How to estimate difference of two normed elements?

Question

Quick and hopefully easy to answer question: If I have an element $a\neq 0$ of a normed space that I can approximate up to $\varepsilon$ with respect to a certain norm by another element $b$, i.e., $||a-b||\leq \varepsilon$, is there an easy estimation for the difference of the normed versions of the two elements, i.e., is there a bound for $||\frac{a}{||a||}-\frac{b}{||b||}||$ in terms of $\varepsilon$?

Answer 1

If $a$ and $b$ are close to each other but their norms are small, multiplying them by the inverse of their norms can make the difference larger, for example: if $\|a\|=\|b\|=\delta$ then $$\left\|\frac a{\|a\|}-\frac b{\|b\|}\right\|=\frac{\|a-b\|}{\delta}\leq \frac\varepsilon\delta.$$ If $\delta$ is small then the value $\varepsilon/\delta$ can be much larger then $\varepsilon$.

Particularly, if $a$ and $b$ are close to $0$ then they can lie on the other sides of zero. This makes the situation even worse. For example, if $b=-a$ and $\|a\|=\varepsilon/2$ then $$ \left\|\frac a{\|a\|}-\frac b{\|b\|}\right\|=2. $$

This is easier to understand if we observe that $x\mapsto x/\|x\|$ is a projection on the unit sphere.

On the other hand, if we know that $a$ and $b$ are separated from zero, i.e. if $\|a\|,\|b\|\geq \delta>0$ then we can find a good estimate. I'll consider the case $\|a\|,\|b\|\geq 1$. It's well known that in Hilbert spaces the function $$\pi(x)=\begin{cases}x/\|x\|;& \|x\|\geq 1,\\ x;&\|x\|\leq 1\end{cases}$$ is a metric projection of the space onto the closed convex set, that is onto closed unit ball -- and therefore it's nonexpansing, that is $\|\pi(x)-\pi(y)\|\leq \|x-y\|$.