Darboux Property of Derivatives

Derivatives Can Be Discontinuous, but They Cannot Jump

A derivative need not be continuous. As the previous tutorial showed, its values can oscillate or even become unbounded near a point. But discontinuity does not mean that every pattern of nearby values is possible. Darboux’s Theorem for Derivatives, whose proof is given in the next tutorial, says that if a derivative takes two values, it must also take every value between them at some intermediate input. This intermediate value property rules out a jump that skips a whole range of values.

The location of the inputs matters. Darboux’s Theorem applies only to points in the interval where the function is differentiable. When testing a claimed derivative near a point \(a\), choose the test inputs inside the domain and as close to \(a\) as needed. We will use that principle to obtain two consequences: a derivative cannot cross from one side of a value to the other while omitting that value nearby, and finite one-sided limits of a derivative must agree.

A Local Test for a Forbidden Crossing

Here is a useful way to apply the intermediate value property. Suppose a function \(g\) is alleged to be a derivative near \(a\). If values of \(g\) on opposite sides of \(a\) lie on opposite sides of some number \(r\), Darboux’s Theorem requires \(g\) to equal \(r\) between those inputs. If \(g\) omits \(r\) throughout a punctured neighborhood of \(a\), the only possible intermediate point left is \(a\) itself. Choosing \(r\ne g(a)\) rules that out too.

Theorem: Let \(I\) be an open interval containing \(a\), and suppose \(g=f'\) on \(I\) for some differentiable function \(f\). There cannot be a number \(r\ne g(a)\) and a \(\delta>0\) such that \(g(x)\ne r\) whenever \(0<|x-a|<\delta\), while there are points \(u,v\in I\) with \(a-\delta<u<a<v<a+\delta\) for which \(g(u)<r<g(v)\) or \(g(v)<r<g(u)\).

Proof. The two values \(g(u)\) and \(g(v)\) lie on opposite sides of \(r\). Darboux’s Theorem for Derivatives therefore gives a point \(c\) strictly between \(u\) and \(v\) such that \(g(c)=r\). If \(c=a\), then \(g(a)=r\), contrary to the hypothesis \(r\ne g(a)\). If \(c\ne a\), then \(0<|c-a|<\delta\), contrary to the hypothesis that \(g\) omits \(r\) throughout that punctured neighborhood. Both possibilities lead to a contradiction. \(\square\)

The requirement \(r\ne g(a)\) is essential to this argument. Without it, Darboux’s Theorem could be satisfied by \(c=a\), even if \(g\) never equals \(r\) at any other nearby input. When the value at \(a\) lies inside the range between the values on either side, one must select a different intermediate value that is omitted near \(a\), if such a value is available.

Worked Example: A Step Function Cannot Be a Derivative Near Its Jump

Define \(g:\mathbb{R}\to\mathbb{R}\) by

g(x)= \begin{cases} -1, & x<0,\\ 2, & x=0,\\ 1, & x>0. \end{cases}

Suppose, for a contradiction, that \(g=f'\) on an open interval \(I\) containing zero. Because \(I\) is open, there is an \(\varepsilon>0\) such that \((-\varepsilon,\varepsilon)\subseteq I\). Choose

u=-\frac{\varepsilon}{2}, \qquad v=\frac{\varepsilon}{2}.

Both inputs lie in \(I\), and direct substitution gives \(g(u)=-1\) and \(g(v)=1\). The value \(r=0\) lies strictly between them and differs from \(g(0)=2\). In fact, \(g\) never equals zero anywhere. Darboux’s Theorem applied to \(u\) and \(v\) would require some \(c\in(u,v)\) with \(g(c)=0\), which is impossible. Thus \(g\) is not a derivative on any open interval containing zero.

The choice of \(\varepsilon\) ensures that both test inputs are in the interval where \(g\) is alleged to be a derivative. No assumption that the interval contains \(-1\) or \(1\) is needed: those numbers are values of \(g\), not inputs to \(g\).

Finite One-Sided Limits Must Match

A jump discontinuity often appears as two different finite one-sided limits. For derivatives, Darboux’s Theorem prevents this. The key is to choose an intermediate value different from the derivative at the point, so the intermediate value cannot be supplied by the point itself.

Theorem: Let \(I\) be an open interval, let \(a\in I\), and let \(f:I\to\mathbb{R}\) be differentiable. If the finite one-sided limits \(\lim_{x\to a,\,x<a}f'(x)=L_-\) and \(\lim_{x\to a,\,x>a}f'(x)=L_+\) both exist, then \(L_-=L_+=f'(a)\).

Proof. Write \(g=f'\). First suppose, seeking a contradiction, that \(L_-<L_+\). There is a number \(r\) strictly between \(L_-\) and \(L_+\) such that \(r\ne g(a)\), since an open interval contains more than one number. The one-sided limits imply that, for all \(x<a\) sufficiently close to \(a\), \(g(x)<r\), and for all \(y>a\) sufficiently close to \(a\), \(g(y)>r\). To see why, choose a positive tolerance smaller than both \(r-L_-\) and \(L_+-r\), and use the definitions of the two limits.

Choose such \(x\) and \(y\) within \(I\). Darboux’s Theorem gives \(c\in(x,y)\) with \(g(c)=r\). Since \(r\ne g(a)\), \(c\ne a\). If \(c<a\), it is sufficiently close to \(a\) on the left, where \(g(c)<r\); if \(c>a\), it is sufficiently close on the right, where \(g(c)>r\). Both conclusions contradict \(g(c)=r\). Thus \(L_-<L_+\) is impossible. Interchanging the roles of the two sides rules out \(L_+<L_-\), so \(L_-=L_+\).

Call their common value \(L\). The two one-sided limits together give the finite punctured limit \(\lim_{x\to a,\,x\ne a}f'(x)=L\). The theorem from Noncontinuous Derivatives states that if a derivative has a finite punctured limit at a point, that limit equals the derivative at the point. Hence \(f'(a)=L\), as required. \(\square\)

This result rules out more than a step-shaped formula. Whenever both one-sided limits exist as finite real numbers, even complicated behavior away from \(a\) cannot make those limits unequal or make their common value differ from \(f'(a)\). The conclusion does not apply if one of the limits fails to exist or is infinite.

Worked Example: Unequal Finite Limits Cannot Describe a Derivative

Consider the function \(g\) given by

g(x)= \begin{cases} -2, & x<0,\\ 1, & x=0,\\ 4, & x>0. \end{cases}

Its left-hand limit at zero is \(-2\), and its right-hand limit is \(4\). The number \(r=0\) lies strictly between these limits and satisfies \(r\ne g(0)\). For any \(\varepsilon>0\), inputs \(u=-\varepsilon/2\) and \(v=\varepsilon/2\) have \(g(u)=-2<0<4=g(v)\), while \(g\) never equals zero. If \(g\) were a derivative on an interval containing zero, Darboux’s Theorem applied to such inputs in that interval would require \(g(c)=0\) somewhere between them. This contradiction shows directly why the two different one-sided limits are forbidden.

A Common Limit Cannot Disagree with the Derivative Value

The preceding theorem has a second implication worth separating from the unequal-limit case. Even if the nearby derivative values approach the same finite number from both sides, the value assigned at \(a\) is not free to differ from that limit. This also follows from the finite-limit theorem in Noncontinuous Derivatives.

Worked Example: A Punctured Limit with the Wrong Point Value

Define \(g(x)=0\) for \(x\ne0\), and \(g(0)=1\). Its punctured limit at zero is zero, whereas \(g(0)=1\). Suppose \(g=f'\) on an open interval containing zero. The finite-limit theorem from Noncontinuous Derivatives would then imply

g(0)=\lim_{x\to0,\,x\ne0}g(x)=0,

contradicting \(g(0)=1\). The intermediate value property gives the same obstruction directly: choose any positive \(v\) in the interval. Then \(g(0)=1\) and \(g(v)=0\), so Darboux’s Theorem requires \(g\) to take the value \(1/2\) somewhere between \(0\) and \(v\). But \(g\) takes only the values \(0\) and \(1\). Therefore \(g\) cannot be a derivative on that interval.

Discontinuous Derivatives Still Take Intermediate Values

The Darboux property does not say that a derivative is continuous. It says that between any two inputs, the derivative takes every value between its values at those inputs. The next example verifies both facts for one derivative: it is discontinuous at zero, yet its oscillating values are consistent with the intermediate value property.

Worked Example: An Oscillating Derivative Without a Jump

Define \(f(0)=0\) and \(f(x)=x^2\sin(1/x)\) for \(x\ne0\). At zero, the difference quotient is

\frac{f(h)-f(0)}{h}=h\sin(1/h), \qquad \left|h\sin(1/h)\right|\leq |h|.

It tends to zero, so \(f'(0)=0\). For \(x\ne0\), the product and chain rules give

f'(x)=2x\sin(1/x)-\cos(1/x).

For positive integers \(n\), set \(x_n=(2\pi n)^{-1}\) and \(y_n=((2n+1)\pi)^{-1}\). Both sequences tend to zero, and substitution gives

f'(x_n)=2x_n\sin(2\pi n)-\cos(2\pi n)=-1, \qquad f'(y_n)=2y_n\sin((2n+1)\pi)-\cos((2n+1)\pi)=1.

Since \(f'(0)=0\), these sequences show that \(f'\) is not continuous at zero. But this is not a jump: for each \(n\), Darboux’s Theorem applied between \(y_n\) and \(x_n\) ensures that \(f'\) takes every value between \(1\) and \(-1\) somewhere in \((y_n,x_n)\). In particular, it takes the value zero there. The derivative oscillates through intermediate values rather than skipping them.

How to Use the Darboux Property

When testing whether a proposed function \(g\) could be a derivative, look for two inputs in its alleged domain whose \(g\)-values straddle a number that \(g\) omits between those inputs. If the inputs lie on opposite sides of a suspected jump, make sure they really are inside the interval under consideration. If the only possible point that might supply the missing value is \(a\), choose an intermediate value \(r\ne g(a)\). These details turn the intermediate value property into a rigorous obstruction.

A common mistake is to infer that every discontinuous function fails to be a derivative. The oscillating example disproves that inference: its derivative is discontinuous but retains the Darboux property. The restriction is specifically on skipped intermediate values. Another mistake is to assume that Darboux’s Theorem makes derivatives continuous. It does not; it constrains how derivative values can change between inputs, not whether they approach a single value as the input approaches a point.

Choose inputs in the domain.
For a local argument at \(a\), first choose a neighborhood contained in the interval where \(g\) is alleged to be a derivative.

Identify a skipped value.
Find \(r\) strictly between two derivative values and check whether \(g\) takes that value anywhere between the corresponding inputs.

Check the point \(a\) separately.
If the inputs straddle \(a\), choose \(r\ne g(a)\) when using a punctured-neighborhood omission argument.

Check Your Understanding

Use the Darboux property and the finite-limit results to answer the following questions.

Why must the inputs in a local Darboux argument lie inside the interval where the function is alleged to be a derivative?
In the local crossing theorem, why is it necessary to choose \(r\ne g(a)\)?
What does the finite one-sided limits theorem imply if \(f'\) has finite left- and right-hand limits at \(a\)?
Why does the function equal to zero off zero and equal to one at zero fail to be a derivative near zero?
How can a derivative be discontinuous at zero and still satisfy the Darboux property?