For $f, g \in C^1$, $fg' - f'g \neq 0$ implies that the zeros interlace

Question

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.

@Alex Zorn: Just curious, have you ever suggested:" use my lemma" ;)? — user99680, Jul 11 '14 at 03:20
Haha, yeah I definitely have! And, in fact, I'm a direct descendant of Max Zorn. — Alex Zorn, Jul 11 '14 at 17:44

Thomas Andrews · Accepted Answer · 2014-07-10T19:08:44.793

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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.

edited Jul 10 '14 at 19:08

answered Jul 10 '14 at 18:26

Thomas Andrews

177,126

Considering both questions is also a hint for the first part, as for each $x_0$, at least one of the quotients is defined and has nonzero derivative in a suitable open neighbourhood of $x_0$. – Hagen von Eitzen Jul 10 '14 at 18:40
Yeah, I just realized that, @HagenvonEitzen. :) – Thomas Andrews Jul 10 '14 at 18:43
Perfect solution, as always! – user182973 Jul 10 '14 at 18:49

score 3 · Answer 2 · answered Jul 10 '14 at 18:43

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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!

answered Jul 10 '14 at 18:43

Gabriel Romon

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You need (1), of course to show that there is such a thing as "successive elements of $A$." – Thomas Andrews Jul 10 '14 at 18:52
@ThomasAndrews yes, sure – Gabriel Romon Jul 10 '14 at 18:56
I do like the way you can avoid $(f/g)'$ to show that $g(x)$ must have a root between any two roots of $f(x)$. That's quite nice. – Thomas Andrews Jul 10 '14 at 19:06
Nice and tricky solution! – Mohammad Khosravi Jul 10 '14 at 19:12

score 1 · Answer 3 · answered Feb 28 '23 at 21:32

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.

For $f, g \in C^1$, $fg' - f'g \neq 0$ implies that the zeros interlace

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