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Nov 14, 2011 at 23:03 comment added Benoit Jubin A simple example showing that "$f \in C^{1,\infty}(\mathbb{R} \times M,M)$" does not imply "$\partial_t f(0,-) \in C^1(M,M)$" is given by taking $M=\mathbb{R}$ and $f(t,x)=t\sqrt{x^2+t^2}$.
Nov 8, 2011 at 16:25 history edited CuriousUser CC BY-SA 3.0
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Nov 8, 2011 at 16:20 history edited CuriousUser CC BY-SA 3.0
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Nov 8, 2011 at 15:30 comment added Paul Siegel Ah, I think I understand now: you're asking if the flow $\phi_X(x,t)$ can have strictly more regularity in $x$ than it does in $t$. Certainly the flow is at least as differentiable as $X$ in both variables, but I can't see any reason why $\phi_X$ couldn't accidentally acquire more smoothness in $x$ than it deserves. Nor can I come up with any useful examples. So I'll politely bow out of the conversation. :)
Nov 8, 2011 at 15:30 answer added Pietro Majer timeline score: 2
Nov 8, 2011 at 14:00 comment added CuriousUser Hi Paul, I'm not sure if I got what you said correctly. Because, as the "$\partial_tf(x,y)|_{t=0}$ is $C^k$ if $f$ is $C^k$ in $x$" is concerned, this is false, at least as I stated it. In fact, just take a function $f(x,t)$ smooth in $x$ and only $C^2$ in $t$, for example $f(x,t)=x+|t|\cdot t\cdot e_1$, or something like that. The point is that such an $f(x,t)$ doesn't come from the flow of a vector field
Nov 8, 2011 at 12:18 comment added Paul Siegel I'll just throw this out there and someone can shoot it down if I'm being stupid... but $X(\phi_X^0)$ is the $t$ derivative of $\phi_X^t$ at $t = 0$, so working in local coordinates I think you are asking if $\partial_t f(x,t)|_{t=0}$ is $C^k$ if $f$ is $C^k$ in $x$. In this case the answer is yes.
Nov 8, 2011 at 11:24 history edited CuriousUser CC BY-SA 3.0
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Nov 8, 2011 at 10:40 history asked CuriousUser CC BY-SA 3.0