How to Derive the Inequality on Page 36 in the Proof of Lemma 11.3?

[Alone]

I am trying to see how to derive the following inequality on page 36 in the proof of Lemma 11.3: https://arxiv.org/pdf/math/0412040.pdf

I.e, of:
$$\| fg \|_{Lip} \le \bigg(1+\ell \sup_{t\in T} |g'(t)|\bigg)\sup_{t\in T}|f'(t)| , \ \ supp \ f(1-g)\subset S^c$$

My thoughts about how to show this, first let's write:
$$|f\cdot g (x)-f\cdot g(y)|/|x-y|$$
For $x\ne y$ and $x,y\in \mathbb{R}$; I think I should estimate the derivative of $f\cdot g$ which is $f'g+fg'$, but I don't see how exactly to use it here.

I hope this question is in the right level for this site.

Cheers!

 

[Opalg]

I am trying to see how to derive the following inequality on page 36 in the proof of Lemma 11.3: https://arxiv.org/pdf/math/0412040.pdf

I.e, of:
$$\| fg \|_{Lip} \le \bigg(1+\ell \sup_{t\in T} |g'(t)|\bigg)\sup_{t\in T}|f'(t)| , \ \ supp \ f(1-g)\subset S^c$$

My thoughts about how to show this, first let's write:
$$|f\cdot g (x)-f\cdot g(y)|/|x-y|$$
For $x\ne y$ and $x,y\in \mathbb{R}$; I think I should estimate the derivative of $f\cdot g$ which is $f'g+fg'$, but I don't see how exactly to use it here.

You are right, you need to estimate $\sup |f'(x)g(x)+f(x)g'(x)|$. You only need to deal with the case $x\in T$, because $g$ (and therefore also $g'$) vanishes outside $T$. So you need to know how large $|f|$ and $|g|$ can be in the interval $T$.

The authors seem to assume that $g$ is equal to $1$ on the inner interval $I$, and then smoothly decreases to $0$ in the region between $I$ and $T$. They don't actually say so, but I think they are assuming that $g$ always lies between $0$ and $1$, so that $\sup_{t\in T}|g(t)| = 1$.

They also assume ("without loss of generality", though I don't see why) that $f$ vanishes at some point of $T$. Say $f(s) = 0$, where $s$ is some point of $T$. Every other point of $T$ is within distance $\ell$ of $s$, and it follows (by the mean value theorem) that $|f(x)| \leqslant \ell\sup_{t\in T} |f'(t)|$ for all $x$ in $T$.

Therefore, for all $x$ in $T$, $$|f'(x)g(x)+f(x)g'(x)| \leqslant \sup_{t\in T}|f'(t)| \sup_{t\in T}|g(t)|+ \sup_{t\in T}|f(t)|\sup_{t\in T}|g'(t)| \leqslant \sup_{t\in T}|f'(t)|\cdot 1 + \ell\sup_{t\in T}|f'(t)| \sup_{t\in T}|g'(t)|,$$ which is what you wanted to prove.