Finding the smallest lipschitz constant

Question

Given is $f(x_1,x_2) = \sqrt {(6x_1)^2 + (9x_2)^2}$

Calculate the smallest Lipschitz constant: $L > 0$

so that: $ |f(x)-f(y)| \leq L||x-y|| $

for all $x,y\in {\Bbb R}^2$ and $||.||$ being the Euclidean norm.

So far the exercises I've been doing have only required me to calculate Lipschitz constants in $f(x)$ and column sum norms. This is why I am totally lost here and would appreciate any help. Thank you.

Answer 1

For shorthand, I will use the notation ${\bf x}=(x_1,x_2)$ and ${\bf y}=(y_1,y_2)$. Let $L\colon\mathbb{R}^2\to\mathbb{R}^2\colon(x_1,x_2)\mapsto(6x_1,9x_2).$ Then $L$ is a bounded linear operator with norm $\|L\|=\|(6,9)\|=\sqrt{36+81}\approx 10.8$. Then $f$ is the composition of the Euclidean norm with $L$, i.e. $f(x,y)=\|L(x,y)\|$. By the reverse triangle inequality and properties of norms,

$$|f({\bf x})-f({\bf y})|=\left|\|L{\bf x}\|-\|L{\bf y}\|\right|\leq \|Lx-Ly\|\leq \|L\|\|x-y\|.$$

You could probably also get another bound using the fact that $\sup_{(x_1,x_2)\in\mathbb{R}^2} \|\nabla f(x_1,x_2)\|$ is a sharp Lipschitz constant.

Answer 2

$$ f(x_1,y_1) = f(x_0,y_0) + f_{x_0}(x_1-x_0)+f_{y_0}(y_1-y_0)+O((x_1-x_0)^2+(y_1-y_0)^2) $$ so

$$ |f(x_1,y_1)- f(x_0,y_0)|\le \sqrt{f_{x_0}^2+f_{y_0}^2}\sqrt{(x_1-x_0)^2+(y_1-y_0)^2} $$

here

$$ \sqrt{f_{x}^2+f_{y}^2} = \sqrt{\frac{9 \left(16 x^2+81 y^2\right)}{4 x^2+9 y^2}} = L $$

hence

$$ 6\le L \le 9 $$