Prove that $f$ has $m+1$ zeros if $\int_{a}^{b} x^nf(x)dx=0$ for all $n\le m$

Question

Let $f$ be a continuous function on closed interval $[a,b]$, $m$ be a nonnegative integer. If $\forall n=0,1,\cdots ,m$, $$\int_{a}^{b}x^nf(x)dx=0$$ Prove that $f$ has at least $m+1$ different zeros on $[a,b]$.

My attempt: I've tried induction and solved the base case $m=0$. For the induction step I tried to use contradiction method and construct a polynomial $g(x)$ based on zeros of $f$, and prove that $g(x)f(x)$ always positive, but not clear about how to proceed on. Thanks for any help.

Answer 1

Suppose $f$ only has $m$ distinct roots (or fewer). Let them be $c_1,\cdots, c_k$. Now, since $f$ is continuous, the sign of $f$ can only change at these points. However, we can construct a polynomial $P(x)$ such that $P(x)f(x)\geq 0$ (can you see how?).

Now, we know that $$\int_a^bP(x)f(x)dx = 0,$$ but this implies either $f(x)=0$ almost everywhere (or is zero since continuous) or there are more than $m$ distinct roots.