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I'm a little bit confused in piecewise continuity of a function. Say, if we have an odd function like $f(x) = x$ defined on the open interval $(0, \pi)$. We then extend it to a period $2\pi$ function and find its sine Fourier series. Can we say that this function is then piecewise continuous, but not piecewise $\mathscr C^1$? Or is it piecewise $\mathscr C^1$?

Would a constant odd function like $f(x)=C$ be piecewise $\mathscr C^1$?

Thanks for helping me clear my gaps.

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This function is piecewise continuous on $\mathbb R$, and it is $\mathscr C^1$ on the continuous parts.

Therefore we can say it is piecewise $\mathscr C^1$ on $\mathbb R$

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  • $\begingroup$ Sorry, which one - f(x) = C or f(x) = x? $\endgroup$ – sequence Jul 23 '15 at 7:42
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    $\begingroup$ This remains valid for any piecewise continuous function : $x \rightarrow x$ and $x \rightarrow C$ are $\mathscr C^{\infty}$ on $\mathbb R$ thus $\mathscr C^1$. Your periodic piecewise continuous function is piecewise $\mathscr C^1$ on $\mathbb R$ $\endgroup$ – BusyAnt Jul 23 '15 at 8:20
  • $\begingroup$ Can we say that if a function is piece $C^1$, it is not necessarily $C^1$ on $\mathbb{R}$, but it can be piecewise $C^1$ on $\mathbb{R}$? $\endgroup$ – sequence Jul 26 '15 at 2:38
  • $\begingroup$ Also, if a function is piecewise continuous, does this imply that it's piecewise $C^1$? $\endgroup$ – sequence Jul 26 '15 at 2:39
  • $\begingroup$ Piecewise continuity is enough to declare that a function is piecewise $\mathscr C^1$ iff the pieces are $\mathscr C^1$ $\endgroup$ – BusyAnt Jul 27 '15 at 5:33
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It is piecewise continuous and piecewise $C^1$. To be derivable at $x$, you must be continuous at $x$ first, so to be piecewise $C^1$, you just need to be piecewise $C^0$ over those same pieces.

A note on what might confuse you: oftentimes in geometry/topology, we work with piecewise $C^1$ paths $[0,1] \to X$. But a path is (among other things) required to be continuous. Thus a map $[0,1] \to X$ which is piecewise $C^1$ but not globally $C^0$ is not a piecewise $C^1$ path because it is not a path at all.

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