For $f, g \in C^1$, $fg' - f'g \neq 0$ implies that the zeros interlace Let $f, g \in C^1$, and suppose that $f(x) g'(x) - f'(x) g(x) \neq 0$ for all $x$.
Show that


*

*The roots of $f$ do not have an accumulation point.

*The roots of $f$ and $g$ interlace, so that if $f(x_0) = f(x_1) = 0$ with $f(x) \neq 0$ for $x \in (x_0, x_1)$, then there exists a unique $y \in (x_0, x_1)$ such that $g(y) = 0$.


I think I am supposed to use the quotient rule somehow, but I cannot get it to work, as $g(x)$ may be $0$, and then $f/g$ is difficult to work with.
 A: It all follows from the following lemma: Between any roots of $f(x)$ there must be a root of $g(x)$, and visa versa.
Proof:  Define $h_1(x)=\frac{f(x)}{g(x)}$. Then if $f(x)$ has two zeros with no zero for $g(x)$ between, what happens to $h_1'(x)$ between the two zeros? Similarly, what happens to the derivative of $h_2(x)=\frac{g(x)}{f(x)}$ if there are two zeros for $g$ in a row?

(1) Now, what if $\{x_i\}$ is a sequence of distinct zeros of $f$ that converges (necessarily to a zero of $f$?) Then we can find between $x_{i},x_{i+1}$ a $y_i$ that is  zero of $g(x)$. And we can easily see that that $\lim y_i=\lim x_i$, which means the accumulation point must be a zero of $g(x)$. But the condition $f'(x)g(x)-f(x)g'(x)\neq 0$ precludes the two functions having a common zero.
(2) Once we know that there are no accumulation points, we can prove that the zeros of $f(x)$ must be "discrete," and therefore that the zeros alternate.
A: I don't know about 1.
Regarding 2., you  can assume WLOG that $\forall x\in (x_0,x_1), f(x) >0$
This implies that $f'(x_0)\geq0$ and $f'(x_1)\leq0$ $(*)$
Set $h=f'g-fg'$
Then $h(x_0)=f'(x_0)g(x_0)$ and $h(x_1)=f'(x_1)g(x_1)$
Since $h$ is continuous and non-zero everywhere, it has constant sign. WLOG $h>0$ $(**)$
$(*)$ and $(**)$ imply that $g(x_0)>0$ and $g(x_1)<0$.
Hence there is some $\chi \in (x_0,x_1)$ such that $g(\chi)=0$.

To get uniqueness of $\chi$, suppose for contradiction that the set $A=\{x\in (x_0,x_1) |g(x)=0\}$ has more than one element.
Consider two successive elements of $A$: that is consider $\alpha<\beta \in A$ such that $(\alpha,\beta)\cap A = \emptyset$.
Now the trick: in the reasoning above, switch $f$ and $g$ , $x_0$ with $\alpha$, $x_1$ with $\beta$. This proves that $f$ becomes $0$ between $x_0$ and $x_1$. Contradiction!
A: HINT:
Assume $f' g - f g'>0$ everywhere, and $a$, $b$ are two consecutive roots of $f$. Then $f'(a)$, $f'(b)$ are of opposite sign, otherwise there would be another root of $f$ between $a$, and $b$. We conclude that $g(a)$, $g(b)$ are of opposite sign.
