Question uniform vs local Lipschitz in arguments

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For Picard's existence and uniqueness theorem, my textbook states we need locally Lipschitz continuous with respect to the second argument and uniformly with respect to the first argument of $f(t,x)$. This is said to be equivalent to the result that for all compact subsets $V \subset U$ where $U \subseteq \mathbb{R}^{n + 1}$ is open we have $$ L := \sup_{(t,x)\neq(t,y) \in V \subset U} \frac{|f(t,x) - f(t,y)|}{|x - y|} < \infty $$

I understand that locally Lipschitz means there is a neighborhood around each point where $f$ is Lipschitz, but what does the uniform part mean with respect to $t$?

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From the first formulation you could get different constants for different times, so $L(t)$. In connecting the times in the Picard iteration and its bounds you want to use the same constant, so you need an uniform bound $L_*$ for $L(t)$. This bound itself is a constant for all the Lipschitz conditions, so you can also use that directly.

More generally one only needs that $L(t)$ is integrable to get a weight function $w(t)=\exp(-2|\int_{t_0}^tL(s)ds|)$ instead of $\exp(-2L|t|)$ for a norm that makes the Picard iteration contracting.