Norm Inequality: Other

Question

I need help proving this inequality to understand a preliminary remark.

$$\Bigg |\Bigg| \dfrac{x}{\|x\|}-\dfrac{y}{\|y\|}\Bigg |\Bigg| \geq \|x-y\| $$
with $x$ and $y$ satisying $\|x\|,\|y\|\leq 1$ and $\|x-y\|\geq 1$

Answer 1

If $\ \| x\| \leqslant \| y \| \ $ , we write:

$\left\|\dfrac{x}{\| x\|}-\dfrac{y}{\| y\|}\right\| = \left\|\dfrac{x}{\| x\|}-\dfrac{y}{\| x\|} + \dfrac{y}{\| x\| } - \dfrac{y}{\| y\|}\right\|$

We use the triangle inequality to obtain:

$\left\|\dfrac{x}{\| x\|}-\dfrac{y}{\| y\|}\right\| \geq \left\|\dfrac{x}{\| x\|}-\dfrac{y}{\| x\|}\right\| - \left\| \dfrac{y}{\| x\| } - \dfrac{y}{\| y\|}\right\|$

Then:

$\left\|\dfrac{x}{\| x\|}-\dfrac{y}{\| y\|}\right\| \geq \dfrac{1}{\| x\|} \|x-y\| - \left( \dfrac{1}{\| x\|} -\dfrac{1}{\| y\| }\right) \| y\|$

We know that $\ \|y\| \leqslant 1 \leqslant \| x-y\| $

So:

$\left\|\dfrac{x}{\| x\|}-\dfrac{y}{\| y\|}\right\| \geq \dfrac{1}{\| x\|} \|x-y\| - \left( \dfrac{1}{\| x\|} -\dfrac{1}{\| y\| }\right) \| x-y\| = \dfrac{\|x-y\|}{\| y\|} \geqslant \|x-y\|$

Answer 2

Let $\|y\|\geq \|x\|$.

\begin{eqnarray*} \|x\|\|x-y\|&=&\|y\|\|x-y\|-(\|y\|-\|x\|)\|x-y\|\qquad {\rm (trivially)}\\\\ &\leq&\|y\|\|x-y\|-(\|y\|-\|x\|)\|y\|\qquad {\rm (as}\,\,\|y\|\leq1 \leq\|x-y\|)\\\\ &=&{\large\|}\|y\|(x-y){\large\|}-{\large\|}(\|y\|-\|x\|)y{\large\|}\qquad {\rm (by\,\, scalability\,\, of\,\, norm)}\\\\ &\leq&{\large\|}\|y\|x-\|x\|y{\large\|}\qquad {\rm (by\,\, triangle\,\, inequality\,\, }\|A\|-\|B\|\leq\|A+B\|) \end{eqnarray*}

Thus $$\|x-y\|\leq \frac{\|x\|\|x-y\|}{\|x\|\|y\|}\leq \left\| \frac{x}{\|x\|}-\frac{y}{\|y\|}\right\|$$