Proving that $y\mapsto \chi(y)f(x,y) $ is globally Lipschitz in variables $y$ with $f$ Locally Lipschitz

Question

Assume $f:[a, b] \times \Bbb R\to \Bbb R $ is continuous and satisfies the following condition: $$\color{red}{\forall \eta \in \Bbb R,~\exists \ \delta = \delta(\eta)~,L(\eta)~\forall \ y_1, y_2 \in [\eta - \delta, \eta + \delta]\\\implies \lvert f(x, y_1) - f(x, y_2) \rvert \leq L\lvert y_1 - y_2 \rvert}~~\forall ~~x\in [a,b]$$ for fixed $\eta\in\Bbb R$ we define $$\chi(y) =\begin{cases} 1&~~\text{if}~~|y-\eta| <\delta/2\\ 2-\frac{2}{\delta}|y-\eta|&~~\text{if}~~\delta/2\le |y-\eta| <\delta \\ 0&~~\text{if}~~|y-\eta|\ge \delta \end{cases}$$

Question: I would like to prove that, $[a, b] \times \Bbb R \ni (x,y)\mapsto \chi(y)f(x,y)$ is Lipschitz in $y$ variables.

Edit

I believe this may help: Is a Lipschitz function differentiable?

What are the $\eta$ and $\delta$ in the definition of $\chi$? — user251257, Oct 16 '17 at 18:25
$\eta$ is fixed and $\delta =\delta(\eta)$ exists such that, the condition in red holds true. — Guy Fsone, Oct 16 '17 at 18:34

user251257 · Answer 1 · 2017-10-16T19:36:16.623

0

I will drop the argument $x$, as the assumption is uniform in $x$ anyway.

Let $g=\chi f$ and $I=[\eta - \delta, \eta + \delta]$. It is sufficient to show that $g$ is Lipschitz in $I$. Notice for example, if $M$ is a Lipschitz constant of $g$, $y_1 \notin I, y_2 \in I, y_1 < y_2$, then $$ |g(y_1) - g(y_2)| = |g(\eta - \delta) + g(y_2) | \le M |\eta - \delta - y_2| \le M|y_1 - y_2|.$$

The short way: Using derivative a.e. and product rule, we obtain a.e. $$ |g’| \le \frac2\delta \sup_{y\in I} |f(y)| + 1 \cdot L < \infty.$$ Thus, $g$ is Lipschitz.

The long way: use the definition.

edited Oct 16 '17 at 19:36

answered Oct 16 '17 at 18:53

user251257

9,417

Why is $g$ differentiable? – Guy Fsone Oct 16 '17 at 19:01
@GuyFsone it isn’t, it is only differentiable almost everywhere (as it is a product of Lipschitz continuous functions). – user251257 Oct 16 '17 at 19:03
Now what happen when $x\in I$ and $y\not\in I$ ? – Guy Fsone Oct 16 '17 at 19:04
@GuyFsone $x$ is in $[a,b]$, not $I$. The assumption and statement is uniform in $x$. So just add $x$ into the proof. It won’t change anything. As for outside of $I$, the function $g$ is zero outside $I$, continuous, and Lipschitz on $I$. It is not hard to prove that $g$ is also Lipschitz on $\mathbb R$. – user251257 Oct 16 '17 at 19:09
Sorry sorry I meant What happens if have $y_1\in I$and $y_2 \not \in I$ – Guy Fsone Oct 16 '17 at 19:27
@GuyFsone I’ve extended the answer. – user251257 Oct 16 '17 at 19:36

Proving that $y\mapsto \chi(y)f(x,y) $ is globally Lipschitz in variables $y$ with $f$ Locally Lipschitz

Edit

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