Lipschitz and Continuously Differentiable Functions Are Absolutely Continuous

Increment Bounds and Absolute Continuity

Absolute continuity controls the total change of a function across a finite collection of disjoint intervals. A useful way to establish it is to find a constant that bounds each increment by the length of its interval. A Lipschitz condition provides exactly such a bound. For a continuously differentiable function on a compact interval, continuity of the derivative makes it bounded, and the mean value theorem then supplies a Lipschitz bound. This tutorial proves both implications and computes the resulting estimates.

Throughout, let $a<b$ be real numbers and let $f:[a,b]\to\mathbb{R}$ be real-valued. A finite collection of intervals $(x_k,y_k)\subseteq[a,b]$ is disjoint if their interiors are pairwise disjoint. Absolute continuity is defined by controlling sums of increments whenever the total interval length is small.

Definition: The function $f:[a,b]\to\mathbb{R}$ is absolutely continuous on $[a,b]$, written $f\in AC[a,b]$, if for every $\varepsilon>0$ there exists $\delta>0$ such that, for every finite pairwise disjoint collection $(x_k,y_k)$ with $a\le x_k<y_k\le b$, $$ \sum_{k=1}^{n}(y_k-x_k)<\delta \quad\Longrightarrow\quad \sum_{k=1}^{n}|f(y_k)-f(x_k)|<\varepsilon. $$

The same $\delta$ must work regardless of the number of intervals and their locations. An estimate of the form $|f(y)-f(x)|\le C|y-x|$, with one fixed finite constant $C$, is especially effective: summing it over any such collection gives a bound by $C$ times the total length.

Lipschitz Functions Are Absolutely Continuous

Definition: A function $f:[a,b]\to\mathbb{R}$ is Lipschitz with constant $L\ge0$ if $$ |f(y)-f(x)|\le L|y-x| \qquad\text{for all }x,y\in[a,b]. $$

Theorem: If $f:[a,b]\to\mathbb{R}$ is Lipschitz with constant $L\ge0$, then $f\in AC[a,b]$. More precisely, for every finite pairwise disjoint collection $(x_k,y_k)$, $$ \sum_{k=1}^{n}|f(y_k)-f(x_k)| \le L\sum_{k=1}^{n}(y_k-x_k). $$

Proof. For each $k$, the Lipschitz condition gives $|f(y_k)-f(x_k)|\le L(y_k-x_k)$, since $x_k<y_k$. Adding these $n$ inequalities proves the displayed increment bound. If $L>0$, fix $\varepsilon>0$ and choose $\delta=\varepsilon/L$. Whenever the total interval length is less than $\delta$, the increment bound gives $\sum_k|f(y_k)-f(x_k)|\le L\sum_k(y_k-x_k)<L\delta=\varepsilon$. If $L=0$, the Lipschitz condition says $f(y)=f(x)$ for every $x,y\in[a,b]$. Every sum of increments is then zero, so any $\delta>0$, for example $\delta=1$, works. Thus $f\in AC[a,b]$ in both cases. $\square$

Worked Example: A Linear Function with a Known Lipschitz Constant

Let $f(x)=3x-2$ on $[-1,2]$. Find a Lipschitz constant, give a $\delta$ that works for $\varepsilon=1/4$, and check the increment estimate on two intervals.

For any $x,y\in[-1,2]$, $$ |f(y)-f(x)|=|(3y-2)-(3x-2)|=3|y-x|. $$ Thus $L=3$ is a Lipschitz constant. For $\varepsilon=1/4$, choose $\delta=\varepsilon/L=(1/4)/3=1/12$. For any finite pairwise disjoint intervals with total length less than $1/12$, the sum of increments is at most three times that total length, hence is less than $3(1/12)=1/4$.

For a numerical check, take $(0,1/48)$ and $(1/2,25/48)$. Their lengths are $1/48$ and $1/48$, so their total length is $1/24<1/12$. The increments are $|f(1/48)-f(0)|=3/48=1/16$ and $|f(25/48)-f(1/2)|=3(1/48)=1/16$. Their sum is $1/8$, while three times the total length is $3(1/24)=1/8$. The bound is attained, and $1/8<1/4$, as required.

Check: Substituting the endpoints directly gives $f(0)=-2$, $f(1/48)=-29/16$, $f(1/2)=-1/2$, and $f(25/48)=-9/16$. The two absolute differences are $3/16$? For the first, $|-29/16-(-2)|=3/16$, which reveals that the interval endpoint calculation must be checked carefully: $f(1/48)=3/48-2=1/16-2=-31/16$, not $-29/16$. Therefore the first difference is $1/16$. For the second, $f(25/48)=75/48-2=25/16-2=-7/16$, so its difference from $-1/2=-8/16$ is $1/16$. The direct evaluations confirm the increments and their sum $1/8$.

Continuously Differentiable Functions on Compact Intervals

Suppose $f\in C^1[a,b]$, meaning $f$ is real-valued and continuously differentiable on $[a,b]$, with the derivative understood on the interior and continuously extended to the endpoints. By the extreme value theorem, the continuous function $|f'|$ has a finite maximum on the compact interval. The mean value theorem converts this bound into an increment estimate.

Theorem: If $f\in C^1[a,b]$ and $M=\max_{t\in[a,b]}|f'(t)|$, then for all $x,y\in[a,b]$, $$ |f(y)-f(x)|\le M|y-x|. $$ Consequently $f\in AC[a,b]$. If $M>0$, $\delta=\varepsilon/M$ works for a given $\varepsilon>0$; if $M=0$, any $\delta>0$ works.

Proof. The maximum $M$ exists and is finite because $|f'|$ is continuous on the compact interval $[a,b]$. If $x\ne y$, the mean value theorem applied between $x$ and $y$ gives a point $c$ strictly between them such that $f(y)-f(x)=f'(c)(y-x)$. Therefore $|f(y)-f(x)|=|f'(c)||y-x|\le M|y-x|$. If $x=y$, both sides are zero, so the same bound holds. For a finite pairwise disjoint collection, add the bound for each interval to obtain $\sum_k|f(y_k)-f(x_k)|\le M\sum_k(y_k-x_k)$. When $M>0$, choosing $\delta=\varepsilon/M$ makes this sum strictly less than $\varepsilon$ whenever the total length is less than $\delta$. When $M=0$, the increment bound makes every sum zero, so any positive $\delta$ works. Thus $f\in AC[a,b]$. $\square$

Worked Example: A Quadratic on a Compact Interval

Let $f(x)=x^2$ on $[-1,2]$. Find a derivative-based increment bound and a valid $\delta$ for $\varepsilon=1/5$.

Here $f'(x)=2x$. On $[-1,2]$, $|2x|\le4$, and equality holds at $x=2$, so $M=4$. The theorem gives $|y^2-x^2|\le4|y-x|$ for every $x,y\in[-1,2]$. For $\varepsilon=1/5$, choose $\delta=(1/5)/4=1/20$. Any finite disjoint family whose total length is less than $1/20$ then has total increment less than $4(1/20)=1/5$.

For a check, use $(-1,-99/100)$ and $(1,101/100)$. Each interval has length $1/100$, so the total length is $1/50<1/20$. The increments are $$ |(-99/100)^2-(-1)^2|=\left|\frac{9801}{10000}-1\right|=\frac{199}{10000}, \qquad |(101/100)^2-1|=\frac{201}{10000}. $$ Their sum is $400/10000=1/25$. The general bound gives at most $4(1/50)=2/25$, and indeed $1/25<2/25$. Also $1/25<1/5$, as the absolute-continuity condition requires.

Check: Direct substitution into $f(x)=x^2$ gives exactly the two displayed increments. The derivative bound is valid but not always sharp for a particular collection; the computed sum $1/25$ is below its upper bound $2/25$.

Worked Example: The Sine Function on a Compact Interval

Let $f(x)=\sin x$ on $[0,\pi]$. Find a derivative-based increment bound and verify it on two specified intervals.

For a separate numerical check of the increment bound, consider $(0,\pi/6)$ and $(\pi/2,2\pi/3)$. Their total length is $\pi/6+\pi/6=\pi/3$. Using the exact values $\sin0=0$, $\sin(\pi/6)=1/2$, $\sin(\pi/2)=1$, and $\sin(2\pi/3)=\sqrt{3}/2$, the sum of the increments is $$ \left|\frac12-0\right|+\left|\frac{\sqrt3}{2}-1\right| =\frac12+1-\frac{\sqrt3}{2} =\frac32-\frac{\sqrt3}{2}. $$ The second absolute value has the stated sign because $\sqrt3<2$. The increment bound predicts at most $\pi/3$; indeed each interval separately has increment no greater than its length by the theorem. This verifies the bound without requiring the two intervals to be adjacent.

Check: The first increment is $1/2$, and the second is $1-\sqrt3/2$, both nonnegative. Their sum is therefore $3/2-\sqrt3/2$, in agreement with direct substitution.

A Common Error: Bounding Each Interval Separately

In the definition of absolute continuity, smallness is required of the sum of all the increments, not merely of each increment individually. A tempting but invalid argument is to choose a $\delta$ so that every interval of length less than $\delta$ has a small increment, and then conclude that the sum is small. The number of intervals is not fixed, so this reasoning provides no bound on the sum.

For a concrete incorrect calculation, take $f(x)=x$ on $[0,1]$ and $\varepsilon=1/10$. The function is $1$-Lipschitz. If one incorrectly uses $\delta=1/10$ and checks only that each interval has length less than $1/10$, the two intervals $(0,9/100)$ and $(1/2,59/100)$ pass that individual test: each has length $9/100<1/10$. But their increments sum to $9/100+9/100=18/100=9/50>1/10$. So the claimed conclusion fails.

The correction is to control the total length. For $f(x)=x$, the increment bound is exactly $\sum_k|f(y_k)-f(x_k)|=\sum_k(y_k-x_k)$. Taking $\delta=1/10$ in the actual definition ensures that this total is less than $1/10$. More generally, the Lipschitz and $C^1$ proofs work because they sum a single uniform increment estimate over all the intervals before applying the condition on total length.

On a compact interval, the conclusion for $C^1$ functions follows from a bounded derivative; no claim that every continuous or uniformly continuous function is absolutely continuous is involved. The essential estimate here is a linear bound on every increment, with the same constant throughout the interval. Compactness is what guarantees the needed derivative bound in the $C^1$ case.

Check Your Understanding

In each question, keep track of whether the estimate concerns one interval or the sum over a finite disjoint family.

If $f$ is Lipschitz with constant $L=0$, explain why it is absolutely continuous without choosing $\delta=\varepsilon/L$.
For $f(x)=5x+1$ on $[0,1]$, find a Lipschitz constant and a value of $\delta$ that works for $\varepsilon=1/2$.
For $f(x)=x^2$ on $[0,3]$, compute $\max_{[0,3]}|f'|$ and state the resulting increment bound.
Explain why a bound on the length of each interval separately does not, by itself, control the sum of increments when the number of intervals can vary.
If $f\in C^1[a,b]$ and $f'$ vanishes everywhere on $[a,b]$, what does the increment bound imply about $f(y)-f(x)$ for arbitrary $x,y\in[a,b]$?

Lipschitz and Continuously Differentiable Functions Are Absolutely Continuous

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