Algebraic Closure Properties of Absolutely Continuous Functions

Absolute Continuity and Algebraic Operations

Absolute continuity is stronger than continuity: it controls the total change of a function across any finite collection of short, disjoint intervals. That control is stable under addition and scalar multiplication. Products require one additional ingredient: the factors must be bounded so that changes in one factor can be controlled while the other factor is held at its size on the whole interval.

Fix a compact interval $[a,b]$, with $a\le b$, and use Lebesgue measure $m$ on $([a,b],\mathcal{L},m)$. All functions in this tutorial are real-valued and defined everywhere on $[a,b]$. The interval length $v-u$ is $m((u,v))$. When $a=b$, every function on this one-point interval is absolutely continuous, and the closure claims below hold trivially. In the proofs and examples involving interval estimates, we may therefore take $a<b$.

Definition: A 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 is a $\delta>0$ such that, for every finite family of intervals $(u_k,v_k)\subseteq[a,b]$ with pairwise disjoint interiors, $$ \sum_k(v_k-u_k)<\delta \quad\Longrightarrow\quad \sum_k|f(v_k)-f(u_k)|<\varepsilon. $$

The definition concerns finite families; the number of intervals is not fixed in advance. We use the standard compact-interval result that an absolutely continuous function is continuous. In particular, each such real-valued function is bounded on $[a,b]$, by the extreme value theorem. For reference, the continuity follows directly from the definition: given a desired bound $\eta>0$, use the definition with $\varepsilon=\eta$; if $|x-y|<\delta$, the single interval with endpoints $x,y$ gives $|f(x)-f(y)|<\eta$. Boundedness then follows from continuity and compactness.

Linear Combinations

Theorem: If $f,g\in AC[a,b]$ and $\alpha,\beta\in\mathbb{R}$, then $\alpha f+\beta g\in AC[a,b]$. The functions are real-valued, and the conclusion is the interval-sum definition of absolute continuity on $[a,b]$.

Proof. Fix $\varepsilon>0$. If $\alpha\ne0$, absolute continuity of $f$ gives a number $\delta_f>0$ such that total interval length less than $\delta_f$ implies that the sum of the $f$-increments is less than $\varepsilon/(2|\alpha|)$. If $\alpha=0$, set $\delta_f=1$; no restriction from $f$ is needed. Define $\delta_g$ in the same way, using $\varepsilon/(2|\beta|)$ when $\beta\ne0$, and $\delta_g=1$ when $\beta=0$. Take $\delta=\min(\delta_f,\delta_g)$.

For any finite family of intervals with total length less than $\delta$, the triangle inequality gives $$ \sum_k|(\alpha f+\beta g)(v_k)-(\alpha f+\beta g)(u_k)| \le |\alpha|\sum_k|f(v_k)-f(u_k)| +|\beta|\sum_k|g(v_k)-g(u_k)|. $$ Each nonzero-coefficient term on the right is less than $\varepsilon/2$, and a zero-coefficient term is zero. If both coefficients are nonzero, the sum is less than $\varepsilon$; if exactly one is nonzero, it is less than $\varepsilon/2$; and if both are zero, it is zero. This proves absolute continuity, including the zero-coefficient cases. $\square$

Taking $(\alpha,\beta)=(1,1)$ proves closure under sums. Taking one function to be zero proves closure under scalar multiplication, including multiplication by zero. More generally, finite linear combinations follow by repeated application of the theorem.

Products: Why Boundedness Matters

Theorem: If $f,g\in AC[a,b]$ are real-valued, then their pointwise product $fg$, defined by $(fg)(x)=f(x)g(x)$ for every $x\in[a,b]$, belongs to $AC[a,b]$.

Proof. If $a=b$, the claim follows from the definition on a one-point interval. Suppose $a<b$. Since $f$ and $g$ are absolutely continuous, they are continuous and bounded on the compact interval. Put $$ A=\sup_{x\in[a,b]}|f(x)|,\qquad B=\sup_{x\in[a,b]}|g(x)|. $$ These are finite nonnegative numbers. If $A=0$, then $f(x)=0$ for every $x$, so $fg$ is the zero function and is absolutely continuous. If $B=0$, the same conclusion follows because $g$ is identically zero. It remains to consider $A>0$ and $B>0$.

Fix $\varepsilon>0$. By absolute continuity of $f$, choose $\delta_f>0$ so that a family of intervals with total length less than $\delta_f$ has total $f$-increment less than $\varepsilon/(2B)$. By absolute continuity of $g$, choose $\delta_g>0$ so that the total $g$-increment is less than $\varepsilon/(2A)$ for every such family of total length less than $\delta_g$. Let $\delta=\min(\delta_f,\delta_g)$.

For each interval with endpoints $u_k,v_k$, use the identity $$ f(v_k)g(v_k)-f(u_k)g(u_k) =f(v_k)\bigl(g(v_k)-g(u_k)\bigr) +g(u_k)\bigl(f(v_k)-f(u_k)\bigr). $$ Since $|f(v_k)|\le A$ and $|g(u_k)|\le B$, the triangle inequality, followed by summation, yields $$ \sum_k|(fg)(v_k)-(fg)(u_k)| \le A\sum_k|g(v_k)-g(u_k)| +B\sum_k|f(v_k)-f(u_k)| < A\frac{\varepsilon}{2A} +B\frac{\varepsilon}{2B} =\varepsilon. $$ This is exactly the definition of $fg\in AC[a,b]$. $\square$

The estimate does not require the intervals to be adjacent, nor does it require the functions to be nonnegative. Absolute values handle changes of either sign. Boundedness is what permits the same constants $A$ and $B$ to control every interval in the family.

Worked Applications

Worked Example: Adding Two Polynomials

On $[0,2]$, let $f(x)=x^2$ and $g(x)=3x$. Verify that their sum $h(x)=x^2+3x$ is absolutely continuous, and check an interval-family estimate.

For $x,y\in[0,2]$, $$ |f(x)-f(y)|=|x-y||x+y|\le4|x-y|, \qquad |g(x)-g(y)|=3|x-y|. $$ Thus each function is Lipschitz. A function satisfying $|q(x)-q(y)|\le C|x-y|$ is absolutely continuous: for any finite family, its increment sum is at most $C$ times the total interval length, so choosing $\delta=\varepsilon/C$ proves the definition when $C>0$. The case $C=0$ is a constant function and is immediate. The linear-combination theorem therefore gives $h\in AC[0,2]$.

Directly, for $x,y\in[0,2]$, $$ |h(x)-h(y)|=|x-y||x+y+3|\le7|x-y|. $$ For $\varepsilon=0.14$, take $\delta=0.02$. Any finite family with total length less than $0.02$ has total $h$-increment less than $7(0.02)=0.14$. For a numerical check, use the intervals $(0,0.01)$ and $(1,1.01)$, whose total length is $0.01+0.01=0.02$; this is not less than $\delta$, so instead take the second interval to be $(1,1.005)$. The total length is $0.01+0.005=0.015<0.02$. The increments are $$ h(0.01)-h(0)=0.0001+0.03=0.0301, $$ and $$ h(1.005)-h(1)=1.010025+3.015-4=0.025025. $$ Their sum is $0.055125<0.14$, as required. The general bound for this family is $7(0.015)=0.105$, which also verifies the estimate.

Worked Example: A Negative Scalar Multiple

On $[-1,1]$, let $f(x)=x^2$ and set $q(x)=-2f(x)=-2x^2$. Verify absolute continuity and compute the increment sum on two intervals.

For $x,y\in[-1,1]$, $|x^2-y^2|=|x-y||x+y|\le2|x-y|$. Hence $f$ is Lipschitz with constant $2$, and so is absolutely continuous. The scalar-multiple theorem gives $q\in AC[-1,1]$. Directly, $$ |q(x)-q(y)|=2|x-y||x+y|\le4|x-y|, $$ so the product of the Lipschitz constant $2$ by the scalar magnitude $2$ gives the bound $4$.

Take $(-1,-0.9)$ and $(0.2,0.3)$. Their total length is $0.1+0.1=0.2$. At the first pair of endpoints, $q(-1)=-2$ and $q(-0.9)=-2(0.81)=-1.62$, so the increment is $0.38$. At the second pair, $q(0.2)=-2(0.04)=-0.08$ and $q(0.3)=-2(0.09)=-0.18$, so the increment is $0.10$. The total is $0.38+0.10=0.48$, while the Lipschitz estimate gives $4(0.2)=0.8$. Thus the computed sum satisfies the bound. The negative sign changes the function values but does not change the absolute-increment estimate.

Worked Example: Multiplying Two Affine Functions

On $[0,1]$, let $f(x)=1+x$ and $g(x)=2-x$. Verify that their product is absolutely continuous using both the general product estimate and a direct bound.

For an independent check, expand $h(x)=f(x)g(x)=2+x-x^2$. Then $$ |h(x)-h(y)|=|x-y||1-x-y|\le|x-y|, $$ because $x,y\in[0,1]$ implies $-1\le1-x-y\le1$. This direct estimate is sharper than the general product bound, but both prove absolute continuity.

For the intervals $(0,0.2)$ and $(0.6,0.9)$, the total length is $0.2+0.3=0.5$. The product values are $h(0)=2$, $h(0.2)=2+0.2-0.04=2.16$, $h(0.6)=2+0.6-0.36=2.24$, and $h(0.9)=2+0.9-0.81=2.09$. The increment sum is $|2.16-2|+|2.09-2.24|=0.16+0.15=0.31$. The direct bound gives $0.5$, and the general product bound gives $4(0.5)=2$; the computed value is below both.

A Common Mistake and the Correct Estimate

A tempting but incorrect step is to estimate a product increment by the sum of the two factor increments alone. For example, one might write $$ |f(v)g(v)-f(u)g(u)|\le |f(v)-f(u)|+|g(v)-g(u)|. $$ This fails even for simple absolutely continuous functions. Let $f(x)=3$ and $g(x)=x$ on $[0,1]$, with $u=0$ and $v=1$. The left side is $|3-0|=3$, while the proposed right side is $0+1=1$. The error is that a change in one factor is multiplied by the size of the other factor.

The correct identity and estimate keep those sizes: $$ |f(v)g(v)-f(u)g(u)| \le |f(v)|\,|g(v)-g(u)|+|g(u)|\,|f(v)-f(u)|. $$ When $f$ and $g$ are bounded by $A$ and $B$, respectively, this is at most $A|g(v)-g(u)|+B|f(v)-f(u)|$. Summing this valid estimate is precisely what allows the absolute-continuity conditions for the two factors to control the product. The constants need not be optimal; they need only be finite and uniform over the whole compact interval.

These closure properties let algebraic expressions built from absolutely continuous functions remain in the same class when they use sums, scalar multiples, and products. The product argument also illustrates why compactness matters: it supplies global bounds for the factors, rather than bounds that might vary from one small interval to another.

Check Your Understanding

For each question, use the interval-sum definition and state any bounds needed in your argument.

Suppose $f,g\in AC[a,b]$. Give an explicit epsilon choice for proving $f-g\in AC[a,b]$ from the linear-combination theorem.
If $f\in AC[a,b]$ and $c=0$, why does the scalar-multiple proof not require a threshold involving $\varepsilon/|c|$?
For $f(x)=2+x$ and $g(x)=1-x$ on $[0,1]$, find bounds $A$ and $B$ and write the resulting product interval-sum estimate.
Use the identity for product increments to explain exactly where boundedness of the factors enters the proof.
Give a concrete choice of constant $f$ and $g(x)=x$ showing why the product increment cannot generally be bounded by the unweighted sum of factor increments.