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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?2012-02-01
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    Yes, the periodic extension is of class $C^3$ on $\mathbb(R)$.2012-02-01

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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

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 $f$.