Analysis: the number of sign changes of an integrable function

Question

Let $a,b \in \mathbb{R}$ with $a < b$ and let $f : [a,b] \rightarrow \mathbb{R}$ be a continuous function which is not identically 0. Suppose for some $n \in \mathbb{N}$ that for all $k \in \{0, ..., n\}$ that

$$\int_a^b t^kf(t)dt = 0$$

We want to show that there are $n+1$ points where $f$ changes sign.

So my first inclination for this problem is that the statement is heavily suggestive of induction on $n$. As such what I want to show is that if $\int_a^b f(t)dt = 0$ and $\int_a^b tf(t)dt = 0$ then $f$ changes sign at least twice - preferably in a way that lends itself to being mimicked in the general case. So I observed that we already know that it changes sign at least once - as $f$ is not identically zero it could not otherwise be the case that $\int_a^b f(t)dt = 0$. So then I thought perhaps to suppose for contradiction that $f$ changes sign exactly once. From here I am not sure where to go.

Answer 1

Note that

$$\int_a^b t^k f(t) dt = 0$$

for all $k =0, 1, \cdots, n$ is the same as saying that

$$(*) \int_a^bP(t) f(t) dt =0$$

for all polynomials of degree less then or equal $n$.

Assume that $f$ only change sign at $x_1< \cdots <x_m$, where $x_j \in (a, b)$ and $m \leq n$. Then define

$$P(t) = (t-x_1)(t-x_2)\cdots(t-x_m)$$

Note that

$$\int_a^b P(t) f(t)dt = \int_a^{x_1} P(t) f(t) dt+ \int_{x_1}^{x_2} P(t) f(t) dt +\cdots + \int_{x_m}^b P(t) f(t) dt$$

Note that if $P(t) f(t)$ is positive/negative on $(a, x_1)$ if and only if it is positive/negative in $(x_1, x_2) \cdots (x_m, b)$.

Thus the integral must be strictly positive or negative. This contradicts to (*).