Let $f\in C^0(I\times \mathbb{R})$,$I=[\xi,\xi+a]$,$a>0$, and suppose that $\max_{x\in I}|f(x,\eta)|=M<\infty$ for some given $\eta\in\mathbb{R}$. Also, suppose $f$ is differentiable with respect to $y$ with $|f_y(x,y)|\leq K|f(x,y)|$ uniformly on $x\in I$. Then for any given $x\in I$, $f(x,\cdot)\in C^1(\mathbb{R})$ is locally Lipshitz, i.e. $$\exists \delta_x>0 \exists L_x\geq 0,(y_1,y_2\in[\eta-\epsilon_x,\eta+\epsilon_x]\rightarrow\\\forall(y_1,y_2)\in\mathbb{R}^2(|f(x,y_1)-f(x,y_2)|\leq L_x|y_1-y_2|)).$$

The question is ,

Is such $L_x$ continuous with respect to $x$?

and I suppose such $L_x$ is taken as a infimum in kind that are possible.

  • 1
    $\begingroup$ The function $f$ given by $f(x,y)=e^y$ satisfies all the conditions (say with $\xi=0,a=1,\eta=0,M=1,K=1$), but is not Lipschitz in $y$. $\endgroup$ Jan 24 '20 at 1:34
  • $\begingroup$ @losif Yes, you are right, I changed my argument into locally Lipschitz. $\endgroup$ Jan 24 '20 at 2:04

The answer is no.

E.g., let $\xi:=-1$ and $a:=2$, so that $I=[-1,1]$. Let $f(x):=1+y(1-|y|/x)^2\,1_{x>0,\,|y|<x}$ for $x\in I$ and real $y$. Let $\eta:=0$, so that $M=0$. Let $K=1$. Then all your conditions on $f$ hold.

Yet, $L_x=1_{x>0}$ is not continuous in $x\in I$.

Here is the graph $\{(x,y,f(x,y))\colon -0.2\le x\le1,\,-1.2\le y\le1.2\}$:

enter image description here

  • $\begingroup$ Perfect, thank you! $\endgroup$ Jan 24 '20 at 3:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.