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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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    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 ), 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 $f$.