound.

Answer 1

You can notice that $$ \left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| = \left|\frac{|x| - |x-y|}{|x-y|}\,\frac{x}{|x|} - \frac{y}{|x-y|}\right| $$ so that by the triangle inequality $$ \left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| \leq \frac{\left||x| - |x-y|\right|}{|x-y|} + \frac{|y|}{|x-y|} \leq \frac{2\left|y\right|}{|x-y|}. $$ Now since $|x| - 2\left|y\right| \geq 0$, you get that $2\left|x-y\right| \geq 2\left|x\right|-2\left|y\right| \geq |x|$, and so $|x-y|^{-1}\leq 2\,\left|x\right|^{-1}$. Therefore, $$ \left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| \leq \frac{2\left|y\right|}{|x-y|} \leq 4 \,\frac{\left|y\right|}{|x|}. $$

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Edit: It seems one can actually get better. Write instead $$ \left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| = \left|\frac{|x|-|x-y|}{\left|x\right|}\,\frac{x-y}{|x-y|} - \frac{y}{|x|}\right| $$ and so by the triangle inequality $$ \left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| \leq \frac{\left||x|-|x-y|\right|}{\left|x\right|} + \frac{|y|}{|x|} \leq 2\,\frac{|y|}{|x|} $$ without needing the condition $|x| \geq 2\left|y\right|$. The constant $2$ is optimal (take $y=\lambda x$ with $\lambda > 1$ converging to $1$).

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Edit 2: When $|x| \geq 2\left|y\right|$, the constant can be even further improved. Since $r+1/r\geq 2$ for any $r>0$, it follows that $$\begin{align} \left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right|^2 &= 2-2\,\frac{x\cdot (x-y)}{\left|x\right|\left|x-y\right|} \\ &\leq \frac{|x|}{|x-y|} + \frac{|x-y|}{|x|} - 2\,\frac{x\cdot (x-y)}{\left|x\right|\left|x-y\right|} = \frac{|y|^2}{\left|x\right|\left|x-y\right|} \end{align}$$ so that without condition on $x$ and $y$ (except of course $x\neq y$ and $x\neq 0$) $$ \left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| \leq \frac{|y|}{\sqrt{\left|x\right|\left|x-y\right|}}. $$ In particular, when $|x| \geq 2\left|y\right|$, then $|x-y|^{-1}\leq 2\,\left|x\right|^{-1}$ and so $$ \left|\frac{x-y}{|x-y|} - \frac{x}{|x|}\right| \leq \sqrt{2}\,\frac{|y|}{\left|x\right|}. $$ Again, the constant $\sqrt{2}$ is optimal (take $x-y = z$ with $z \cdot x = 0$ and $|z|\to 0$).