The function $f'+f'''$ has at least $3$ zeros on $[0,2\pi]$.

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.

Solution 1:

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.

Solution 2:

I guess we should start from this: Set $g=f'+f'''$. Then (integration by parts) $$ \int_0^{2\pi} g(x) dx=0,\quad \int_0^{2\pi} g(x)\sin x dx=0,\quad \int_0^{2\pi} g(x)\cos x dx=0,\quad $$ The first relation implies that $g$ (if not identically zero) takes both positive and negative values and hence has at least two zeros. The second and third relation should somehow imply that there are at least two more zeros. (This is yet more of a comment, but I posted it here because it is too large).

Solution 3:

A possible direction:

Write out the Fourier series: $$f(x)= a_0 + \sum\left( a_n \cos nx + b_n \sin nx\right)$$ $$g(x)\equiv f'(x)+f'''(x) = -\sum n (n^2-1) \left(b_n \cos (n x)-a_n \sin (n x)\right)$$

Clearly, when $n=1$, $g(x)=0$, meaning that the lowest order in $g(x)$ is $n=2$. Now, since there is no constant term, by the mean value theorem $g(x)$ must cross zero at at least one point, $x_0$.

Intuitively, any function with no frequencies lower than two, must cross zero at least $4$ times in the interval $[0,2\pi]$, though I'm not sure how to complete this proof.