If $f(x)=0 \implies f'(x)>0$, is the zero set of $f$ a single point?

Question

Let $f:\mathbb R \to \mathbb R$ be a real-valued differentiable function. Suppose that $f'(x)>0$ for every $x$ such that $f(x)=0$. Does it follow that the number of zeroes of $f$ is at most one?

This sounds quite reasonable to me: it seems intuitive that if $x_1, x_2$ are two different zeroes of $f$, then a third zero with negative derivative should lie between them. I can't seem to adapt this argument to a solid proof though.

Is there any way to prove (or disprove!) this fact in a quick fashion?

Answer 1

Assume $a < b$ and $f(a) = f(b) = 0.$ By the existence of the derivative at $a$ and its positivity, there is a small neighborhood $(a, a + \delta)$ on which $f(x) > 0.$

Define the set $C$ as the set of all $a < x < b$ such that $f(x) > 0.$ $C$ is not empty. On the other hand, there is a small neighborhood $(b - \varepsilon, b)$ on which $f(x) < 0.$ Therefore $\sup C \neq b.$

Note supremum is the same as the least upper bound. Original Latex has, among "log-like functions," $\sup$ but does not have LUB for least upper bound. I could make one with operatorname, $\operatorname{lub}$

Let $c = \sup C.$ We know $c < b.$ By continuity, $f(c) = 0.$ However, $f'(c) \leq 0$ since there are points $x$ with $f(x) > 0$ arbitrarily close to $c$ but $x<c.$

In case anyone is wondering why $f$ is positive immediately to the right of $a,$ we are told that there is some $f'(a) = A > 0.$ The limit that defines the derivative has $$ A = \lim_{t \rightarrow 0} \frac{f(a+t) - f(a)}{(a+t - a)} = \lim_{t \rightarrow 0} \frac{f(a+t) }{t}. $$ Since $A > 0$ is the limit, there is some $\delta > 0$ such that $$ 0 < t < \delta \Longrightarrow \; \; \; \; \; \; \frac{f(a+t) }{t} > \frac{A}{2}, $$ or $$ 0 < t < \delta \Longrightarrow \; \; \; \; \; \; f(a+t) > \left( \frac{A}{2} \right) \; t > 0. $$

Answer 2

Suppose that $f$ has at least two zeros.

Assume first that $a<b$ are two consecutive zeros of $f$ (which means that $f(x)\not=0$ for all $x\in(a,b)$). Since $f'(a)>0$ then there is $x_1\in (a,b)$ such that $f(x_1)>0$. In a similar way, $f'(b)>0$ implies that there is $x_2\in (a,b)$ such that $f(x_2)<0$. Hence, by continuity, $f$ has zero in $(a,b)$. Contradiction.

If there are no "consecutive" zeros, then between two distinct zeros there are infinite zeros. Here we can find a strictly monotone sequence of zeros $x_n$ which converges to some point $x_0$. By continuity $0=\lim_{n\to\infty}f(x_n)=f(x_0)$. Finally we get a contradiction $$0=\frac{f(x_n)-f(x_0)}{x_n-x_0}\to f'(x_0)>0.$$

Answer 3

Let $a<b$ be two zeros. Since $f'(a)>0$, there exists $c\in (a,b)$ such that $f(c)>f(a)=0$. Since $f'(b)>0$, there exists $d\in (c,b)$ such that $f(d)<f(b)=0$. Since $f$ is continuous, it have a zero in $(c,d)$.

In such a way we can construct an infinite sequence $(x_n)_{n\in\mathbb N}$ such that $n\neq m\Rightarrow x_n\neq x_m$ and $f(x_n)=0$. Without restricting the generality (by taking a subsequence) you can suppose that $x_n$ converges towards $x$. We have $f(x)=0$ and $f'(x)=0$. Contradiction.