Function with Always-Zero Derivative Is Constant

Suppose a function $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f'(x)=0$ for all $x$ in $(a,b)$, what would that mean for $f(x)$ graphically?

Great! So if $f'(x)=0$ for all $x$ in $(a,b)$, it must mean that at any point in the interval, the tangent line of the curve must be horizontal.

This shows us that geometrically, if $f'(x)=0$ for all $x$ in $(a,b)$, then $f(x)$ is constant.

Now, before we give it away, try to prove it algebraically on your own first! (Hint: You should try using the Mean Value Theorem.)

To prove this lemma, let's choose a point $p$ in the interval $(a,b)$. By the Mean Value Theorem, there exists $r$ in $(a,p)\subset (a,b)$ such that $$f'(r)=\frac{f(p)-f(a)}{p-a}.$$ Since $(a,p)$ is a subset of $(a,b)$, it is given that $f'(r)=0$, which means that $$0=\frac{f(p)-f(a)}{p-a}\Longrightarrow f(p)-f(a)=0\Longrightarrow f(p)=f(a).$$ So, for any $p$ in the interval $(a,b)$, we just showed that $f(p)=f(a)$. Thus, we can conclude that $f(x)$ is constant over the entire interval.

Steven Kwon already gave a nice explanation with good intuition. Here we'll give a (nonstandard?) proof avoiding the mean value theorem, inspired by a blog post of Tim Gowers.

Suppose $f$ is differentiable on $[0,1]$, with $f'(t) = 0$ for all $t\in[0,1]$. We want to show $+f(1)-f(0) = 0$. (The same reasoning will show that $f$ is constant on $[0,1]$: do you see why?)

The difficulty (and also why this fact is interesting) is that the condition "$f'(t) = 0$" (always-zero derivative) is local, while the desired statement $+f(1)-f(0) = 0$ is a more global expression.

Note that the "global" change in $f$ has a nice "telescoping" property: for example, $$[+f(1)-f(0)] = [+f(0.5)-f(0)] + [+f(1)-f(0.5)].$$

The opposite $+,-$ signs of the $f(0.5)$ terms cancel each other out. So this leads to a natural idea: if we keep breaking up larger intervals (like $[0,1]$) into smaller intervals (like $[0,0.5]$ and $[0.5,1]$), then "eventually" we'll be able to use the local zero-change/sensitivity/derivative condition $f'(t)=0$ "better and better".

OK, so for any $x,y$ in the big interval $[0,1]$, let's write $\Delta(x,y) = +f(y) - f(x)$ for convenience. Then we can nicely rewrite the "telescoping" property above as $$\Delta(a,b) = \Delta(a,c)+ \Delta(c,b).$$ We want to keep splitting into smaller and smaller intervals like this, until we can use the always-$0$-derivative condition. For example, if we had $a=0$, $b=1$, then we could decompose $$\Delta(0,1) = \Delta(0.0,0.1)+\Delta(0.1,0.2)+\cdots+\Delta(0.9,1.0).$$

We want to be able to say that each of these $10$ (or however many we end up decomposing into) terms are small, and this is where the derivative condition comes into play.

The technical details

"Smallness" is relative, so let's fix a concrete threshold $\epsilon>0$ that we can refer to. Recall that $f'(t) = 0$ just means that $\Delta([t,x]) = o(|x-t|)$, so for our $\epsilon>0$, there exists $\delta>0$ such that $|\Delta([t,x])| < \epsilon|x-t|$ for all $x$ within $\delta = \delta(t)$ of $t$. (Here $\delta$ may depend on $t$ and $\epsilon$.)

Let $B_t$ be the interval ("ball") where we can apply the inequality in the previous paragraph: that is, $B_t = (t-\delta(t),t+\delta(t))$. Of course, as we vary over all $t\in[0,1]$, all of $[0,1]$ is "covered" by the $B_t$. In fact, the "completeness/compactness" (this is technical: don't worry about it) of the reals tells us that we only need finitely many of these balls: the interval $[0,1]$ can be covered by finitely many balls $B_{t_1},B_{t_2},\ldots,B_{t_n}$, say with $0 \le t_1 < t_2 < \cdots < t_n \le 1$, and $B_{t_i}$ having radius $r_i = \delta(t_i)$.

In other words, every point in $[0,1]$ is now under the control/jurisdiction (in the territory of) one of these $n$ balls.

Now we can assume without loss of generality that these balls form a "chain from $0$ to $1$" going from left to right, so that $0$ lies in $B_{t_1}$, $B_{t_1}$ intersects $B_{t_2}$ somewhere between $t_1$ and $t_2$, $B_{t_2}$ intersect $B_{t_3}$ somewhere between $t_2$ and $t_3$, etc., and $1$ lies in $B_{t_n}$. This is hopefully intuitive, so don't worry about the details, but the idea is that the only way two neighboring balls could ever not intersect is if one of the balls were "redundant" (contained in a larger ball, so then we could remove the redundant/smaller ball).

Now that we know all the balls "interweave" in this nice fashion, let $u_1$ be any point in both the neighboring balls $B_{t_1},B_{t_2}$ (between $t_1$ and $t_2$). Similarly define $u_2,u_3,\ldots,u_{n-1}$. Also, for convenience, let $u_0 = 0$ and $u_n = 1$; by assumption we know $u_0$ lies in $B_{t_1}$ and $u_n$ lies in $B_{t_n}$.

Implementing the key "telescoping" idea

Instead of breaking up the interval $[0,1]$ at the points $0.1,0.2,\ldots,0.9$, we'll break it up at the points we just constructed: $$0=u_0,t_1,u_1,t_2,u_2,t_3,\ldots,t_{n-1},u_{n-1},t_n,u_n=1.$$ Then neighboring points are either of the form $u_{i-1},t_i$ or $t_i,u_i$. In the former case, we have $$|\Delta(u_{i-1},t_i)| < \epsilon|u_{i-1} - t_i| = \epsilon(t_i - u_{i-1}),$$ since $u_{i-1}$ lies in the territory $B_{t_i}$ of $t_i$, and $u_{i-1} < t_i$. Similarly, in the latter case, we have $$|\Delta(t_i,u_i)| < \epsilon|u_i - t_i| = \epsilon(u_i - t_i),$$ since $u_i$ also lies in the territory $B_{t_i}$ of $t_i$, and $t_i < u_i$.

But we also have the telescoping property $$\Delta(0,1) = \Delta(0=u_0,t_1) + \Delta(t_1,u_1) + \cdots + \Delta(u_{n-1},t_n) + \Delta(t_n,u_n=1).$$ Finally, by the triangle inequality, we have $$\begin{align*} \small{|\Delta(0,1)|} & \le |\Delta(0=u_0,t_1)| + |\Delta(t_1,u_1)| + \cdots + |\Delta(u_{n-1},t_n)| + |\Delta(t_n,u_n=1)| \\ & < \epsilon(t_1-u_0) + \epsilon(u_1-t_1) + \cdots + \epsilon(t_n-u_{n-1}) + \epsilon(u_n-t_n) \\ & = \epsilon(u_n - t_n + t_n - u_{n-1} + \cdots + u_1 - t_1 + t_1 - u_0) \\ &= \epsilon (u_n - u_0) = \epsilon (1-0) = \epsilon. \end{align*}$$

(Intuition: the $t_1-u_0$, $u_1-t_1$, etc. terms are just the lengths of the intervals we broke up $[0,1]$ into, so their total length is just the length $1-0$ of $[0,1]$. In fact, this is another "telescoping" sum, with the $t_1,u_1,\ldots,u_{n-1},t_n$ all canceling out: everything but the endpoints $u_n=1$ and $u_0=0$.)

So $|\Delta(0,1)| < \epsilon$. But remember how we said smallness was relative? Well, $\epsilon>0$ was an arbitrary reference threshold: we can make it as small as we like, so $\Delta(0,1) = 0$, as desired.

Here is an interesting problem in elementary calculus:(taken from a problem solving group on Facebook.)

$f:[0,1]\rightarrow \textbf{R}$ is continuous on $[0,1]$ and differentiable on $(0,1)$ so that $f(0)=0$ and $0\leq f'(x)\leq k f(x)$, where $k\in \textbf{R}^{+}, \ \forall x \in (0,1)$.

Prove that $f(x)=0 \ \forall x \in (0,1)$.