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Assuming $a<b$, function f(x) is continous at [a,b], and we have

$\int_a^bf(x)dx=\int_a^bxf(x)dx=\int_a^bx^2f(x)dx=0$

Prove that $\exists \ x_1,x_2,x_3\text {(different from each other)} \in(a,b)$ satisfying $f(x_1)=f(x_2)=f(x_3)=0$.

I have proved the existence of one zero point by differential mean value theorem, but have no idea about going on.

Thanks for any solution or hint.

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Using mean value theorem for integrals we have $$\int_{a}^{b}f(x)\,dx = (b - a) f(x_{1})$$ for some $x_{1} \in (a, b)$. Hence $\int_{a}^{b}f(x)\,dx = 0$ implies that $f(x_{1}) = 0$ for some $x_{1} \in (a, b)$. This gives us first root of $f$ in $(a, b)$. Next let $F(x) = \int_{a}^{x}f(t)\,dt$ so that $F'(x) = f(x)$ and $F(a) = F(b) = 0$. Next we can use integration by parts to get $$\int_{a}^{b}xf(x)\,dx = bF(b) - aF(a) - \int_{a}^{b}F(x)\,dx$$ and thus we can see that $\int_{a}^{b}F(x)\,dx = 0$ and hence by mean value theorem there is a $d \in (a, b)$ for which $F(d) = 0$. Now $F(a) = F(d) = F(b) = 0$ gives us two distinct roots of $F'(x) = f(x)$ in $(a, b)$.

Next we need to make use of the condition $\int_{a}^{b}x^{2}f(x)\,dx = 0$. Clearly if we use integration by parts we can see that $$\int_{a}^{b}x^{2}f(x)\,dx = b^{2}F(b) - a^{2}F(a) - 2\int_{a}^{b}xF(x)\,dx$$ so that we have $\int_{a}^{b}xF(x)\,dx = 0$. So we can see that the function $F(x)$ satisfies the same first two conditions which are satisfied by $f(x)$ and hence by applying the previous logic to $F(x)$ it follows that there are two distinct roots of $F(x)$ in $(a, b)$ and let's call them $d, e$. From $F(a) = F(d) = F(e) = F(b) = 0$ and mean value theorem there are clearly three distinct roots of $ F'(x) = f(x)$ in $(a, b)$.

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  • $\begingroup$ My initial argument was bit lengthy and unnecessary at some places. I have simplified the proof to make it more understandable. $\endgroup$ – Paramanand Singh Jul 14 '13 at 17:54
  • $\begingroup$ The integration by parts is great! It's fascinating to see the similarity of f(x) and F(x). Thanks a lot. $\endgroup$ – Syrtis Major Jul 15 '13 at 6:34
  • $\begingroup$ By the way it was a nice question in the first place. Thanks to you too, substructure. $\endgroup$ – Paramanand Singh Jul 15 '13 at 7:17
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Here's a simpler way to see this: If $f$ has exactly $n$ zeros say $a_1 < a_2 < \dots < a_n$, then $(x - a_1)(x - a_2) \dots (x - a_n)f(x)$ has same sign throughout the interval. But the integral of this should have been zero - Contradiction.

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    $\begingroup$ A impressive solution! $\endgroup$ – Syrtis Major Jul 19 '13 at 2:23

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