Let $f : \mathbb R \rightarrow \mathbb R : x \mapsto f(x)$ a smooth non constant function such that $f(x) = 0 \implies f'(x) > 0.$ Does $f$ have a unique root ?
I think this is true but I can't prove it. Here what I've done so far.
Case 1. Let $x_1 < x_2$ be two roots with no other root in $(x_1,x_2)$. Since $f(x_i) = 0$ we have $f'(x_i) > 0$ so by continuity of $f'$ we have that $f' > 0$ on a small interval $(x_i - \delta,x_i + \delta)$ around $x_i$. Therefore $f$ is strictly increasing on a neighbourhood of each root. In particular $f>0$ on $(x_i, x_i + \delta)$ and $f< 0$ on $(x_i-\delta,x_i).$
Using the intermediate value theorem we can find another root $c$ somewhere between $x_1$ and $x_2$, a contradiction.
In particular $f(x) = 0$ as infinitely many solutions.
Here's where I'm not so sure :
Case 2. If there are roots between any given given roots $x_1 < x_2$ then we can apply the above reasoning to $x_1,c$ and $c,x_2$ to find new roots between $x_1$ and $x_2$ so we can find infinitely many roots between any given roots $x_1$ and $x_2$.
In don't know how to proceed. It feels like $f$ should be equal to $0$ on some interval which would then contradict $f(x) = 0 \implies f'(x)>0$
Can someone find a counter example or finish the proof ?
Case 1 is done.
Now let us consider case 2. In this case, take two distinct roots $a < b$. As you mentioned, the function has infinitely many roots in the segment $[a,b]$. By Bolzano-Weierstrass, there exists a sequence $(x_k)$ of roots in $[a,b]$ converging to a certain $x$. By continuity, $x$ is a root and thus $$ f’(x) = \lim_{k \to \infty} \frac{f(x_k) - f(x)}{x_k - x} = 0 $$ which contradicts the hypothesis that the derivative of $f$ is positive at all roots of $f$.
Hope this helps!
ADDENDUM: Note that you don't even need Bolzano-Weiterstrass theorem. Indeed, you can construct an increasing sequence of roots by induction. There exists a root $x_1 \in ]a,b[$, and therefore a root $x_2 \in ]x_1,b[$, and so on. You get an increasing sequence of roots $(x_k)$ bounded by $b$ so by monotone convergence the sequence converges to a certain $x$.