0

Let $f$ be a continuous function on closed interval $[0,1]$. Prove that if $n$ is a positive integer and $\int_{0}^{1}x^{k}f(x)\text{d}x=0$ for all $k=0,1,...,n$ then $f$ has $n+1$ different zeros on $[0,1]$.

I got $\int_{0}^{1}P(x)f(x)\text{d}x=0$ for all polynomial $P(x)$ of degree less or equal than $n$. Assume for a contradiction that $f(x)$ has $n$ distinct zeros $a_{1},...,a_{n}$ (or fewer), then I want to construct a polynomial $P(x)$ such that $P(x)f(x)\geq 0$ for all $x\in [0,1]$. Maybe $P(x)$ should be constructed based on the zeros of $f$, but I'm stuck. Can anybody provide me a hint? Thanks

Batominovski
  • 49,629
vutuanhien
  • 111

1 Answers1

3

If $f$ has infinitely many roots in $[0,1]$, then there is nothing to prove. Suppose from now on that $f$ has finitely many roots in $[0,1]$. We say that a root $a$ of $f$ switches signs if, for some open interval $I$ containing $a$, either of the following requirements is met:

  1. $f(x)<0$ for all $x\in I$ such that $x<a$, whereas $f(x)>0$ for all $x\in I$ such that $x>a$;

  2. $f(x)>0$ for all $x\in I$ such that $x<a$, whereas $f(x)<0$ for all $x\in I$ such that $x>a$.

Now, take $P(x)$ to be the product of $x-a$, where $a$ is a root of $f$ in $[0,1]$ that switches signs.

Remark: If a boundary point $0$ or $1$ is a root of $f$, then it is optional whether you will take it to be a root that switches signs. This choice does not affect the proof.

Batominovski
  • 49,629