If $f(t)$ converges as $t \to \infty$ and $f'(t)$ is uniformly continuous then $f'(t) \to 0$ as $t \to \infty$ where $t$ is time.

Question

I came across this result while I was reading mathematical epidemiology. What I interpreted from the statement of the result is that if a function is ultimately becoming constant then it's derivative will ultimately get arbitrarily close to $0$. Is this the correct interpretation? If yes, then I don't understand why uniform continuity of $f'$ is needed. If no, then a sketch of proof would be highly appreciated. Thanks in advance.

Answer 1

As mentioned in the comment, it is not necessary that $f$ is eventually constant.
The existence of $\displaystyle\lim_{t\to\infty}f(t)$ just tells us that for any $\epsilon > 0$, the function $f$ is eventually within $\epsilon$ distance from the limit.

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You do require uniform continuity of $f'$. For otherwise, consider the function $f:[1, \infty)\to\Bbb R$ defined as $$f(t) = \dfrac{\sin(t^2)}{t}.$$ It is easy to see that $f(t) \to 0$ as $t\to\infty$ but $f'(t)\not\to0$ as $t\to\infty$.
In fact, the latter limit does not even exist. (You can simply calculate $f'$ and see this.)

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In the case of uniform continuity of $f'$, we do have the result. A proof is as follows:

Let $\epsilon > 0$ be given. We need to show that there exists $X \in \Bbb R$ such that $|f'(t)| < \epsilon$ for all $t > X$.

Since $f'$ is uniformly continuous, there exists $\delta > 0$ such that whenever $|x - y| < \delta$, we have $$|f'(x) - f'(y)| < \epsilon/2.\quad(\star)$$

Let $L = \displaystyle\lim_{t\to\infty}f(t)$. (Which exists, by hypothesis.)
By definition of limit, we see that there exists $M \in \Bbb R$ such that $|f(t) - L| < \epsilon\delta / 8$ for all $t > M$.
In particular, we have $$|f(x) - f(y)| < \epsilon\delta/4 \quad \text{for all } x, y > M. \quad (*)$$ (Use triangle inequality.)

Consider the following sequence of points: $$x_n = M + n\dfrac{\delta}{2} \quad n \ge 1.$$ In particular, note that $x_n > M$ and $x_{n+1} - x_n = \delta/2$ for all $n \in \Bbb N$.

By Lagrange's mean value theorem, there exists $y_n \in (x_n, x_{n+1})$ such that $$f'(y_n) = \dfrac{f(x_{n+1}) - f(x_n)}{x_{n+1}-x_n} = \dfrac{f(x_{n+1}) - f(x_n)}{\delta/2}.$$ Using $(*)$, we can estimate the absolute value as $$|f'(y_n)| < \dfrac{\epsilon\delta/4}{\delta/2} = \dfrac{\epsilon}{2}.$$

Now, the sequence $(y_n)$ has the following properties:

1. $|f'(y_n)| < \epsilon/2$ for all $n \in \Bbb N$.
2. $y_n \in (x_n, x_{n+1})$.
3. In particular, 2 implies the following crucial fact:
   Given any $t > x_1$, there exists $n \in \Bbb N$ such that $$|t - y_n| < \delta.$$

We are now almost done.

Let $X = x_1$ and let $t > X$ be arbitrary. By 3., we see that there exists $n \in \Bbb N$ such that $|t - y_n| < \delta$.
From $(\star)$, we have that $$|f'(t) - f'(y_n)| < \epsilon/2.$$ The above implies that $$-\epsilon/2 + f'(y_n) < f'(t) < f'(y_n) + \epsilon/2.$$ Using 1., we see that $-\epsilon/2 < f'(y_n) < \epsilon/2$ and this gives us the desired inequality that $$-\epsilon < f'(t) < \epsilon.$$