Let $a<b$ $a \in R$ and $f: [a,b] \to R$ be a continuous function such that for all $k = 0,1,...,n$ we have $\int_a^bt^kf(t)dt=0$. Find the minimum number of elements of the set $\{x\in (a,b)|f(x)=0\}$
I came across this question in one of my homework. Intuitively I think that the solution of this question must be n+1.
Assume the interval is [-1,1], then it is obvious that for $\int_a^bt^0f(t)dt=0$ we need at least 1 $x$ such that $f(x)=0$ because the integral of the function that is above f(x)=0 must be equal to the integral of the function that is below it. Now, multiplying this by t, the interval [-1,0) will shift signs and therefore there must be another point with $f(x)=0$, so to get $\int_a^bt^1f(t)dt=0$ and $\int_a^bt^0f(t)dt=0$ we need at least 2 points in the interval [-1,1] such that $f(x)=0$. Continuing inductively, I assume that the answer to this question must be $n+1$, however, I am having trouble showing my solution in a mathematically correct way.
Thank you!
The polynomials of degree $\le n+1$ form a vector space $V_{n+1}$ of dimension $n+2$. The linear map $T_n: f \mapsto [\int_a^b x^0 f(x)\; dx, \ldots, \int_a^b x^{n} f(x)\; dx]$ takes $V_{n+1}$ into $\mathbb R^{n+1}$. Since $n+2 > n+1$ it must have nontrivial kernel, i.e. there is some nonzero $f \in V_{n+1}$ such that $\int_a^b x^j f(x)\; dx = 0$ for all $j = 0 \ldots n$. Since it has degree $\le n+1$ and is not identically $0$, such $f$ has at most $n+1$ zeros in the interval.
On the other hand, suppose $f$ has fewer than $n+1$ zeros in the interval. Let $p_i$ be the points in the interval where $f$ changes sign. There are at most $n$ of them. Thus $P(x) = \prod_i (x - p_i)$ is a polynomial of degree $\le n$ such that $P(x) f(x)$ is either always $\ge 0$ or always $\le 0$ for $x$ in the interval. Thus $\int_a^b P(x) f(x)\; dx \ne 0$. Conclude that $\int_a^b x^j f(x)\; dx \ne 0$ for some $j \in \{0,1,\ldots,n\}$.