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AlonsoMcLaren
Nov11-11, 10:33 AM
Give me a function that is piecewise continuous but not piecewise smooth

arildno
Nov11-11, 10:43 AM
The Weierstrass function should fit the bill, in being everywhere continuous, but nowhere differentiable.

jmm
Nov13-11, 01:08 AM
|x| is defined piecewise as abs(x)={-x,x<0; x,x>0} which is continuous but not smooth at x=0. And is perhaps a little bit easier to visualize than a fractal :D

micromass
Nov13-11, 09:12 AM
|x| is defined piecewise as abs(x)={-x,x<0; x,x>0} which is continuous but not smooth at x=0. And is perhaps a little bit easier to visualize than a fractal :D

Yeah, but this IS piecewise smooth.

pwsnafu
Nov13-11, 05:20 PM
|x| is defined piecewise as abs(x)={-x,x<0; x,x>0} which is continuous but not smooth at x=0. And is perhaps a little bit easier to visualize than a fractal :D

jmm, the domain of of the function you wrote down is the reals minus 0. So it is irrelevant what the function is doing at the origin.

Further as micromass has noted, even if you define abs(0) = 0, the function will still be piecewise smooth. This is because you can split the domain into (-\infty,0) \cup [0,\infty), and so abs has one sided derivatives of all orders at x = 0.

lavinia
Nov13-11, 08:40 PM
The path of a particle following continuous Brownian motion is nowhere differentiable. You can think of this as a continuous random walk Many stochastic models for practical problems use continuous Brownian motion.

For a weird example, I was told that there is a continuous map of the sphere into the plane that is everywhere distance preserving. Such a map can not be differentiable anywhere.