TITLE: The function $f'+f'''$ has at least $3$ zeros on $[0,2\pi]$.
QUESTION [15 upvotes]: Show that if $f\in \mathcal{C}^3$ and $2\cdot\pi$ periodic then the function $f'+f'''$ has at least $3$ zeros on $[0,2\pi]$.

My attempt :
f is $2\pi$ periodic and $\mathcal{C}^3$, we have :
$$\lim_{h\rightarrow 0, h>0} \frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0, h>0} \frac{f(2\pi+h)-f(2\pi)}{h}\Rightarrow f'(0)=f'(2\pi)$$
After that I tried to compute differently the limit 
$$
\lim_{h\rightarrow 0, h>0}\frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0, h<0}\frac{f(h)-f(0)}{|h|}=\lim_{h\rightarrow 0, h<0}-\frac{f(2\pi+h)-f(0)}{h}\Rightarrow f'(0)=-f'(2\pi) 
$$
wich is clearly false because I get $f'(0)=0$ ( $f(x)=\sin(x)$ is a counterexample).
For $2$ zeros is relatively easy but I am stuck for the additional zero.
EDIT : I find this exercice (as usual) here : Revue de la filière Mathématique. This was asked during an oral examination of École normale supérieure rue d'Ulm.

REPLY [14 votes]: Assume that $g := f' + f'''$
has only finitely many zeros in ${\bf R} / 2\pi{\bf Z}$.
Then it has an even number of sign changes.  We show that there must be
more than $2$, and thus that there are at least $4$ sign changes
in each period interval, so a fortiori at least $4$ zeros.
Vladimir already noted that $\int_0^{2\pi} g(x) \, dx = 0$,
which implies at least two sign changes, and also
$$
\int_0^{2\pi} g(x) \sin x \, dx = 0,
\quad
\int_0^{2\pi} g(x) \cos x \, dx = 0.
$$
(This can be proved either by integration by parts,
as Vladimir suggested, or by Fourier expansion
as suggested by nbubis.)   Therefore
$$
\int_0^{2\pi} g(x) \, (A + B \sin x + C \sin x) = 0
$$
for all $A,B,C$.  But suppose that
$g$ had only two sign changes in each period,
say at $x_1$ and $x_2$.  Then we could find reals $A,B,C$
such that $t(x) = A + B \sin x + C \cos x$ has sign changes
at the same $x_1$ and $x_2$ and nowhere else.  Then
$g(x) t(x)$ is either everywhere $\geq 0$ or everywhere $\leq 0$,
but is not everywhere zero; this contradicts $\int_0^{2\pi} g(x) t(x) \, dx = 0$,
and we're done.
This is a known technique, used for instance to prove that
all the roots of an orthogonal polynomial are real.
To contruct $A,B,C$ we can consider the points
$(\sin x_1, \cos x_1)$ and $(\sin x_2, \cos x_2)$
on the circle $s^2 + t^2 = 1$, and join them by a line
$A+Bs+Ct = 0$, which meets the circle at those two points
and thus nowhere else.