Prove that between every two roots of f there is a root of g and vice versa

Question

Suppose $f$ and $g$ are two continuous and differentiable functions such that $f' = g$ and $g' = -f$. Prove that between every two consecutive roots of $f$, there is a root of $g$ and between every two consecutive roots of $g$, there is a root of $f$.

Answer 1

First we need to show that the problem is well defined.

Consider $W(f,g)(x) = \det \begin{bmatrix} f(x) & g(x) \\ f'(x) & g'(x)\end{bmatrix} = \det \begin{bmatrix} f(x) & g(x) \\ g(x) & -f(x)\end{bmatrix} = -(f(x)^2+g(x)^2)$.

Note that $W(f,g)'(x) = 0$, hence $W(f,g)(x)$ is constant.

If $W(f,g)(0) = 0$, then $f=g = 0$. In this case, all points are zeros of $f,g$ hence the notion of consecutive zeros is ill defined. However, it is clear that between any two zeros of $f$ ($g$) that $g$ ($f$) has a zero.

So, suppose $W(f,g)(0) \neq 0$.

This implies that if $f(x) = 0$ then $f'(x) = g(x) \neq 0$, and similarly for $g$. Hence the zeros of $f,g$ are isolated and never collocated, so the problem is well defined.

Suppose $x_1,x_2$ are two consecutive zeros of $f$. Then $f$ has a maximum or minimum at some $x^* \in (x_1,x_2)$, and $f'(x^*) = 0$, hence $g(x^*) = f'(x^*) = 0$.

The same sort of analysis applies to $g$.