Darboux's Theorem

A derivative keeps the intermediate value property even when discontinuous. Watch a wildly oscillating derivative step on every level.

A derivative cannot jump across values. To get from one value to another it must pass through everything between. Even a discontinuous derivative obeys this rule.

For example, a 🌡️ thermometer sliding from −1° to 1° passes through every temperature between. A derivative likewise never jumps over an intermediate value.

Examples & Exercises

Prove Darboux's theorem: if $f$ is differentiable on $[a,b]$ , then $f'$ takes every value between $f'(a)$ and $f'(b)$ . (Hint: use the Extreme Value Theorem scene and Fermat's interior-extremum lemma.)

g(x) = f(x) - cx,\qquad g'(a) < 0 < g'(b)\ \Longrightarrow\ \exists\, x_0 \in (a,b):\ g'(x_0) = 0

g'(x_0) = 0\ \Longrightarrow\ f'(x_0) = c

Zoom 70 (window

[\pm 0.004]

), test level

y = 0.4

: the discontinuous

f'

meets the horizontal line 704 times in this window (at grid resolution) — the Darboux property in action.

Show proof

First Fermat's lemma: if $g$ has a local max (or min) at an interior point $x_0 \in (a,b)$ and $g'(x_0)$ exists, then $g'(x_0) = 0$ .
For a local min, $\frac{g(x_0+h)-g(x_0)}{h} \geq 0$ for small $h > 0$ and $\leq 0$ for $h < 0$ ; the two-sided limit defining the derivative is forced to 0.
A local max is the same with signs flipped.
Assume $f'(a) < f'(b)$ (apply the argument to $-f$ if $>$ ; the equal case is trivial).
Fix $c$ between them and put $g(x) = f(x) - cx$ : then $g$ is differentiable, hence continuous, on $[a,b]$ , and $g'(x) = f'(x) - c$ gives $g'(a) < 0 < g'(b)$ .
By the Extreme Value Theorem (the EVT scene), $g$ attains its minimum at some $x_0 \in [a,b]$ .
The point $x_0$ is interior: since $g'(a) < 0$ , $\frac{g(a+h)-g(a)}{h} \to g'(a) < 0$ , so $g(a+h) < g(a)$ for small $h > 0$ — $a$ is not the minimizer.
Likewise $g'(b) > 0$ gives $g(b-h) < g(b)$ for small $h > 0$ — $b$ is not the minimizer either.
Hence $x_0 \in (a,b)$ .
Applying Fermat's lemma at the interior minimizer $x_0$ gives $g'(x_0) = 0$ , hence $f'(x_0) = c$ . Since $c$ was arbitrary between $f'(a)$ and $f'(b)$ , $f'$ has the intermediate value property — even though it may be discontinuous.

Prove that $f(x) = x^2\sin(1/x)$ ( $x \neq 0$ ), $f(0) = 0$ is differentiable everywhere, but its derivative $f'$ is discontinuous at $0$ . (Get $f'(0)$ from the definition and $f'(x)$ for $x \neq 0$ by the product and chain rules.)

f'(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} h\sin(1/h) = 0

f'(x) = 2x\sin(1/x) - \cos(1/x) \quad (x \neq 0)

a_k = \frac{1}{2k\pi}:\ f'(a_k) = -1, \qquad b_k = \frac{1}{(2k+1)\pi}:\ f'(b_k) = 1 \quad (a_k, b_k \to 0)

Zoom 100 (window

[\pm 0.0005]

y = 0

: no amount of zooming calms the

-\cos(1/x)

oscillation — 1579 crossings of

f'

with the level in this window (at grid resolution).

Show proof

Get $f'(0)$ from the definition: $\frac{f(h)-f(0)}{h} = \frac{h^2\sin(1/h)}{h} = h\sin(1/h)$ , and $|h\sin(1/h)| \leq |h|$ , so the squeeze theorem gives the limit $0$ .
The numbers agree: $\approx -0.00506$ at $h = 0.01$ and $\approx -3.5 \times 10^{-7}$ at $h = 10^{-6}$ .
Hence $f'(0) = 0$ .
For $x \neq 0$ , the product and chain rules give $f'(x) = 2x\sin(1/x) + x^2\cos(1/x)\cdot(-1/x^2) = 2x\sin(1/x) - \cos(1/x)$ . So $f$ is differentiable at every point, including $0$ .
But $\lim_{x \to 0} f'(x)$ does not exist.
On $a_k = \frac{1}{2k\pi} \to 0$ we have $\sin(1/a_k) = 0$ and $\cos(1/a_k) = 1$ , so $f'(a_k) = -1$ ; on $b_k = \frac{1}{(2k+1)\pi} \to 0$ we have $\sin(1/b_k) = 0$ and $\cos(1/b_k) = -1$ , so $f'(b_k) = 1$ (e.g. exactly $-1.000000$ and $1.000000$ for $k = 5, 10, 20$ ).
If the limit existed it would have to be both $-1$ and $1$ — a contradiction.
Since $f'(0) = 0$ but $\lim_{x \to 0} f'(x)$ does not exist, $f'$ is discontinuous at $0$ — the derivative of a differentiable function need not be continuous. Yet by Darboux's theorem (problem 1), this discontinuous $f'$ still has the intermediate value property.

Prove as a corollary of Darboux's theorem: no function can have the Heaviside step function $H$ ( $0$ for $x < 0$ , $1$ for $x \geq 0$ ) as its derivative.

F' = H,\quad F'(-1) = 0 < \tfrac12 < 1 = F'(1)\ \xrightarrow{\text{Darboux}}\ \exists\, \xi \in (-1,1):\ F'(\xi) = \tfrac12 \notin \{0,1\}\ \Rightarrow\Leftarrow

Zoom 0 (window

[\pm 0.5]

y = 0.5

: even the wildly oscillating

f'

passes this level 46 times (at grid resolution) — a derivative's discontinuity can only be oscillation, never a jump.

Show proof

Suppose some $F$ , differentiable on all of $\mathbb{R}$ , had $H$ as its derivative.
Take $a = -1$ , $b = 1$ : then $F'(-1) = H(-1) = 0$ and $F'(1) = H(1) = 1$ .
Since $0 < \frac{1}{2} < 1$ , Darboux's theorem (problem 1) gives some $\xi \in (-1,1)$ with $F'(\xi) = \frac{1}{2}$ .
But $H$ takes only the values $0$ and $1$ , so $F'(\xi) = H(\xi) = \frac{1}{2}$ is impossible — a contradiction. Hence no function has $H$ as its derivative.
In general: a derivative cannot have a jump discontinuity.
At a point where the one-sided limits exist but differ, Darboux forces the derivative to take the values strictly between them — which a step function never does.
A derivative's discontinuities can only be essential oscillation, like the $-\cos(1/x)$ of problem 2.

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Examples & Exercises