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I am interested in a relationship (if any) between the number of critical points of a periodic function $f$ of class $C^3([0,T])$ and the number of critical points of $f''$ in $[0,T]$.

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What do you mean by "of class $C^3([0,T])$"? Do you mean that the periodic extension should also be three times continuously differentiable? – Dirk Feb 1 '12 at 10:25
Yes, the periodic extension is of class $C^3$ on $\mathbb(R)$. – user24158 Feb 1 '12 at 12:30

Consider a $C^2$ function $F\colon\mathbb{R}\to\mathbb{R}$ periodic of period $T>0$ and assume that $F$ has $N$ distinct zeroes $\{x_1,\dots,x_N\}\subset[0,T]$. By Rolle's theorem, $F'$ has at least $N-1$ zeroes in $(0,T)$, one in each interval $(x_i,x_{i+1})$, $1\le i\le N$.

  • If $x_1=0$ (and hence $x_N=T$ ), then $F'$ may have exactly $N-1$ zeroes.
  • If $x_1>0$ (and hence $x_N<T$ ), then $F'$ has at least one zero between $x_N$ and $x_1+T$; call it $\xi$. Then either $\xi\in(x_N,T]$ or $\xi-T\in(0,x_1)$. Conclude that $F'$ has at least $N$ zeroes. If $\xi=T$, then $F'$ has at least $N+1$ zeroes.

Applying the above argument to $F'$ shows that $F''$ has at least $N-1$ zeroes. For $F''$ to have exactly $N-1$ zeroes it must be that $F(0)=F(T)=0$.

Returning to your original question, $f''$ has at least as many critical points as $f$, except when $0$ and $T$ are critical points, in which case $f''$ may have one less critical point than $f$.

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Thank you very much!!!! – user24158 Feb 1 '12 at 15:44

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