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Let a function $f: I\times J \rightarrow \mathbb R$, where $I,J$ are intervals in $\mathbb R$, be Lipschitz with respect of each variable separately. Is it then $f$ continuous with respect of both variables?

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1 Answer 1

It depends on what you mean with Lipschitz with respect to each variable separately.

Consider the function $f\colon\mathbb R^2\to \mathbb R$, $(x,y)\mapsto xy$. Then for fixed $y$, the function $x\mapsto f(x,y)$ is Lipschitz (with Lipschitz constant $y$ depending on $y$) and similarly $y\mapsto f(x,y)$ for fixed $x$. However $f$ itself is not Lipschitz as $f(t,t)=t^2$ shows.

On the other hand, if there is a single Lipschitz constant $L$ working for all functions of the form $x\mapsto f(x,y)$ and $y\mapsto f(y,x)$, then $f$ is Lipschitz with constant $L\sqrt 2$.

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Yes, by Lipschitz with respect to each variable I understand that for each fixed $y \in J$ the function $f(\cdot,y)$ is Lipschitz and for each $x\in I$ the function $f(x,\cdot)$ is continuous. But the function $f(x,y)=xy$ is still continuous as a function of $x,y$. –  Richard Oct 20 '12 at 10:16

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