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Does the following claim hold?

Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f'(x)\neq 0$ for all $x\in \mathbb R$. Then $f'(x)>0 \, \forall x\in \mathbb R$ or $f'<0\, \forall x\in \mathbb R$.

Note that $f$ is not assumed to be continuously differentiable.

Pavel Kocourek
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2 Answers2

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The claim is true. (And perhaps trivial for anyone who remembers the Darboux's theorem, which I must admit I learned only recently.)

Proof. For contradiction, suppose that $f:\mathbb R \to \mathbb R$ does not have a zero derivative anywhere, but there are points $a,b\in \mathbb R$ such that $f'(a)<0<f'(b)$. Applying the Darboux's theorem it follows that there exists $x\in [a,b]$ such that $f'(x)=0$. Contradiction. $\tag*{$\Box$}$

Pavel Kocourek
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  • I learned only recently --- You might be interested in the MSE question How irregular can $f'$ be beyond Darboux's Theorem? Also the 3rd paragraph of my 21 May 2008 sci.math post, altered only by changing ASCII math symbols to LaTeX math symbols (for the references I've cited, see the original sci.math post and here for Clarkson's 1947 paper): Derivatives also satisfy (continued) – Dave L. Renfro Jan 13 '23 at 10:18
  • stronger versions of the intermediate value property. For example, $f'$ having the intermediate value property implies that the set $E(a,b) = {x \in J: ; a < f'(x) < b }$ is either empty or has cardinality of the continuum. Denjoy (1916) (later rediscovered by J. A. Clarkson in 1947) improved this by showing that each of the sets $E(a,b)$ is either empty or has positive measure. A further strengthening of the IVP for derivatives, stronger than this positive measure result, was made by Zygmunt Zahorski [3]. (continued) – Dave L. Renfro Jan 13 '23 at 10:18
  • Then Clifford E. Weil [4] came up with a property satisfied by all derivatives that is strictly stronger than Zahorski's property and, after this, Peter S. Bullen and D. N. Sarkhel [5] came up with a property satisfied by all derivatives that is strictly stronger than Weil's property. (This last paper might not the strongest result so far published, by the way.) – Dave L. Renfro Jan 13 '23 at 10:18
  • @DaveL.Renfro Thank you for supplying the historical context and list of possible extensions. – Pavel Kocourek Jan 13 '23 at 14:11
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Suppose this function is not continuously differentiable. If $f$ is continuously differentiable then all is well and the result follows without difficulty.

Suppose that were not the case, $f'$ is discontinuous. We know that there can be no "simple" discontinuities of $f'$, one where both left hand and right hand limits of $f'$ exist but do not satisfy the criteria for continuity. A proof can be found, for example, in Rudin.

Since it is not possible for $f'$ to intersect the x-axis or "jump" across it, it should either be strictly positive or strictly negative.

LaxHat1376
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    A derivative has no simple discontinuities because it has the intermediate value property, and that is because of Darboux's theorem. – Martin R Jan 13 '23 at 08:43