Prove the exsistence of 3 zero points of a function

Solution 1:

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 $(a, b)$.

Solution 2:

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.