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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.

tictac
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    What have you tried? – Chickenmancer Jun 19 '20 at 20:14
  • i differentiated the function in terms of $x_1 , x_2$ and got

    $fx = 36x / \sqrt { 36x_1^2 + 81x_2^2} $ and $fy = \dfrac{81y}{\sqrt{81y^2+36x^2}} $

    i'm swinging in the dark here. what i've usually done is i differentiated $f(x)$ and always had a value I could compare it with in order to get L. but in this exercise I don't know how to proceed

     – tictac Jun 19 '20 at 20:30
  • @tictac Good work; now can you try to bound the derivatives above? for instance, can you bound their denominator so you might get out $|x-y|$ as a coefficient? – Zim Jun 19 '20 at 20:48
  • sorry i'm really not seeing the light – tictac Jun 19 '20 at 21:04
  • @Zim should i try subtracting the two deravites from each other? $fx - fy$ ? – tictac Jun 19 '20 at 21:50
  • @tictac Not quite, so the derivative stuff is motivated by the fact that if $\sup_{(x_1,x_2)\in\mathbb{R}^2} |\nabla f(x_1,x_2)|$ exists, then it is a sharp Lipschitz constant, i.e. you've found $L$ and you're done. So finding an upper bound on the norm of your gradient is your next task -- i.e. you'll need to stack $[fx, fy]$ in a vector and see if you can bound it above. – Zim Jun 20 '20 at 14:49

2 Answers2

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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.

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

$$ 6\le L \le 9 $$

Cesareo
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