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If $f$ is a continuous function on $[a,b]$ , and $\displaystyle\int_a^bf(x)dx=\int_a^bxf(x)dx=0$ .

How to show that there are at least two roots for $f$ on $[a,b]$ ?

By mean value theorem,

$\qquad\displaystyle\exists c\in(a,b)\ni f(c)=\frac{\int_a^bf(x)dx}{b-a}=0$

so we know $x=c$ is a root for $f$.

But how to find another root for $f$?

Thanks for the answer.

yhliu
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    Compare https://math.stackexchange.com/q/1733548/42969. I am reluctant to close the question as a duplicate because the answer to that one is very short ... – Martin R Apr 26 '23 at 08:42
  • Here is a similar question: https://math.stackexchange.com/q/443266/42969 – Martin R Apr 26 '23 at 08:44
  • I was going to suggest that if there was only 1 root $c$, to then look at $ 0 = \int (x-c) f(x) , dc $. This is a (specific) solution in one of Martin's links. – Calvin Lin Apr 26 '23 at 08:48
  • Thank you, Martin, Calvin. – yhliu Apr 27 '23 at 04:51

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HINT: integrate by parts with an antiderivative of $f$. Can you find roots of this antiderivative? Can you then conclude something about the stationary points of this antiderivative? The mean value theorem idea is good, it just isn't strong enough (I don't think?) to use on $f$ alone.

$\newcommand{\d}{\,\mathrm{d}}$We are given a nondegenerate compact interval $[a,b]\subset\Bbb R$ and a continuous $f:[a,b]\to\Bbb R$ satisfying: $$\int_a^b f(x)\d x=0=\int_a^b xf(x)\d x$$

Define $F:[a,b]\to\Bbb R$, $x\mapsto\int_a^x f(t)\d t$.

We have $F(b)=0=F(a)$. Integrating by parts: $$0=\int_a^b xf(x)\d x=bF(b)-aF(a)-\int_a^bF(x)\cdot(1)\d x\\therefore\int_a^bF(x)\d x=0$$

Since $F$ is continuous, it follows that there exists a root $\xi\in(a,b)$ of $F$. $F(a)=F(\xi)=F(b)=0$ implies by Rolle's theorem (and differentiability of $F$) that $F$ has at least two stationary points, one between $a$ and $\xi$, the other inbetween $\xi$ and $b$.

Stationary points of $F$ are exactly roots of $f$, since $F'=f$ on $(a,b)$.

FShrike
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