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$$f(t,x)=\begin{cases}0,&& t\le0\\2t, && t>0 \ \wedge \ x<0\\ 2t-\frac{4x}{t}, && t>0 \ \wedge \ 0\le x\le t^2\\ -2t, && t>0 \ \wedge \ t^2<x\end{cases}$$

How can I show that $f$ is not locally Lipschitz continuous in its 2nd argument?

Locally Lipschitz:

$G\subseteq \mathbb R \times \mathbb R^n, f: G \to \mathbb R^n, (t,x)\mapsto f(t,x)$. $f$ is called locally Lipschitz with regard to $x$ iff for every point $a \in G$ there is a closed ball $\bar B_r(a)$ with radius $r$ around $a$ and a constant $L=L(a)>0$ such that

$|f(t,x)-f(t,y)|\le L|x-y|\ \ \ \forall (t,x),(t,y)\in \bar B_r(a) \cap G$

user657391
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1 Answers1

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Let $a=(0, 0)$. For any $L$, $r$, take $0<t\leq \min\{\frac{1}{L}, \frac r2, 1\}$, $x=\frac{t^2}{4}$, $y=\frac{3t^2}{4}$, we see $ |f(t, x)-f(t, y)|=2t$, while $|x-y|=\frac{t^2}{2}$, thus $\frac{|f(t, x)-f(t, y)|}{|x-y|}=\frac 4t >L$.

Yuval
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