Given that [itex]x\left(t\right)[/itex] is locally absolutely continuous, and [itex]\dot{x}=f\left(x,t\right)[/itex] exists almost everywhere, is it possible to place a condition on [itex]f\left(x,t\right)[/itex] to allow us to show that [itex]x\left(t\right)[/itex] is uniformly continuous? Perhaps that [itex]\dot{x}=f\left(x,t\right)[/itex] is locally uniformly bounded in t? But I'm not sure how to prove this and have not been able to find any specific resources that point to this fact.(adsbygoogle = window.adsbygoogle || []).push({});

I understand that absolutely continuous functions are by nature uniformly continuous, and if [itex]x\left(t\right)[/itex] is differentiable everywhere and [itex]\dot{x}=f\left(x,t\right)[/itex] is bounded everywhere, the [itex]x\left(t\right)[/itex] is UC. However, I am at a loss as to how to extend UC to these measure variations.

Thanks for any help!

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# Condition for local absolute continuity to imply uniform continuity

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