Jtdylktuy

I am stuck on the following problem:

Let $f:mathbb{R} to mathbb{R}$ be continuous and $2pi$-periodic and $n$ be a positive integer. If for any integer $p in [0,n-1]$,
$$int_{0}^{2pi} f(t) cos(pt),mathrm{d}t=int_{0}^{2pi} f(t) sin(pt),mathrm{d}t=0,$$

then is it true that $f$ has at least $2n$ roots on $[0,2pi]$?

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 :
$$ int_{0}^{2pi} f(t) e^{ipt} mathrm{d}t=0$$ ,
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 $nle m$

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 $R_{n-1}[X]$ ,$0leq a_{1}<a_{2}<... <a_{m}leq pi leq a_{m+1}<..<a_{p}$ the zeros of $f$ and suppose that $pleq n-1$ .

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 : $forall x in [0,2pi]-J : f(x) Pi_{i=1}^{p} (cos(x)-d_{i}) > 0$ 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 : $int_{0}^{2pi} f(t)P(t)=0$ hence $f(x)=0$ For any $x in [0,2pi]-J$ , 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 .

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 $2pi$ periodic with $Min mathbb{N}$ 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_{ellgeq 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_{ellgeq 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 $Mgeq 2n$.

Definition:

Given a continuous $2pi$-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^{2pi} 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_{ellgeq 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 $|a_{ell}|le frac1{sqrt{2pi}}int_0^{2pi}|cos(ell x)f(x)|dxle frac1{sqrt{2pi}}int_0^{2pi}|f(x)|dx$ and the same bound for $|b_{ell}|$, $forallell$. The terms are bounded as follows
begin{align*}
left|sum_{ellgeq n+1}
frac{a_ell}{ell^{4L}} cos(ell t)
+frac{b_ell}{ell^{4L}} sin(ell t)right|
&leq
2max_{ell>n }{|a_ell|,|b_ell|}
sum_{ellgeq n+1}
frac{1}{ell^{4L}} \
&leq
2max_{ell>n }{|a_ell|,|b_ell|}
+
left(
frac{1}{(n+1)^{4L}}
+
sum_{ellgeq n+2}
frac{1}{ell^{4L}}
right)\
&leq
2max_{ell>n }{|a_ell|,|b_ell|}
left(
frac{1}{(n+1)^{4L}}
+
int_{n+1}^{infty}
x^{-4L}
right)\
&=
2max_{ell>n }{|a_ell|,|b_ell|}
left(
frac{1}{(n+1)^{4L}}
+
frac{1}{(4L-1)(n+1)^{4L-1}}
right)\
&leq
4max_{ell>n }{|a_ell|,|b_ell|}
frac{1}{(n+1)^{4L-1}}
end{align*}
Now, we just have to show that
begin{equation*}
4max_{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{8nmax_{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 $Lrightarrow infty$. Hence the result is true for some $L$.

Proposition:
Suppose $f$ has $Min mathbb{N}$ 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 $Min mathbb{N}$ 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.

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
$$
g(x)=left(sin frac{x-x_1}{2}right)cdotleft(sin frac{x-x_2}{2}right)cdotsleft(sin frac{x-x_{2k}}{2}right).
$$
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}cosfrac{x_1}{2}-cosfrac{x}{2}sinfrac{x_1}{2}right)cdot left(sin frac{x}{2}cosfrac{x_2}{2}-cosfrac{x}{2}sinfrac{x_2}{2}right)\
&cdots left(sin frac{x}{2}cosfrac{x_{2k}}{2}-cosfrac{x}{2}sinfrac{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_1sin x + a_0tag{3}label{3}
end{equation}
begin{equation}
f_2(cos x) = b_k cos kx + b_{k-1}cos (k-1)x+cdots+b_1cos x + b_0.tag{4}label{4}
end{equation}
You can prove it by induction(see the notation below). If degree $kleq n$, 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 $kgeq n+1$ 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_1sin x + a_0
end{equation}
begin{equation}
f_2(cos x) = b_k cos kx + b_{k-1}cos (k-1)x+cdots+b_1cos x + b_0
end{equation}
for all $ f_1, f_2in mathbb{R}_k[x]$.
Then for all $ g_1, g_2inmathbb{R}_{n}[x]$ , you can extract a $cos x$ and write them as
$$
g_1(cos x)sin x = cos x(a_{n-1} sin (n-1)x + a_{n-2}sin (n-2)x+cdots+a_1sin x + a_0)
$$
$$
g_2(cos x) = cos x(a_{n-1} cos (n-1)x + a_{n-2}cos (n-2)x+cdots+a_1cos x + a_0).
$$
Then use the formula
$$
cosalphasinbeta=frac{1}{2}left[sin(beta+alpha)+sin(beta-alpha)right]
$$
and
$$
cosalphacosbeta=frac{1}{2}left[cos(beta+alpha)+cos(beta-alpha)right].
$$