3
$\begingroup$

This is a follow up to this answer that I came across while researching this.

As an example, I have a function $f: \mathbb{R}^2 \to \mathbb{R}$ that is Lipschitz continuous in both $x$ and $y$ separately, so if I fix $y$, I get

\begin{align} |f(x_1, y) - f(x_2, y)| &\leq L ||(x_1, y) - (x_2, y)|| \\ &= L|x_1 - x_2| \end{align}

Fixing $x$ gives me \begin{align} |f(x, y_1) - f(x, y_2)| &\leq L ||(x, y_1) - (x, y_1)|| \\ &= L|y_1 - y_2| \end{align}

I tried the telescoping sum \begin{align} |f(x_1, y_1) - f(x_2, y_1)| &= |f(x_1, y_1) - f(x_2, y_1) + f(x_2, y_1) - f(x_2, y_1)| \\ &\leq |f(x_1, y_1) - f(x_2, y_1)| + |f(x_2, y_1) - f(x_2, y_1)| \\ &\leq L|x_1 - x_2| + L|y_1 - y_2| \end{align}

but I'm not sure where to go from there because $||(x_1, y_1) - (x_2, y_2)|| \leq |x_1 - x_2| + |y_1 - y_2|$, so I can't automatically conclude from the previous step that $|f(x_1, y_1) - f(x_2, y_1)| \leq L||(x_1, y_1) - (x_2, y_2)||$, which is what I need.

Did I miss something?

$\endgroup$
  • 2
    $\begingroup$ Note $|x_1-x_2| \le \|(x_1,y_1)-(x_2,y_2)\|,$ etc. $\endgroup$ – zhw. Apr 27 '15 at 0:33
  • $\begingroup$ @zhw. I can't believe I missed that. So $f$ is Lipschitz and $L = 2M$. Thank you! $\endgroup$ – M T Apr 27 '15 at 0:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.