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Let \begin{align} f(x,y)&=2x-4\frac{y}{x} \\ f(0,0)&=0, \end{align} for $|x|\leq 1$, $0\leq y\leq x^2$. Show that $f$ is not Lipschitz.

I calculated its derivative and it is unbounded, but I don't know hot to prove it by definition since every path I chose, like $(0,0)$ and $(x,x^2)$ is Lipschitz with constant $2$, thanks.

Silvia Ghinassi
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José
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  • I'm just an undergrad student so I may be in the wrong here, but we learned that $f$ is Lipschitz on $S \subset \mathbb{R}^2$ iff it's derivative (with respect to $x$ or $y$) is bounded on $S$. Having proved that the derivative is not bounded leads to the desired result. – implicati0n Sep 04 '15 at 15:23
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    @Toni1234 that is only true, if $f$ is differentiable (and $S$ is convex or at least connected). – user251257 Sep 04 '15 at 15:54

1 Answers1

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The function $f$ is Lipschitz continuous (w.r.t. to the Euclidean norm) on the domain $$ D=\{ (x,y) : |x| \le 1, 0\le y \le x^2 \} $$
if and only if there exists some $L$ such that for every $(x,y), (\tilde x, \tilde y)\in D$ it follows $$ |f(x,y) - f(\tilde x, \tilde y)| \le K \|(x,y) - (\tilde x, \tilde y)\|. $$ That is, the quotient $$ \frac{|f(x,y) - f(\tilde x, \tilde y)|}{\|(x,y) - (\tilde x, \tilde y)\|} $$ is bounded for $(x,y) \ne (\tilde x, \tilde y)$.

Let $x\in(0,1]$. Then, we have $$ \frac{|f(x,x^2) - f(x, 0)|}{\sqrt{|x-x|^2 + |x^2 - 0|^2}} = \frac{4x}{x^2} = \frac4x \to \infty$$ for $x\to 0$. So $f$ is not Lipschitz continuous.

user251257
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