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Locally Lipschitz with respect to a variable uniformly to another implies Lipschitz for every compact subset

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Let $$f:A\subset{\mathbb{R}^{n+1}}\to{\mathbb{R}^{n}}$$$$(t,\mathbf{y})\to{f(t,\mathbf{y})}$$ with $A$ open, $t\in{\mathbb{R}}$ and $\mathbf{y}\in{\mathbb{R}^{n}}$.

The function $f$ is said to be locally Lipschitz in A with respect to $\mathbf{y}$ uniformly in $t$, if for every $(t,\mathbf{y})\in{A}$ there is an open neighbourhood $U$ of $(t,\mathbf{y})$ and a number $L\in{\mathbb{R}}$ such that $$(s,\mathbf{x}),(s,\mathbf{z})\in{U}\implies{|f(s,\mathbf{x})-(s,\mathbf{z})|\le{L|\mathbf{x}-\mathbf{z}|}}$$

The function $f$ is said to be Lipschitz in A with respect to $\mathbf{y}$ uniformly in $t$, if there is $L\in{\mathbb{R}}$ such that $$(t,\mathbf{x}),(t,\mathbf{z})\in{A}\implies{|f(t,\mathbf{x})-(t,\mathbf{z})|\le{L|\mathbf{x}-\mathbf{z}|}}$$

Prove the following:

If $f$ is locally Lipschitz in $A$ with respect to $\mathbf{y}$ uniformly in $t$, then for every compact subset $K\subset{A}$, $f$ is Lipschitz in K with respect to $\mathbf{y}$ uniformly in $t$.

I know and understand the proof of this fact when we consider "normal" Lipschitz continuity -i.e. for $$f:A\subset{\mathbb{R}^n}\to{\mathbb{R}^m}$$$$|f(x)-f(y)|\le{L|x-y|}$$ and there are also some question about that on this site. What I don't understand is how to extend the proof when we require the Lipschitz condition to hold with respect to a variabile uniformly to the other.

In my understanding, the proof of the proposition for the "normal" case relies on the fact that the local Lipschitz condition implies continuity everywhere and that the continous image of a compact set is also compact.

In this case however, the local Lipschitz condition does not imply continuity but only continuity with respect to $\mathbf{y}$ so we can not use the same line of reasoning.

Any suggestions?


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