Algebra Of Functions Of Bounded Variation

Variation and Pointwise Algebra

Bounded variation is preserved by the basic pointwise algebraic operations, but the estimates have different forms. Addition is controlled by the sum of the two variations, scalar multiplication scales variation exactly, and a product requires both variation and size bounds. The product estimate is useful precisely because a function of bounded variation on a compact interval is bounded.

Let $a\le b$ be real numbers. All functions below are real-valued or complex-valued and are defined at every point of $[a,b]$. All variation inequalities and identities are pointwise statements about these functions, not almost-everywhere statements; no measure space is involved. A partition of $[a,b]$ is a finite sequence $a=x_0<x_1<\cdots<x_n=b$. When $a=b$, the only interval has no nontrivial increments, and its variation is defined to be zero.

Definition: For $a<b$, the total variation of $f:[a,b]\to\mathbb{C}$ (or $\mathbb{R}$) is $$ V_a^b(f)=\sup_{P}\sum_{i=1}^{n}|f(x_i)-f(x_{i-1})|, $$ where the supremum is over all partitions $P=(x_0,\ldots,x_n)$ of $[a,b]$. The function $f$ is of bounded variation, written $f\in BV[a,b]$, if $V_a^b(f)<\infty$. For $a=b$, set $V_a^a(f)=0$. The supremum norm is $\|f\|_\infty=\sup_{x\in[a,b]}|f(x)|$, when this is finite.

The absolute value in the definition is the usual modulus for complex values. In particular, the definition applies without change to real-valued functions. We will use the exact variation of a monotone real-valued function: if $f$ is nondecreasing on $[a,b]$, then $V_a^b(f)=f(b)-f(a)$; if it is nonincreasing, then $V_a^b(f)=f(a)-f(b)$. Indeed, each increment in the monotone direction has the same sign, so the absolute increments telescope to the endpoint difference for every partition.

Variation Bounds for Sums and Scalars

Theorem: Let $f,g\in BV[a,b]$, and let $c\in\mathbb{C}$ (or $c\in\mathbb{R}$ when the functions are real-valued). Then $f+g\in BV[a,b]$ and $$ V_a^b(f+g)\le V_a^b(f)+V_a^b(g), \qquad V_a^b(cf)=|c|V_a^b(f). $$ These statements hold also when $a=b$.

Proof. First suppose $a<b$, and fix any partition $P=(x_0,\ldots,x_n)$. For each $i$, the triangle inequality gives $$ |(f+g)(x_i)-(f+g)(x_{i-1})| \le |f(x_i)-f(x_{i-1})|+|g(x_i)-g(x_{i-1})|. $$ Summing over $i$ bounds the partition sum for $f+g$ by the partition sum for $f$ plus the partition sum for $g$. Each of the latter sums is at most its total variation. Taking the supremum over $P$ proves the addition inequality, and its finite right-hand side proves $f+g\in BV[a,b]$. For a scalar $c$, every partition sum for $cf$ is $|c|$ times the corresponding sum for $f$. Taking suprema gives the scalar identity if $|c|>0$; if $c=0$, every increment of $cf$ is zero, and both sides are zero. When $a=b$, all variations are zero by definition, so both claims hold. $\square$

The addition inequality need not be an equality: increments of the two functions can cancel. By contrast, scalar multiplication gives an exact identity, including for the zero scalar. A constant function has variation zero, since every one of its increments is zero.

Worked Example: Cancellation in a Sum and Scaling

On $[0,1]$, let $f(x)=x$ and $g(x)=1-x$. Compute their variations, the variation of their sum, and the variation of $-2f$.

Both functions are monotone: $f$ is nondecreasing from $0$ to $1$, and $g$ is nonincreasing from $1$ to $0$. Thus $$ V_0^1(f)=1-0=1,\qquad V_0^1(g)=1-0=1. $$ Their sum is $f(x)+g(x)=1$ at every $x$, so $V_0^1(f+g)=0$. For any partition, the sum's increments are all zero, which checks this value directly. The addition bound reads $0\le1+1=2$, and is not an equality.

For the scalar multiple, the exact identity gives $V_0^1(-2f)=|-2|V_0^1(f)=2$. Directly, on a partition $0=x_0<\cdots<x_n=1$, the variation sum is $$ \sum_{i=1}^{n}|-2x_i+2x_{i-1}| =2\sum_{i=1}^{n}(x_i-x_{i-1}) =2. $$ The partition increments telescope to $1$, so this calculation verifies the answer for every partition. Check: the endpoint difference for the monotone function $-2x$ has absolute value $|-2-0|=2$, agreeing with the partition calculation.

The Product Estimate

A variation estimate for a product cannot depend only on the variations: even a constant function has zero variation while multiplying by it can rescale the other function. The required size information is available here because every function in $BV[a,b]$ is bounded.

Proposition: If $f\in BV[a,b]$, then $f$ is bounded and $$ \|f\|_\infty\le |f(a)|+V_a^b(f). $$

Proof. If $a=b$, the function has just one value on its domain, so it is bounded and $\|f\|_\infty=|f(a)|$. Suppose $a<b$. For any $x\in(a,b]$, the two-point partition $a<x$ gives $|f(x)-f(a)|\le V_a^b(f)$. The same bound holds at $x=a$, since its left side is zero. The triangle inequality therefore gives $|f(x)|\le |f(a)|+V_a^b(f)$ for every $x\in[a,b]$. Taking the supremum proves the claimed bound and boundedness. $\square$

Theorem: If $f,g\in BV[a,b]$, then their pointwise product $fg$ belongs to $BV[a,b]$, and $$ V_a^b(fg)\le \|f\|_\infty V_a^b(g)+\|g\|_\infty V_a^b(f). $$ The supremum norms are finite by the proposition.

Proof. When $a=b$, the product has variation zero and the right-hand side is zero. Suppose $a<b$. Fix a partition $P=(x_0,\ldots,x_n)$. For each adjacent pair, add and subtract $f(x_i)g(x_{i-1})$ to obtain $$ f(x_i)g(x_i)-f(x_{i-1})g(x_{i-1}) =f(x_i)\bigl(g(x_i)-g(x_{i-1})\bigr) +g(x_{i-1})\bigl(f(x_i)-f(x_{i-1})\bigr). $$ The triangle inequality and the definitions of the supremum norms imply $$ |f(x_i)g(x_i)-f(x_{i-1})g(x_{i-1})| \le \|f\|_\infty|g(x_i)-g(x_{i-1})| +\|g\|_\infty|f(x_i)-f(x_{i-1})|. $$ Sum over $i$. The two sums of increments are bounded by $V_a^b(g)$ and $V_a^b(f)$, respectively. Thus every partition sum for $fg$ is at most the stated finite right-hand side. Taking its supremum proves the estimate and shows $fg\in BV[a,b]$. $\square$

The choice of which endpoint value appears in the algebraic identity is immaterial to the bound: each value of $f$ and $g$ is controlled by its supremum norm. The estimate does not assert that the upper bound is attained.

Worked Example: A Product of Two Increasing Functions

On $[0,1]$, take $f(x)=x$ and $g(x)=1+x$. Compute the individual bounds and the exact variation of their product.

The functions are nondecreasing, so their variations are their endpoint increases: $$ V_0^1(f)=1-0=1,\qquad V_0^1(g)=2-1=1. $$ Their supremum norms are $\|f\|_\infty=1$ and $\|g\|_\infty=2$, since both are nonnegative and increasing. The product estimate gives $$ V_0^1(fg)\le 1\cdot1+2\cdot1=3. $$ But $f(x)g(x)=x+x^2$, which is nondecreasing: for $0\le x\le y\le1$, its increment is $(y-x)+(y^2-x^2)=(y-x)(1+x+y)\ge0$. Its exact variation is therefore its endpoint increase: $$ V_0^1(fg)=(1+1^2)-(0+0^2)=2. $$ For any partition, the increments of this nondecreasing product telescope to $2$, so the endpoint computation is also a direct check. The general bound is valid but not sharp in this example.

Worked Example: A Jumping Factor and a Product

On $[0,1]$, define $f(x)=1$ for $0\le x<\tfrac12$ and $f(x)=2$ for $\tfrac12\le x\le1$, and set $g(x)=x$. Compute the variations and check the product estimate.

The function $f$ is nondecreasing, with endpoint values $1$ and $2$, so $V_0^1(f)=2-1=1$. Also $g$ is nondecreasing from $0$ to $1$, giving $V_0^1(g)=1$. Their supremum norms are $\|f\|_\infty=2$ and $\|g\|_\infty=1$. Hence the product estimate gives $V_0^1(fg)\le2\cdot1+1\cdot1=3$.

To compute the product exactly, write $h=fg$. For $x<\tfrac12$, $h(x)=x$; for $x\ge\tfrac12$, $h(x)=2x$. On either part, $h$ is nondecreasing. If $x<\tfrac12\le y$, then $h(x)=x<\tfrac12$ and $h(y)=2y\ge1$, so $h(x)\le h(y)$. Thus $h$ is nondecreasing on the whole interval, including its specified value at $\tfrac12$. Its variation is $$ V_0^1(h)=h(1)-h(0)=2-0=2. $$ Every partition sum equals this endpoint difference because $h$ is nondecreasing; this verifies the exact value even for partitions that do not include $\tfrac12$. The bound $2\le3$ is satisfied. Check: the product's value at the jump point is $h(\tfrac12)=2\cdot\tfrac12=1$, consistent with the nondecreasing behavior used in the calculation.

Algebraic Consequences and a Common Pitfall

The estimates show that $BV[a,b]$ is closed under addition, scalar multiplication, and pointwise multiplication. In particular, any finite sum of bounded-variation functions is again of bounded variation: apply the addition inequality repeatedly, or use induction on the number of summands. The product conclusion applies repeatedly as well, since each product obtained is again in $BV[a,b]$ and hence is bounded. These are closure statements; they do not say that the resulting variation equals the sum or product of the original variations.

Common mistake: omitting the size factors in the product estimate. One might incorrectly claim $V_0^1(fg)\le V_0^1(f)V_0^1(g)$. Take $f(x)=x$ and $g(x)=1$ on $[0,1]$. Then $V_0^1(f)=1$, $V_0^1(g)=0$, but $fg=f$ and $V_0^1(fg)=1$. The incorrect bound would say $1\le1\cdot0=0$, which is false. The correction is to include the supremum norms: here the valid estimate gives $$ V_0^1(fg)\le \|f\|_\infty V_0^1(g)+\|g\|_\infty V_0^1(f) =1\cdot0+1\cdot1=1. $$ Another mistake is to replace the sum inequality by equality. The first worked example shows why that fails: $f+g$ is constant even though each summand has variation $1$.

Check Your Understanding

In each question, variations are taken on the stated closed interval and refer to the pointwise functions.

For $f(x)=3x$ and $g(x)=2-x$ on $[0,1]$, compute $V_0^1(f)$, $V_0^1(g)$, and the variation of $f+g$. Is the addition bound an equality?
If $V_a^b(f)=0$, show directly from the definition that $f$ is constant on $[a,b]$, including the case $a=b$.
For $f(x)=x$ and $g(x)=2+x$ on $[0,1]$, compute the product estimate and the exact variation of $fg$.
Explain why the bound for $V_a^b(fg)$ must involve information about the sizes of $f$ and $g$, and test the proposed bound $V_0^1(fg)\le V_0^1(f)V_0^1(g)$ with $f(x)=x$, $g(x)=1$.
For a complex scalar $c$, why does $V_a^b(cf)=|c|V_a^b(f)$ remain true when $c=0$?

Algebra Of Functions Of Bounded Variation

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