Picard-Lindelöf and the choice of the metric So the theorem of Picard-Lindelöf ensures that an ordinary differential equation $\frac{dy}{dt}=f(y)$ has a unique solution if $f$ is Lipschitz-continuous. The general form of Lipschitz continuity is given if $\forall x_1,x_2 \in X : d_Y(f(x_1),f(x_2)) \le L \cdot d_X(x_1,x_2)$ with any two metric spaces $(X,d_X)$ and $(Y,d_Y)$.
But what if we have a function $f$ that is Lipschitz-continuous for some metrics $d_X,d_Y$ but not for others? I currently can neither prove nor disprove that such a function might exist, so I am assuming that PL should be applicable even in that case.
I could imagine the following situations:

*

*This is no problem at all because PL only provides a sufficient condition for existence and convergence. So essentially providing any pair of metrics for which $f$ is Lipschitz is enough.

*The solution of the ODE actually somehow depends on the metric and I currently fail to see how.

*Picard-Lindelöff actually depends on some specific subset of metrics for which $f$ must be Lipschitz, but this is sometimes omitted when talking about PL.

*This situation is impossible, that is there is some proof that a function that is Lipschitz under one pair of metrics is Lipschitz under all pairs of metrics.

Which of these four situations is the correct one? I am currently tending towards 1, because this would imply very weak conditions (just find any pair of metrics that make it work) for PL.
 A: Consider the function $f(x) = x$ from $\mathbb{R}$ to $\mathbb{R}$. This function is trivially Lipschitz continuous with respect to the usual metrics on both copies of $\mathbb{R}$ with Lipschitz constant $L = 1$. But we can replace the metric $d_X$ on the domain copy of $\mathbb{R}$ with a topologically equivalent metric (in the sense that it induces the same topology) which is bounded, for example $\text{min}(d_X, 1)$. Now $f(x)$ is no longer Lipschitz continuous because $d_X \le 1$ but $d_Y$ can still get arbitrarily large.
This argument shows more generally that a surjective function from a bounded metric space to an unbounded metric space can't be Lipschitz. So Lipschitz continuity absolutely depends on the choice of metrics, even if we only limit ourselves to topologically equivalent metrics.
For Picard-Lindelof the implicit choice of metric, on both copies of $\mathbb{R}^n$, is the Euclidean metric. It could be replaced with any metric coming from a norm, e.g. any $\ell^p$ norm, because these are all Lipschitz equivalent and so have the same Lipschitz functions. If you look at the proof you'll see that the place where the Lipschitz condition is used is to prove that a certain operator called Picard iteration is a contraction in order to apply the Banach fixed point theorem. Over the course of the proof the fact that the Euclidean metric is induced by a norm is used constantly. Perhaps the proof could be rewritten to not depend on this assumption but it seems like a lot of work for an unclear payoff.
So the situation is #3. I disagree with Didier in the comments who seems to be ignoring metrics not induced by a norm.
