Sharp upper bound for distance between projections onto boundary of unit sphere in $\mathbb{R}^n$

Question

In Stein’s book on singular integrals he claims that given $x,y \in \mathbb{R}^n$ satisfying $|x| \geq 2|y|$ there exists a constant $c$ independent of $x,y$ or $n$ such that the inequality $$\left|\frac{x-y}{|x-y|} - \frac{x}{|x|} \right| \leq c \frac{|y|}{|x|}$$ This is true and its not hard to find that $c = 4$ is a suitable constant. My question is what is the sharpest constant possible here?

Answer 1

After working on this for a few hours I believe I’ve come up with a bound that is not in any of the literature that I’ve read. I don’t think anyone REALLY cares what the sharpest bound is since it just gets absorbed into an implicit constant in the context in which this result is useful. My professor had told be that he read somewhere the optimal bound was 2 so I worked hard and managed to beat that bound and obtain a slightly more general result.

Let $\alpha > 1$, and suppose $|x| \geq \alpha |y|$ then take $\theta = \arcsin(1/\alpha)$. Then we can write $\frac{|y|}{|x|} = \sin(\psi)$, for some $|\psi| \leq \theta$. Now let $\varphi$ denote the angle between $x-y$ and $x$, then projecting down to a (probably the only) plane containing the vectors $x$,$y$ we see by simple geometry that $|\varphi|$ can take any positive value less than $|\psi|$. Thus given $x,y$ satisfying $|x| \geq \alpha |y|$, there exists unique $|\varphi| \leq |\psi| \leq \theta$ such that

$$\left|\frac{x-y}{|x-y|}-\frac{x}{|x|}\right| \frac{|x|}{|y|} = \frac{2 \sin(\varphi/2)}{\sin(\psi)}$$ Then we may take $$c_\alpha = \sup_{|\varphi| \leq |\psi| \leq \theta} \frac{2 \sin(\varphi/2)}{\sin(\psi)}$$ Clearly the numerator is maximized by taking $\varphi = \psi$ and so $$c_\alpha = \sup_{|\psi| \leq \theta}\frac{2 \sin(\psi/2)}{\sin(\psi)} = \sup_{|\psi| \leq \theta}\frac{1}{\cos(\psi / 2)}. = \frac{1}{\cos(\theta/2)} = \sqrt{\frac{2}{1-\cos(\theta)}} = \sqrt{\frac{2}{1-\sqrt{1 - \frac{1}{\alpha^2}}}}$$ Taking $\alpha = 2$ we see that $$c_2 = \sqrt{\frac{2}{1 + \frac{\sqrt{3}}{2}}}= 2\sqrt{2-\sqrt{3}} \approx 1.03527...$$ is the optimal constant for this problem, this is because we achieve this value whenever we choose $|x| = \alpha|y|$ and $\varphi = \arcsin(1/\alpha)$. I end with noting that we could get a uniform bound for $|x| \geq |y|$ by taking limits as $\alpha \to 1$, which gives the bound $c_1 = \sqrt{2}$.