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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.

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  • $\begingroup$ It is not clear what you say is false. The assertion is that $f'+f'''$ has at least three zeroes. The sine function certainly satisfies that, so I don't see a counterexample to what. $\endgroup$ – Martin Argerami Jun 9 '14 at 14:16
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    $\begingroup$ @clark: yes, that's because what he is doing is wrong (if anyone missed it, he arbitrarily included an absolute value that negates the equality). I fail to see the point of including it in the question. $\endgroup$ – Martin Argerami Jun 9 '14 at 14:41
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    $\begingroup$ Where have you found this problem? The context could give a clue about which techniques can be useful. $\endgroup$ – ajotatxe Jun 9 '14 at 16:39
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    $\begingroup$ As the problem stands. From periodicity is suffices to prove that there are $3$ roots for any interval of the form $[x_0, x_0 + 2\pi]$. The function $ G = f + f'' $ has $ G(0) = G(2 \pi)$ from Rolle's theorem we get $ G'(\xi) = 0$, since $G$ is periodic. Now $ G'( \xi) = G'(\xi + 2 \pi)$. Now applying Rolle's theorem in $ [\xi ,\xi + 2 \pi ]$ for $G$ we get another root $G'$ on $ (\xi ,\xi + 2 \pi )$ so in total $3$ roots for $ f' +f'''$ on $ [\xi ,\xi + 2 \pi ]$. $\endgroup$ – clark Jun 9 '14 at 17:56
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    $\begingroup$ @clark But when going from $[\xi,\xi+2\pi]$ to $[0,2\pi]$, both $\xi$ and $\xi+2\pi$ can get to the same point (that is $\xi$). $\endgroup$ – Luiz Cordeiro Jun 9 '14 at 18:08
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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.

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  • $\begingroup$ You're right; fixed now. Thanks. It wouldn't work with $f'-f'''$ (because $f(x) = \sin x$ would be a counterexample). $\endgroup$ – Noam D. Elkies Jun 14 '14 at 20:34
  • $\begingroup$ In fact, one can compute that $t(x)=\sin(x_2-x)+\sin(x-x_1)-\sin(x_2-x_1)$. My related, recent MSE question at math.stackexchange.com/questions/834321/… shows that $x_1$ and $x_2$ are the only zeroes $\endgroup$ – Ewan Delanoy Jun 14 '14 at 21:25
  • $\begingroup$ That's a nice formula. More generally, a "trigonometric polynomial" of degree at most $d$ (that is, a linear combination of terms $\sin(mx+a)$ and $\cos(nx+b)$ with $|m|,|n| \leq d$) is either identically zero or has at most $2d$ roots per $2\pi$ period, and any $2d$ points (with multiplicity) can arise. In this $d=1$ case the intersection of a line with the unit circle also gives a geometrical visualization. $$ $$ (BTW my previous comment acknowledges a typo correction from Daniel Fischer, who then chose to delete the comment that I responded to.) $\endgroup$ – Noam D. Elkies Jun 14 '14 at 23:06
  • $\begingroup$ @Noam It is customary to delete comments that point to a typo, usually to avoid noise and since they become irrelevant after the edit. $\endgroup$ – Pedro Tamaroff Jun 15 '14 at 4:59
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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).

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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.

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