Dini Derivatives And Covering Estimates

One-Sided Slopes at Every Scale

A derivative records the limiting slope of secant lines. When that limit does not exist, the largest and smallest limiting slopes can still describe local behavior. The upper and lower Dini derivatives formalize these extremes by taking a limsup or liminf of one-sided difference quotients. For a monotone function, a covering estimate then shows that points with very large upper slope occupy little Lebesgue measure. This estimate is a key input to almost-everywhere differentiation results, but does not itself assert differentiability.

Fix a compact interval $[a,b]$, where $a<b$, and a finite, nondecreasing function $f:[a,b]\to\mathbb{R}$. Nondecreasing means $f(x)\le f(y)$ whenever $x\le y$. The definitions below also make sense for any real-valued function on $[a,b]$, but the covering estimate will require monotonicity. At an interior point, take positive increments $h$ tending to zero while keeping the quotient inside the interval.

Definition: For $a\le x<b$, the upper and lower right Dini derivatives are $D^+f(x)=\limsup_{h\downarrow0,\ x+h\le b}\frac{f(x+h)-f(x)}{h}$ and $D_+f(x)=\liminf_{h\downarrow0,\ x+h\le b}\frac{f(x+h)-f(x)}{h}$. For $a<x\le b$, the upper and lower left Dini derivatives are $D^-f(x)=\limsup_{h\downarrow0,\ x-h\ge a}\frac{f(x)-f(x-h)}{h}$ and $D_-f(x)=\liminf_{h\downarrow0,\ x-h\ge a}\frac{f(x)-f(x-h)}{h}$. These limits are taken over positive $h$ small enough to remain in the domain; their values may be extended real numbers. The right derivatives are defined at $a$, and the left derivatives at $b$, but not in the unavailable direction.

The orientation of the left quotient matters: it is the change in function value from $x-h$ to $x$, divided by the positive interval length $h$. Thus both left and right quotients are nonnegative for a nondecreasing function. At every point where the ordinary two-sided derivative exists and is finite, all four available Dini derivatives equal that derivative. Indeed, the ordinary difference quotients tend to the derivative from both sides, so their limsup and liminf have that same value. In contrast, a Dini derivative may be infinite or the upper and lower values may differ.

Worked Example: A Linear Function

Let $f(x)=3x+2$ on $[0,2]$. For every admissible $h>0$ and $x$, the right quotient is $\frac{3(x+h)+2-(3x+2)}{h}=\frac{3h}{h}=3$. The left quotient is $\frac{3x+2-(3(x-h)+2)}{h}=\frac{3h}{h}=3$. Therefore all four Dini derivatives equal $3$ wherever their respective directions are defined.

For the strict right upper-slope set at threshold $\lambda=2$, every $x\in[0,2)$ belongs, since $3>2$, and $x=2$ is excluded because a right derivative is not defined there. Thus the set is $[0,2)$, of Lebesgue measure $2$. The function’s total increase is $f(2)-f(0)=8-2=6$, so the covering bound below gives $2\le 6/2=3$. Check: the interval length is $2-0=2$, and the bound is $3$, as computed directly.

Large Upper Slopes and a Covering Estimate

For $\lambda>0$, define the strict right upper-slope set $E_\lambda=\{x\in[a,b):D^+f(x)>\lambda\}$. The strict inequality is important: points where the upper Dini derivative equals $\lambda$ are not included. The estimate bounds the Lebesgue measure $m(E_\lambda)$ in terms of the total increase of $f$, without assuming that $f$ is continuous or differentiable.

Theorem (Covering estimate for a monotone function): Let $f:[a,b]\to\mathbb{R}$ be finite and nondecreasing, and let $\lambda>0$. Then $E_\lambda$ is Lebesgue measurable and $$ m(E_\lambda)\le \frac{f(b)-f(a)}{\lambda}. $$ The corresponding left-sided set $\{x\in(a,b]:D^-f(x)>\lambda\}$ satisfies the same bound.

We first verify measurability so that the measure in the conclusion is well-defined. A monotone real-valued function is Borel measurable: for each real $c$, its strict superlevel set $\{x:f(x)>c\}$ is an interval, possibly with an endpoint included or omitted, and hence is Borel. For fixed $r>0$, the map $x\mapsto f(x+r)-f(x)$ is Borel on $[a,b-r]$. The rational increments below detect a strict upper Dini derivative. In fact, if a quotient at some $h>0$ exceeds $c>0$, then $f(x+h)-f(x)>ch$. By monotonicity, $f(x+r)\ge f(x+h)$ for $r>h$, and choosing rational $r>h$ sufficiently close to $h$ gives $f(x+r)-f(x)>cr$. The increment can be chosen arbitrarily small. Conversely, rational increments are among the allowed increments in the limsup.

Consequently, for $c>0$, the condition that quotients exceed $c$ at arbitrarily small scales is expressible as a countable intersection over positive integers $n$ of countable unions over rational $r$ with $0<r<1/n$ of the Borel conditions $x\le b-r$ and $f(x+r)-f(x)>cr$. The strict upper-level set has the representation $$ E_\lambda=\bigcup_{j=1}^{\infty}\left\{x:\text{quotients exceed }\lambda+\frac1j\text{ at arbitrarily small scales}\right\}. $$ Indeed, if $D^+f(x)>\lambda$, some $\lambda+1/j$ is strictly below $D^+f(x)$; conversely, exceeding $\lambda+1/j$ arbitrarily close to zero implies $D^+f(x)\ge\lambda+1/j>\lambda$. This proves the asserted measurability.

Proof. If $f(b)-f(a)=0$, monotonicity gives $f(x)=f(a)$ throughout the interval. Every right quotient is zero, so $E_\lambda$ is empty and the result follows. Otherwise, consider, for each $x\in E_\lambda$, the closed intervals $[x,x+h]$ with $h>0$ small enough that $x+h\le b$ and $$ \frac{f(x+h)-f(x)}{h}>\lambda. $$ Since $D^+f(x)>\lambda$, there are such $h$ arbitrarily close to zero. These intervals therefore form a Vitali cover of the measurable set $E_\lambda$: each point of the set belongs to intervals in the family of arbitrarily small length.

Apply the Vitali covering lemma to this bounded measurable set and this family of intervals. It gives a finite or countable disjoint family in the sense of disjoint interiors, say $[u_i,v_i]$, whose union covers $E_\lambda$ up to a Lebesgue null set. The lemma applies because $E_\lambda$ is measurable and bounded, and the family contains arbitrarily short intervals at each of its points. For every selected interval, its defining slope condition gives $$ \lambda(v_i-u_i)<f(v_i)-f(u_i). $$ For any finite subfamily ordered from left to right, monotonicity implies $$ \sum_i\bigl(f(v_i)-f(u_i)\bigr)\le f(b)-f(a). $$ To see this, write the intervals in increasing order. Between consecutive intervals, $f(u_{i+1})-f(v_i)\ge0$; adding these nonnegative gaps to the sum of the increments gives $f(v_{\mathrm{last}})-f(u_{\mathrm{first}})\le f(b)-f(a)$. The same bound for the countable family follows by taking limits of its finite partial sums, whose terms are nonnegative. Hence $\lambda\sum_i(v_i-u_i)\le f(b)-f(a)$. Since the intervals cover $E_\lambda$ except for a null set, countable subadditivity of Lebesgue measure yields $m(E_\lambda)\le\sum_i(v_i-u_i)$. Combining the inequalities proves the right-sided estimate.

For the left-sided estimate, reflect the interval by setting $g(t)=-f(a+b-t)$ for $t\in[a,b]$. This is finite and nondecreasing. At $t=a+b-x$, its right quotient is $$ \frac{g(t+h)-g(t)}{h} =\frac{f(x)-f(x-h)}{h}. $$ Reflection preserves Lebesgue measure, so applying the right-sided estimate to $g$ proves the stated bound for the left-sided set. $\square$

The only covering result used here is the Vitali covering lemma, in its standard form for a measurable set and a family of intervals containing arbitrarily short intervals at each point. The estimate itself follows from monotonicity: increments over intervals with disjoint interiors cannot add up to more than the full increase $f(b)-f(a)$. No absolute continuity or finiteness of a derivative is assumed.

Worked Example: The Square-Root Function

Let $f(x)=\sqrt{x}$ on $[0,1]$, and take the threshold $\lambda=1$. For $x>0$ and admissible $h>0$, the right quotient is $$ \frac{\sqrt{x+h}-\sqrt{x}}{h} =\frac{1}{\sqrt{x+h}+\sqrt{x}}. $$ Letting $h\downarrow0$ gives $D^+f(x)=1/(2\sqrt{x})$. At $x=0$, the quotient is $\sqrt{h}/h=1/\sqrt{h}$, which tends to $+\infty$; hence $D^+f(0)=+\infty$. Thus $D^+f(x)>1$ exactly when $x=0$ or $0<x<1/4$, since $1/(2\sqrt{x})>1$ is equivalent to $x<1/4$. The strict level set is $[0,1/4)$, so its measure is $1/4$.

The total increase is $f(1)-f(0)=1-0=1$, and the estimate gives $m(E_1)\le1/1=1$. Check: $m([0,1/4))=1/4\le1$. At the boundary point $x=1/4$, the derivative equals $1$, so that point is correctly excluded by the strict condition $D^+f(x)>1$.

Worked Example: A Jump Seen from the Left

Define $f(x)=0$ for $0\le x<1/2$ and $f(x)=1$ for $1/2\le x\le1$. At $x=1/2$, every admissible right quotient equals $(1-1)/h=0$, so $D^+f(1/2)=0$. Every admissible left quotient equals $(1-0)/h=1/h$, which tends to $+\infty$; therefore $D^-f(1/2)=+\infty$.

For any fixed $\lambda>0$, the left upper-slope set contains $1/2$, but it contains no other point: away from the jump, sufficiently small left increments stay within a constant part of the function and have quotient zero. Its measure is therefore zero. The total increase is $f(1)-f(0)=1-0=1$, so the left-sided estimate reads $0\le1/\lambda$. Check: a singleton has Lebesgue measure zero, while the right quotient at the jump is zero; the two directions give different answers because the function’s value at the jump is its right-hand value.

Why the Covering Step Matters

A common mistake is to add the lengths or function increments of every interval that witnesses a large slope. Those intervals may overlap, so their increments can count the same part of the function more than once. For example, for $f(x)=x$ on $[0,1]$, the intervals $[0,3/4]$ and $[1/4,1]$ each have increment $3/4$. Adding them gives $3/2$, which exceeds the total increase $f(1)-f(0)=1$. The calculation is wrong as a bound on the total increase because the intervals overlap on $[1/4,3/4]$. The Vitali selection is the correction: use intervals with disjoint interiors, for which monotonicity bounds the sum of increments by the full increase.

The estimate is useful because it turns a pointwise condition into a measure bound. If $\lambda$ increases while $f(b)-f(a)$ stays fixed, the set of points with $D^+f(x)>\lambda$ must shrink in measure. This is a quantitative control of exceptional large slopes. It does not say that the Dini derivatives are finite everywhere, as the square-root example shows at $0$, nor does it establish that the upper and lower Dini derivatives agree almost everywhere. Those are distinct questions.

Check Your Understanding

Use the definitions with the direction and strictness of each condition stated explicitly.

For $f(x)=x^2$ on $[0,1]$, compute the right and left Dini derivatives at an interior point $x$, and find the strict right upper-slope set for a fixed $\lambda>0$.
Explain why the right derivative at the jump point in the step-function example is zero while the left derivative is infinite.
For a finite nondecreasing $f:[a,b]\to\mathbb{R}$ with $f(a)=f(b)$, show directly that every strict right upper-slope set with $\lambda>0$ is empty.
In the covering proof, where is monotonicity used to bound the sum of increments over intervals with disjoint interiors?
Why does a point where $D^+f(x)=\lambda$ not belong to $E_\lambda$, and why is that distinction relevant to the stated level-set definition?

Dini Derivatives And Covering Estimates

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