Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $[0,T]$.

share|improve this question
1  
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

1 Answer 1

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 $f$.

share|improve this answer
    
Thank you very much!!!! –  user24158 Feb 1 '12 at 15:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.