A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$; consequently two such functions with the same derivative differ by a constant

Let $I\subseteq \mathbb{R}$ be order-convex (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) and let $f:I\to \mathbb{R}$ be continuous on $I$ (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point) and differentiable at every point of $I$ that is interior to $I$ (Interior, closure, boundary and exterior of a subset of $\mathbb{R}$, The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set), with

$f^{\mathrm{\prime }}(x)=0\text{at every interior point }x\text{ of }I\mathrm{.}$

Then $f$ is constant on $I$: there is a real $k$ with $f(x)=k$ for every $x\in I$.

Consequently, if $f,g:I\to \mathbb{R}$ are both continuous on $I$ and both differentiable at every interior point of $I$, with $f^{\mathrm{\prime }}(x)=g^{\mathrm{\prime }}(x)$ at every interior point $x$, then there is a real $k$ with

$f(x)=g(x)+k\text{for every }x\in I\mathrm{.}$

Order-convexity of $I$ is essential and is not a convenience. The conclusion is false on a domain that falls into separate pieces, since a function may be constant on each piece with different constants; nothing in the proof would survive, because the mean value theorem is applied to the segment joining two points of the domain and that segment must lie in the domain.

The hypothesis is imposed only at interior points. At an endpoint of $I$ nothing is asked at all: $f$ need not be differentiable there, and the proof never evaluates a difference quotient at an endpoint, since it applies the mean value theorem on a segment $[u,v]\subseteq I$ and uses the derivative only at points of $(u,v)$, all of which are interior to $I$. What is not meant is that the derivative at an endpoint is free to be nonzero: once $f$ is known to be constant its difference quotient at an endpoint is constantly $0$, so wherever $f^{\mathrm{\prime }}$ exists at an endpoint it is $0$ too. That is a consequence of the theorem, not a hypothesis of it.

• What is really being used. Only that any two points of $I$ are joined by a segment inside $I$, and that on such a segment the mean value theorem turns a vanishing derivative into a vanishing increment. Both facts are about $I$, not about $f$, which is why order-convexity is the hypothesis and not, say, openness or connectedness in some other sense.

• The second claim is the uniqueness half of antidifferentiation. It says that a function on an interval is determined by its derivative up to one additive constant. It says nothing about existence: that some given function is a derivative is a separate question, settled by different machinery, and this page does not address it.

• A vanishing derivative at every interior point is far stronger than a vanishing derivative somewhere. The theorem consumes the hypothesis at every point of a segment at once; a single stationary point carries no information about $f$ anywhere else, which is what Fermat's interior extremum theorem: if $f$ has a local extremum at a point $c$ interior to its domain and is differentiable at $c$, then $f^{\mathrm{\prime }}(c)=0$ already made clear from the other side.