Let $f:\mathbb{R}\times\mathbb{R}\longrightarrow\mathbb{R}$ with $(t,x)\mapsto t|x|^{a}$ be given. Let $a\in[0,\infty)$.
- Prove that for $a=1$ $f$ is not Lipschitz continuous (wrt 2nd argument)
- Find $a$ for which $f$ is Lipschitz continuous (wrt 2nd argument)
- Find $a$ for which $f$ is locally Lipschitz continuous (wrt 2nd argument)
My answers so far:
Let $t\in\mathbb{R}$ and $x,y\in\mathbb{R}$ be arbitrary. Then the following holds by the reversed triangle inequality: $|f(t,y)-f(t,x)|=|\,t|y|-t|x|\,|\leq |t|\,|y-x|\overset{t\rightarrow\infty}{\longrightarrow}\infty$. So there is no constant $K>0$ such that for all $t\in\mathbb{R}$ and $x,y\in\mathbb{R}$ we have $|f(t,y)-f(t,x)|=|\,t|y|-t|x|\,|\leq\,K|y-x|$. We thus conclude that $f$ is not Lipschitz continuous.
The only choice is $a=0$. Then $f$ is linear and thus Lipschitz continuous with constant $K=2$ for example.
I think that $a\geq\,1$ gives a locally Lipschitz continuous function. For $a<1$ we get a problem at $x=0$.
My questions:
- Is this proof correct?
- Is it correct that this is the only choice?
- Is my choice correct? And how can I prove my choice?
Thanks in advance!