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Suppose that the function $h:\mathbb{R}\to\mathbb{R}$ is differentiable and that there is a positive number $c$ such that $h'(t)\geq c$, for all points $t\in\mathbb{R}$. Prove that there is exactly one number $t$ at which $h(t)=0$.

Proof (BWOC)

Suppose that there are exactly $n$ numbers, $\{x_1,x_2,\dots,x_n\}\in\mathbb{R}$, $x_1<x_2<\dots<x_n$, for which $h(x_1) = h(x_2) = \cdots = h(x_n) = 0$.

Because $h$ is continuous, $h$ is either positive or negative for all $t_1\in (x_1,x_2)$, all $t_2\in(x_2,x_3)$, $\cdots$, all $t_{n-1}\in(x_{n-1},x_n)$.

Additionally, because $h'(t)$ is defined for all $t\in\mathbb{R}$, if $h$ is positive in $(-\infty, x_1)$, $h$ must become negative in $(x_1,x_2)$, else the derivative will not be defined at $x_1$. Likewise, $h$ must become negative in $(x_2,x_3)$.

Because $h$ transitioned from positive to negative to positive, $h'$ transitioned from negative to positive respectively. By the Intermediate Value Theorem, $h'(t^*) = 0$ for some $t^*$ in $(x_1,x_2)$. There are $n-1$ such $t^*$'s.

This is a contradiction, because $h'(t)>0\forall t\in\mathbb{R}$.

Therefore $h$ attains exactly one zero, so that the number of points at which $h'(t) = 0$ is exactly $0$.

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    $\begingroup$ I'm not sure why you say Lipschitz in the title, since that isn't an assumption of the problem. $\endgroup$ – Nate Eldredge May 1 at 16:31
  • $\begingroup$ Sorry, its been a while. but doesnt lipschitz continuity imply $h'(t) \leq c~\forall t \in \mathbb{R}$ not the other way around? in which case if $h'(t) \leq c$ you have h is bounded by $|h(x)-h(y)| \leq c|x-y| < M|x-y||$ assuming c is the best Lipchitz constant. what i would want to do in this instance is see if theres a way to show that c$\in [0,1]$ in order to call the contraction mapping and then perhaps show that this point is zero. but i'm not sure. i just wanted to bring to your attention a potential mistake (apologies if i'm actually wrong) $\endgroup$ – Vaas May 1 at 17:00
  • $\begingroup$ @NateEldredge and Vaas, you are both correct. I forgot that Lipschitz was the other way around. I mean what I wrote, disregarding the Lipschitz mistake. $\endgroup$ – MurderOfCrows May 3 at 16:13
  • $\begingroup$ Why can this prove that there is at least one zero? (In your proof, it is perfectly fine to have no zeros at all) $\endgroup$ – user12986714 May 3 at 16:18
  • $\begingroup$ @user12986714 That's a good point. I suppose the bound on the derivative is important for that. For example, the exponential function has no zeros, but it also has no minimum derivative. I'm not sure how to handle that, exactly. $\endgroup$ – MurderOfCrows May 3 at 16:34
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The intermediate value theorem for derivatives is a bit much. If $h(x_1) = h(x_2) = 0$, then by Rolle's theorem there exists $t \in (x_1,x_2)$ satisfying $h'(t) = 0$, contrary to hypothesis.

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By the Mean Value Theorem, $h(y)-h(x)\ge c\cdot (y-x)$ for all real $x<y$. Deduce that $h(t)=0$ for at most one $t$. Examine the inequality for large $y$ ($x$ fixed) and then for small $x$ ($y$ fixed) to see that the range of $h$ is all of $\Bbb R$.

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