L1 norm of (L1 normalized vector minus original vector) less than or equal to 1 minus the L1 norm of the original vector

Question

I am reading one paper (https://arxiv.org/pdf/2208.09407.pdf). There is one step in the proof of Proposition B.1 that I can't understand. To simplify the notations, I restate the equation as follows where $x,y \in [0, 1]^d$ for some dimension $d$:

$$ ||\frac{x}{||x||_1} - y||_1 \leq \big(\frac{1}{||x||_{1}} - 1\big)||x||_1 + ||x-y||_1 \le 1 - ||x||_1 + ||x-y||_1 $$

The authors said the above step is by triangle inequality, which I believe is by $$ ||\frac{x}{||x||_1} - y||_1 = ||\frac{x}{||x||_1} -x + x - y||_1 \le ||\frac{x}{||x||_1} -x||_1 + ||x - y||_1 $$

However, it seems the authors also utilized $$ ||\frac{x}{||x||_1} -x||_1 \le \big(\frac{1}{||x||_{1}} - 1\big)||x||_1 $$ which I am not sure why. Is the above inequality always true for any vector within range $[0,1]^d$?

Answer 1

$ ||\frac{x}{||x||_1} - y||_1 \leq \big(\frac{1}{||x||_{1}} - 1\big)||x||_1 + ||x-y||_1$ is false when $x=y$ and $\|x||_1=2$.

Everything looks fine when $\|x\|_1 \leq 1$. In that case, $ ||\frac{x}{||x||_1} -x||_1 \le \big(\frac{1}{||x||_{1}} - 1\big)||x||_1 $ follows from the fact that $\frac{x}{||x||_1} -x=(\frac 1 {\|x\|_1}-1) x$ and $\frac{1}{||x||_{1}} - 1 \geq 0$.

Answer 2

They seem to have taken the following steps. \begin{align*} ||\frac{x}{||x||_1} -x||_1 = ||(\frac{1}{||x||_1} -1)x||_1 = \big|\frac{1}{||x||_{1}} - 1\big| ||x||_1 \end{align*} but as @MPW already mentioned I can not justify why they have normal brackets around the $\frac{1}{||x||_{1}} - 1$ unless $||x||_1 \leq 1$