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I am stuck on the following problem:

Let $f:\mathbb{R} \to \mathbb{R}$ be continuous and $2\pi$-periodic and $n$ be a positive integer. If for any integer $p \in [0,n-1]$, $$\int_{0}^{2\pi} f(t) \cos(pt)\,\mathrm{d}t=\int_{0}^{2\pi} f(t) \sin(pt)\,\mathrm{d}t=0,$$
then is it true that $f$ has at least $2n$ roots on $[0,2\pi]$?

I tried to prove the problem using induction. $p=0$ is easy , but $p=1$ is giving me a hard time.

Edit: After the comment of Dear Dunham , it's seems that we need both integral to be equal to zero , in other to work , in that case we have the form : $$ \int_{0}^{2\pi} f(t) e^{ipt} \mathrm{d}t=0$$ , Now it looks like the trigonometric version of Prove that $f$ has $m+1$ zeros if $\int_{a}^{b} x^nf(x)dx=0$ for all $n\le m$

Edit: I made some progress with the new version of the question in the case $f$ is not identicaly $0$ , and proved $M \geq n$ , where $M$ is the number of zeros of $f$ as follow :
Let $(T_{n})$ the sequence of Tchebychev polynomial , $\cos(nt)=T_{n}(\cos(t))$ and $\deg(T_{n})=n$ . hence $(T_{0},...,T_{n-1})$ is a basis of $R_{n-1}[X]$ ,$0\leq a_{1}<a_{2}<... <a_{m}\leq \pi \leq a_{m+1}<..<a_{p}$ the zeros of $f$ and suppose that $p\leq n-1$ .
For $i \in [1,m-1]$ such that $a_{i}<x<a_{i+1}$ since cosinus is decreasing in $[0,\pi[$ then $b_{i+1}=\cos(a_{i+1}) < \cos(x) < b_{i}=\cos(a_{i})$ ,
on the other hand, for $j \in [m ,p-1]$ we have $c_{j}=\cos(a_{j}) < \cos(x) < c_{j+1}=\cos(a_{j+1})$ since here cosine is increasing in $[\pi ,2 \pi]$. We have then $(d_{n})$ constructed from $(b_{n})$ and $(c_{n})$ such that : $\forall x \in [0,2\pi]-J : f(x) \Pi_{i=1}^{p} (\cos(x)-d_{i}) > 0$ where $J=\{a_{1},a_{2},....,a_{p}, d_{1},...,d_{p}\}$ , the polynomial $P(x)=\Pi_{i=1}^{p}(x-d_{i})$ is of degree at most $n-1$ , hence we have $(t_{i})$ such that: $P(\cos(t))=\sum_{i=1}^{n-1} t_{i}T_{i}(\cos(t))$ by linearity of the integral, we conclude that : $\int_{0}^{2\pi} f(t)P(t)=0$ hence $f(x)=0$ For any $x \in [0,2\pi]-J$ , hence by continuity $f=0$ every where, contradiction .

I feel that I can improve this solution , by a better choice of the sequence of polynomial , one that also include the orthoganility with the familly $\sin(pt)$ $p \in [[0,n-1]]$ this information is critical to improve the bound , any help ? thank you a lot .

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  • $\begingroup$ I probably do not understand your statement correctly. Your induction should go on $n$, right? For $n=1$ (then you only have $p=0$) you should, according to the statement have at least $2$ zeros. But the only thing you get from the condition is $\int_0^{2\pi}f(t)\,dt=0$ which enforces only one zero. What do I miss? $\endgroup$
    – mickep
    Commented Feb 14, 2018 at 19:34
  • $\begingroup$ @mickep , thank your interest , here is the details of my argument for $n=1$ , because $f$ is $2\pi$ periodic and continuous $f$ must vanish at least $2$ times ; suppose that there is only one $u \in [0, 2\pi[$ such that $f(u)=0$ , then $f$ have different signe in $[0,u[$ and $[u,2\pi]$ by continuity ; then Wlog $f(0)> 0$ and $f(2\pi) < 0$ but this a contradiction with the fact that $f(0)=f(2\pi)$ $\endgroup$
    – uvw
    Commented Feb 15, 2018 at 3:50
  • $\begingroup$ I see. Did not think about the periodicity. You should probably add that $f$ is continuous, though. $\endgroup$
    – mickep
    Commented Feb 15, 2018 at 5:30
  • $\begingroup$ oh yes i am sorry , didn't notice that I missed it . I edited thank you $\endgroup$
    – uvw
    Commented Feb 15, 2018 at 5:43
  • $\begingroup$ Similar one math.stackexchange.com/questions/2650153/… $\endgroup$
    – rtybase
    Commented Feb 17, 2018 at 17:42

2 Answers 2

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original problem

If $f(t)=\cos(t)+\sin(t)$ then there are only 2 zeros, yet the conditions are true for any value of $n$.

Revised problem: proof sketch

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be continuous and $2\pi$ periodic with $M\in \mathbb{N}$ zeros. Also, suppose $\int f(t)e^{ipt}dt=0$ for $|p|<n$.

Define $f_k$ to be the $k$th zero mean antiderivative of $f$.

Since $f_1$ has $M$ critical values, $f_1$ has at most $M$ zeros. Similarly, $f_k$ has at most $M$ zeros for all $k$.

Now, $f_1$ has a convergent Fourier series \begin{equation} f_1(t) = \sum_{\ell\geq n} a_\ell \cos(\ell t) +b_\ell \sin(\ell t) \end{equation}

Then $f_{4L+1}$ has Fourier series \begin{equation} f_{4L+1}(t) = \sum_{\ell\geq n} \frac{a_\ell}{\ell^{4L}} \cos(\ell t) +\frac{b_\ell}{\ell^{4L}} \sin(\ell t) \end{equation}

For $L$ sufficiently large, all terms except the first \begin{equation} \frac{a_n}{n^{4L}} \cos(n t) +\frac{b_n}{n^{4L}} \sin(n t) \end{equation} are negligible. This first term has $2n$ zeros, so $f_{4L+1}$ has at least $2n$ zeros. Thus $M\geq 2n$.

Definition:

Given a continuous $2\pi$-periodic function $g:\mathbb{R}\rightarrow \mathbb{R}$ with mean 0, we can find an antiderivative \begin{equation*} \widetilde G(x) = \int_0^x g(t)dt. \end{equation*} Then the mean zero antiderivative of $g$ is \begin{equation*} G(x) = \widetilde G(x) - \int_0^{2\pi} \widetilde{G}(t)dt. \end{equation*} By the $k$th mean zero antiderivative, we mean $f_1$ is the mean zero antiderivative of $f$, $f_2$ is the mean zero antiderivative of $f_1$, and so on.

Proposition: In the series, \begin{equation} f_{4L+1}(t) = \sum_{\ell\geq n} \frac{a_\ell}{\ell^{4L}} \cos(\ell t) +\frac{b_\ell}{\ell^{4L}} \sin(\ell t) \end{equation} for sufficiently large $L$, all terms except the first are neglibible.

Proof: WLOG assume $a_n$ or $b_n$ is nonzero. Note that \begin{equation} a_n \cos(n t) +b_n \sin(n t) \end{equation} has $2n$ zeros and $2n$ extrema. The absolute value of the extrema is $\sqrt{|a_n|^2+|b_n|^2}$. Therefore, we shall make $L$ large enough so that the sum of the other terms is at most $\frac{\sqrt{|a_n|^2+|b_n|^2}}{2n^{4L}}$. We have $|a_{\ell}|\le \frac1{\sqrt{2\pi}}\int_0^{2\pi}|\cos(\ell x)f(x)|dx\le \frac1{\sqrt{2\pi}}\int_0^{2\pi}|f(x)|dx$ and the same bound for $|b_{\ell}|$, $\forall\ell$. The terms are bounded as follows \begin{align*} \left|\sum_{\ell\geq n+1} \frac{a_\ell}{\ell^{4L}} \cos(\ell t) +\frac{b_\ell}{\ell^{4L}} \sin(\ell t)\right| &\leq 2\max_{\ell>n }\{|a_\ell|,|b_\ell|\} \sum_{\ell\geq n+1} \frac{1}{\ell^{4L}} \\ &\leq 2\max_{\ell>n }\{|a_\ell|,|b_\ell|\} + \left( \frac{1}{(n+1)^{4L}} + \sum_{\ell\geq n+2} \frac{1}{\ell^{4L}} \right)\\ &\leq 2\max_{\ell>n }\{|a_\ell|,|b_\ell|\} \left( \frac{1}{(n+1)^{4L}} + \int_{n+1}^{\infty} x^{-4L} \right)\\ &= 2\max_{\ell>n }\{|a_\ell|,|b_\ell|\} \left( \frac{1}{(n+1)^{4L}} + \frac{1}{(4L-1)(n+1)^{4L-1}} \right)\\ &\leq 4\max_{\ell>n }\{|a_\ell|,|b_\ell|\} \frac{1}{(n+1)^{4L-1}} \end{align*} Now, we just have to show that \begin{equation*} 4\max_{\ell>n }\{|a_\ell|,|b_\ell|\} \frac{1}{(n+1)^{4L-1}} \leq \frac{\sqrt{|a_n|^2+|b_n|^2}}{2n^{4L}} \end{equation*} or equivalently \begin{equation*} \frac{8n\max_{\ell>n } \{|a_\ell|,|b_\ell|\}} {\sqrt{|a_n|^2+|b_n|^2}} \leq \left( \frac{n+1}{n} \right)^{4L-1} \end{equation*} The left-hand side is constant, while the right-hand side diverges to $\infty$ as $L\rightarrow \infty$. Hence the result is true for some $L$.

Proposition: Suppose $f$ has $M\in \mathbb{N}$ zeros. Then $f_k$ cannot have more zeros than $f$ for all $k$.

Proof: Consider $f_1$ as a differentiable function on the circle $\mathbb{T}$. The derivative of $f_1$ is $f$, which has $M\in \mathbb{N}$ zeros. Since $f$ is continuous, the derivative of $f_1$ is defined everywhere, so the only critical values of $f_1$ are the zeros of $f$. Let $t_0$ and $t_1$ be any two consecutive zeros of $f$. Then $f_1$ is either strictly increasing or strictly decreasing on $[t_0,t_1]$. Hence $f_1$ has at most one zero on this interval. There are $M$ pairs of consecutive zeros of $f$. Hence $f_1$ has at most $M$ zeros. A similar argument applies to $f_2$ and so on.

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    $\begingroup$ you are right thank you a lot , it's seems that we need both integral to be equal to zero . $\endgroup$
    – uvw
    Commented Feb 17, 2018 at 20:40
  • $\begingroup$ Dear @Dunham , do you mean by the k-th zero mean antiderivative , $f_{k+1}(x)=\int_{0}^{x} f_{k}(x) dx$ ? $\endgroup$
    – uvw
    Commented Feb 22, 2018 at 5:43
  • $\begingroup$ @ oty I added a comment to clarify $\endgroup$
    – Dunham
    Commented Feb 22, 2018 at 17:15
  • $\begingroup$ "Since $f_1$ has $M$ critical values, $f_1$ has at most $M$ zeros." This is false. $f_1$ has at most $M+1$ zeros. So it does not follow that $M\ge 2n$. $\endgroup$
    – Hans
    Commented Feb 24, 2018 at 2:02
  • $\begingroup$ @hans can you provide a counterexample? $\endgroup$
    – Dunham
    Commented Feb 24, 2018 at 2:21
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I provide a new solution here for someone who is interested.

WLOG, assume $f(-\pi)=f(\pi)\neq 0$(if not, just find a $x_0$ s.t. $f(x_0)\neq 0$ and do the transform $g(x)=f(x-\pi+x_0$).

Since $f(x)$ is continuous and periodic, there are only even points to change the sign of $f(x)$. Let these points to be $$ x_1<x_2<\cdots<x_{2k}. $$ Then construct a function $$ g(x)=\left(\sin \frac{x-x_1}{2}\right)\cdot\left(\sin \frac{x-x_2}{2}\right)\cdots\left(\sin \frac{x-x_{2k}}{2}\right). $$ Then we get \begin{equation} \int_{-\pi}^\pi f(x)g(x)\mathrm{d} x>0. \tag{1}\label{1} \end{equation} It can be rewritten as \begin{align}g(x)= &\left(\sin \frac{x}{2}\cos\frac{x_1}{2}-\cos\frac{x}{2}\sin\frac{x_1}{2}\right)\cdot \left(\sin \frac{x}{2}\cos\frac{x_2}{2}-\cos\frac{x}{2}\sin\frac{x_2}{2}\right)\\ &\cdots \left(\sin \frac{x}{2}\cos\frac{x_{2k}}{2}-\cos\frac{x}{2}\sin\frac{x_{2k}}{2}\right)\\ =&f_1(\cos x)\sin x+f_2(\cos x)\tag{2}\label{2} \end{align} where $f_1$ and $f_2$ are polynomials and $\text{deg}f_1=k-1, \text{deg}f_2=k$. These two terms can be written as \begin{equation} f_1(\cos x)\sin x = a_k \sin kx + a_{k-1}\sin (k-1)x+\cdots+a_1\sin x + a_0\tag{3}\label{3} \end{equation} \begin{equation} f_2(\cos x) = b_k \cos kx + b_{k-1}\cos (k-1)x+\cdots+b_1\cos x + b_0.\tag{4}\label{4} \end{equation} You can prove it by induction(see the notation below). If degree $k\leq n$, we get $$ \int_{-\pi}^\pi f(x)g(x)\mathrm{d}x=0 $$ by using the condition of the problem. It controdicts to the equation \eqref{1}. Hence $k\geq n+1$ and the number of roots of $f(x)$ in $[-\pi,\pi]$ is at least $2n+2$.


Notation: Assume for any $k<n$, we have \begin{equation} f_1(\cos x)\sin x = a_k \sin kx + a_{k-1}\sin (k-1)x+\cdots+a_1\sin x + a_0 \end{equation} \begin{equation} f_2(\cos x) = b_k \cos kx + b_{k-1}\cos (k-1)x+\cdots+b_1\cos x + b_0 \end{equation} for all $ f_1, f_2\in \mathbb{R}_k[x]$. Then for all $ g_1, g_2\in\mathbb{R}_{n}[x]$ , you can extract a $\cos x$ and write them as $$ g_1(\cos x)\sin x = \cos x(a_{n-1} \sin (n-1)x + a_{n-2}\sin (n-2)x+\cdots+a_1\sin x + a_0) $$ $$ g_2(\cos x) = \cos x(a_{n-1} \cos (n-1)x + a_{n-2}\cos (n-2)x+\cdots+a_1\cos x + a_0). $$ Then use the formula $$ \cos\alpha\sin\beta=\frac{1}{2}\left[\sin(\beta+\alpha)+\sin(\beta-\alpha)\right] $$ and $$ \cos\alpha\cos\beta=\frac{1}{2}\left[\cos(\beta+\alpha)+\cos(\beta-\alpha)\right]. $$

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