# Prove the exsistence of 3 zero points of a function

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.

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 $(a, b)$.

• My initial argument was bit lengthy and unnecessary at some places. I have simplified the proof to make it more understandable. – Paramanand Singh Jul 14 '13 at 17:54
• The integration by parts is great! It's fascinating to see the similarity of f(x) and F(x). Thanks a lot. – Syrtis Major Jul 15 '13 at 6:34
• By the way it was a nice question in the first place. Thanks to you too, substructure. – Paramanand Singh Jul 15 '13 at 7:17

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.

• A impressive solution! – Syrtis Major Jul 19 '13 at 2:23