Zero derivative implies constant function (No MVT, Rolle's Theorem, etc.)

Question

I'm working through Foundations of Mathematical Analysis by Johnsonbaugh. The following problem is given before MVT is proven.

However, we do know about compactness, completeness, continuity, intermediate value theorem, metric spaces, sequences, etc.

I am trying to find a proof that does not use MVT. This question seems similar to mine, but they end up using something equivalent to Rolle's theorem in the answer.

Problem:

Suppose $f'(x) = 0$ for all $x \in (a,b)$. Prove that $f$ is constant on $(a,b)$.

Attempt:

Let $x,y \in (a,b)$. We have

$$f'(x) = \lim_{z \to x} \frac{f(z)-f(x)}{z-x} = 0$$

and

$$f'(y) = \lim_{z \to y} \frac{f(z)-f(y)}{z-y} = 0$$

Answer 1

Note that since $f$ is differentiable on $(a,b),$ $f$ is continuous on $(a,b).$

Suppose $a<c<d<b.$ We'll show $f(c)-f(d)=0,$ which implies $f$ is constant. Let $\epsilon>0.$ Define

$$E=\{x\in [c,d]:|f(x)-f(c)|\le \epsilon(x-c)\}.$$

Then $c\in E,$ so $E$ is nonempty. Let $s= \sup E.$ Then there is a sequence $x_n$ in $E$ such that $x_n\uparrow s.$ The continuity of $f$ then shows $s\in E.$

Suppose $s<d.$ Because $f'(s) = 0,$ there exists $x\in (s,d)$ such that $|f(x)-f(s)|\le \epsilon(x-s).$ We then have

$$|f(x)-f(c)|\le |f(x)-f(s)| + |f(s)-f(c)| \le \epsilon(x-s)+\epsilon(s-c) = \epsilon(x-c).$$

Thus $x\in E$ and $x> s,$ contradiction. Therefore $s=d,$ and since we know $s\in E,$ we have $d\in E.$ That means $|f(d)-f(c)|\le \epsilon(d-c).$ Since $\epsilon$ was arbitrary, we have $f(d)=f(c)$ as desired.

Answer 2

Let $a < x < y < b$. We will prove $f(y)=f(x)$ by proving $|f(y)-f(x)| < \epsilon(y-x)$ for every $\epsilon>0$. So fix $\epsilon > 0$. By the hypothesis on vanishing of derivatives, for every $t \in [x,y]$ there is $\delta_t > 0$ such that for $|z-t|<\delta_t$, $|f(z)-f(t)|<\epsilon|z-t|$.

Now here is a minor little trick: The intervals $(t-\delta_t/2,t+\delta_t/2)$ (of radius $\delta_t/2$, rather than $\delta_t$) cover $[x,y]$. For now say an interval of the form $(t-\delta_t/2,t+\delta_t/2)$ is a narrow interval and one of the form $(t-\delta_t,t+\delta_t)$ is a wide interval. By compactness there is a finite subcover by narrow intervals, centered at say $t_1,\dotsc,t_s$, with $x \leq t_1 < t_2 < \dotsc < t_s \leq y$. We can reduce to an irredundant cover, i.e., so that no narrow interval centered at $t_i$ is contained in the narrow interval centered at $t_j$ for $j \neq i$. This implies that for each $i$, the narrow intervals centered at $t_i$ and $t_{i+1}$ overlap: otherwise, at least one point in the gap between them must be covered by the interval centered at some $t_j$ with $j<i$ or $j>i+1$; and either $t_i$ is redundant, or $t_{i+1}$ is redundant. Next, this implies that for each $i$, $|t_{i+1}-t_i| < \max(\delta_{t_{i+1}},\delta_{t_i})$. Indeed, if a point $u$, $t_i < u < t_{i+1}$, lies in the intersection of the narrow intervals, then $$ t_{i+1}-t_i = t_{i+1}-u+u-t_i < \delta_{t_{i+1}}/2 + \delta_{t_i}/2 \leq 2 \max(\delta_{t_{i+1}}/2,\delta_{t_i}/2). $$ It follows that for each $i$, $t_{i+1}$ lies in the wide interval centered at $t_i$ or $t_i$ lies in the wide interval centered at $t_{i+1}$ (or both). In particular, for each $i$, $|f(t_{i+1})-f(t_i)| < \epsilon |t_{i+1}-t_i|$.

We claim that the same relation holds between $x$ and $t_1$, and between $y$ and $t_s$. Indeed, $x$ lies in the narrow interval centered at $t_1$, hence also in the wide interval. So $|f(t_1)-f(x)| < \epsilon |t_1-x|$. Similarly for $y$ and $t_s$.

Putting it all together, we have: $$ \begin{split} |f(y)-f(x)| &\leq |f(y)-f(t_s)| + |f(t_s)-f(t_{s-1})| + \dotsb + |f(t_1)-f(x)| \\ &< \epsilon|y-t_s| + \epsilon|t_s-t_{s-1}| + \dotsb + \epsilon|t_1-x| \\ &= \epsilon(y-t_s) + \epsilon(t_s-t_{s-1}) + \dotsb + \epsilon(t_1-x) \\ &= \epsilon(y-x), \end{split} $$ as claimed.

Notes: (1) Thanks for several very helpful comments about simplifications, gaps in the proof, etc! (2) I wonder if an application of Cousin's theorem might simplify the argument.