Prove a twice differentiable fucntion can only have at most two zeros [duplicate]

Solution 1:

We assume that for all $u$, $q$ is continuous, second differentiable and not uniformly zero on any interval.

Assume that there exist $x,y$ such that $q(x)=q(y)=0$.

Since $q$ is continuous, second differentiable and not uniformly zero on $(x,y)$ it is either the case that there is a

1. local maximum or a

2. local minimum point $(z,q(z))$ of $q$ on the open interval $(x,y)$.

In either case, $q^\prime(z)=0$, so it follows that

$$ z^2q^{\prime\prime}(z)=(z^2+1)q(z) $$

So both $q(z)$ and $q^{\prime\prime}(z)$ must have the same sign.

But if $(z,q(z))$ is a global maximum $q(z)>0$ and $q^{\prime\prime}(z)<0$ and it it is a global minimum then $q(z)<0$ and $q^{\prime\prime}(z)>0$.

This is a contradiction.

So $q$ cannot have distinct zeros unless $q$ is uniformly zero on some interval.