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Prove that $\exists a<b$ s.t. $f(a)=f(b)=0$ when $\int_0^1f(x)dx=\int_0^1xf(x)dx=0$

Suppose that $f:[0,1]\to \mathbb{R}$ is continuous, and that $$\int_{0}^{1} f(x)=\int_{0}^{1} xf(x)=0.$$

How does one prove that $f$ has at least two distinct zeroes in $[0,1]$?

Well, if not say $\forall a,b\in [0,1] \ni a<b$, but $f(a)\neq f(b), f(a)>0$ then there will be a neighborhood of $a$ say $(a-\epsilon,a+\epsilon)$ where $f(x)>0$ and hence the integral will not be equal to $0$, but I don't know where I am using the other integral condition. am I wrong anywhere in my proof? please help.


marked as duplicate by Ragib Zaman, Did, Gerry Myerson, t.b., Zev Chonoles Jul 8 '12 at 15:22

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  • 3
    $\begingroup$ nbd? pla? $ $ $ $ $\endgroup$ – Did Jul 7 '12 at 20:38
  • $\begingroup$ could you tell me where I am wrong making the negation of the fact that I need to show? $\endgroup$ – Marso Jul 7 '12 at 20:38
  • $\begingroup$ The mean value theorem for integrals will get you at least one such point. $\endgroup$ – Dylan Moreland Jul 7 '12 at 20:40
  • $\begingroup$ $\int_{a}^{b}f(x)dx= f(c)(b-a)$,for some $c\in (a,b)$ This one? $\endgroup$ – Marso Jul 7 '12 at 20:44
  • $\begingroup$ yah so $\int_{0}^{1}f(x)=f(a)=0$ and $\int_{0}^{1}xf(x)=bf(b)=0$ for some $a\,b in(0,1)$ but how to show $a<b$? $\endgroup$ – Marso Jul 7 '12 at 20:49

Define $F(x):= \int_0^xf(t)\,dt$ for $x \in [0,1]$. Then the second integral tells us $$ \begin{aligned} 0 = \int_0^1xf(x)\,dx \, & = \, xF(x)\Bigr|_0^1-\int_0^1F(x)\,dx \\ & = 1\,F(1) - 0\,F(0) - \int_0^1F(x)\,dx \\ & = \int_0^1f(x)dx - \int_0^1F(x)\,dx \\ & = -\int_0^1F(x)\,dx. \end{aligned} $$

By the mean value theorem and continuity of $F$, which follows from continuity of $f$, this tells us that there is $c \in (0,1)$ such that $F(c)=0$. That is, $$ \int_0^cf(x)\,dx = 0. $$ It follows that $$ \int_c^1f(x)\,dx = \int_0^1f(x)\,dx-\int_0^cf(x)\,dx = 0 - 0 = 0 $$ as well.

We then apply the mean value theorem two more times, using continuity of $f$, for the intervals $[0,c]$ and $[c,1]$ to find $a \in (0,c)$ and $b \in (c,1)$ respectively such that $f(a)=f(b)=0$.


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