# Banach space inequality

I'm looking to prove the following inequality

\begin{align} ||\frac{u}{||u||}-\frac{v}{||v||}|| \leq 2||u-v|| \end{align}

where $$u$$ and $$v$$ are elements of a Banach space such that $$||u||$$ and $$||v||$$ are greater than $$1$$.

I know that $$||\frac{u}{||u||}||=1$$ and have also tried using the triangle inequality.

• HAve you tried with the inverse triangle inequality? – Tito Eliatron Apr 11 at 18:43
• @TitoEliatron Thanks for the hint. The reverse triangle inequality implies that $2|| ||u||-||v|| || < 2||u-v||$, which looks promising. Not sure where to go from there. – Bartolo Colon Apr 11 at 19:05

Assume w.l.o.g that $$\|y\|\leq \|x\|$$. For convenience set $$\|\frac{x}{\|x\|}-\frac{y}{\|y\|}\|=[x,y]$$.
First consider the case when $$\|y\|\leq (1-\frac{1}{2}[x,y])\|x\|.$$ Then $$\|x\|\leq\|x-y\|+\|y\|\leq\|x-y\|+(1-\frac{1}{2}[x,y])\|x\|,$$ implying, along with our assumption, that $$[x,y]\leq2\|x-y\|/\|x\|\leq2\|x-y\|.$$ If $$\|y\|\geq (1-\frac{1}{2}[x,y])\|x\|$$, we have: \begin{align*} [x,y]\|x\|&=\|x-\frac{\|x\|y}{\|y\|}\|\\ & =||x-y+\frac{y}{\|y\|}(\|y\|-\|x\|)\|\\ &\leq \|x-y\|+\|y(\frac{\|x\|}{\|y\|}-1)\|\\ &=\|x-y\|+\|x\|-\|y\|\\ &\leq \|x-y\|+\|x\|+(\frac{1}{2}[x,y]-1)\|x\|. \end{align*} Rearranging this inequality and using the fact that $$\|x\|\geq 1$$ completes the proof.