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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.

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    $\begingroup$ 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 $y$. $\endgroup$ Commented Jan 24, 2020 at 1:34
  • $\begingroup$ @losif Yes, you are right, I changed my argument into locally Lipschitz. $\endgroup$ Commented Jan 24, 2020 at 2:04

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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

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  • $\begingroup$ Perfect, thank you! $\endgroup$ Commented Jan 24, 2020 at 3:24

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