Total Variation From Partitions

Measuring Total Change on an Interval

The difference between a function’s endpoint values records its net change, but it can miss substantial movement in between. Total variation measures all the changes seen along the interval: choose finitely many ordered points, add the absolute changes between consecutive values, and then take the supremum over all such choices. The supremum may be infinite, so this definition applies to every real-valued function on a closed bounded interval, not only to functions already known to have bounded variation.

Here the object is a pointwise real-valued function $f:[a,b]\to\mathbb{R}$, where $a,b\in\mathbb{R}$ and $a\le b$. The definition uses no measure space or integral. It depends on the values of $f$ at points, so changing a function on even a single point can change its variation.

Definition: If $a<b$, a finite partition of $[a,b]$ is a list $P=(x_0,x_1,\ldots,x_n)$ with $n\ge1$ and $a=x_0<x_1<\cdots<x_n=b$. The variation sum of $f$ along $P$ is $$ S_P(f)=\sum_{i=1}^{n}|f(x_i)-f(x_{i-1})|. $$ The total variation of $f$ on $[a,b]$ is $$ V_a^b(f)=\sup\{S_P(f):P\text{ is a finite partition of }[a,b]\}\in[0,\infty]. $$ For the degenerate interval $a=b$, set $V_a^a(f)=0$. The function $f$ is of bounded variation on $[a,b]$, written $f\in BV[a,b]$, if and only if $V_a^b(f)<\infty$.

Each partition gives just one candidate value; the definition asks for the least upper bound of all candidate values. It does not say that some partition must attain that bound. Absolute values are essential: an increase and a decrease both contribute positively, even if their signed changes later cancel. Because the partition includes the endpoints, every variation sum includes the net endpoint change as a lower bound, but the interior points can make the sum larger.

Constants and Affine Functions

Worked Example: A Constant Function

Let $f(x)=c$, where $c\in\mathbb{R}$, on $[a,b]$, with $a<b$. For any finite partition $P=(x_0,\ldots,x_n)$, every increment is zero: $$ |f(x_i)-f(x_{i-1})|=|c-c|=0. $$ There are $n$ such terms, so $S_P(f)=\sum_{i=1}^n0=0$ for every partition. The supremum of these sums is therefore $V_a^b(f)=0$. If $a=b$, the defining convention also gives zero. This calculation includes $c=0$ and nonzero constants.

Check: using the endpoint partition $P=(a,b)$ gives $S_P(f)=|c-c|=0$, consistent with the value for every partition.

Proposition: If $f(x)=mx+c$ on $[a,b]$, where $m,c\in\mathbb{R}$ and $a\le b$, then $V_a^b(f)=|m|(b-a)$.

Proof. If $a=b$, both sides equal zero. Suppose $a<b$, and let $P=(x_0,\ldots,x_n)$ be any finite partition. For each $i$, $|f(x_i)-f(x_{i-1})|=|m(x_i-x_{i-1})|=|m|(x_i-x_{i-1})$, because $x_i-x_{i-1}>0$. Summing and using $x_0=a$ and $x_n=b$ gives $$ S_P(f)=|m|\sum_{i=1}^n(x_i-x_{i-1})=|m|(b-a). $$ Thus every partition sum has this value, so its supremum does too. $\square$

Worked Example: A Decreasing Affine Function

Take $f(x)=-3x+2$ on $[-1,2]$. The proposition gives $V_{-1}^{2}(f)=|-3|(2-(-1))=3\cdot3=9$. To check directly, the endpoint values are $f(-1)=3+2=5$ and $f(2)=-6+2=-4$, so the endpoint partition has sum $|-4-5|=9$. For the three-subinterval partition $(-1,0,1,2)$, the function values are $5,2,-1,-4$. Its sum is $$ |2-5|+|-1-2|+|-4-(-1)|=3+3+3=9. $$ In fact, for any partition, each absolute increment is $3$ times the corresponding interval length, and the lengths sum to $2-(-1)=3$. Thus every partition gives $3\cdot3=9$, as the proposition asserts. This also checks the sign: the function decreases, but its variation is nonnegative.

Telescoping in Specific Monotone Examples

When the values in a particular example move in only one direction, the absolute increments can be evaluated by removing their signs in a consistent way. The resulting sum telescopes. The calculations below establish the variation of these specific functions directly from the definition; they do not require a general theorem about all monotone functions.

Worked Example: The Square Function on [0, 1]

Let $f(x)=x^2$ on $[0,1]$. For a partition $0=x_0<x_1<\cdots<x_n=1$, the points are nonnegative and increasing, so $x_i^2-x_{i-1}^2=(x_i-x_{i-1})(x_i+x_{i-1})\ge0$. Consequently each absolute increment equals $x_i^2-x_{i-1}^2$, and $$ S_P(f)=\sum_{i=1}^n(x_i^2-x_{i-1}^2)=x_n^2-x_0^2=1^2-0^2=1. $$ This holds for every partition, so $V_0^1(f)=1$. As a numerical check, the partition $(0,\tfrac12,1)$ gives values $0,\tfrac14,1$ and sum $|\tfrac14-0|+|1-\tfrac14|=\tfrac14+\tfrac34=1$. The value agrees with the endpoint change $f(1)-f(0)=1-0=1$.

Worked Example: The Cube Function on [-1, 2]

Let $f(x)=x^3$ on $[-1,2]$. For $u<v$, factor the difference: $$ v^3-u^3=(v-u)(v^2+vu+u^2). $$ The second factor is nonnegative, since $v^2+vu+u^2=(v+\tfrac12u)^2+\tfrac34u^2\ge0$, and $v-u>0$. Hence $v^3-u^3\ge0$ whenever $u<v$. For any partition $-1=x_0<\cdots<x_n=2$, every increment of $f$ is nonnegative, and telescoping gives $$ S_P(f)=\sum_{i=1}^n(x_i^3-x_{i-1}^3)=2^3-(-1)^3=8-(-1)=9. $$ Therefore $V_{-1}^{2}(f)=9$. For an independent check, the one-interval partition gives $|f(2)-f(-1)|=|8-(-1)|=9$. For the partition $(-1,0,2)$, the values are $-1,0,8$, and the sum is $|0-(-1)|+|8-0|=1+8=9$.

A Turning Point Requires More Than the Endpoint Difference

Worked Example: The Absolute-Value Function on [-2, 3]

Let $f(x)=|x|$. The endpoint values are both nonnegative, and the net change is $f(3)-f(-2)=3-2=1$. That number is not the total variation. To obtain the variation, first bound every partition sum. For real numbers $u,v$, the reverse triangle inequality gives $||v|-|u||\le|v-u|$. If $P=(-2=x_0<\cdots<x_n=3)$, then $$ S_P(f)=\sum_{i=1}^n\big||x_i|-|x_{i-1}|\big| \le\sum_{i=1}^n|x_i-x_{i-1}| =\sum_{i=1}^n(x_i-x_{i-1})=3-(-2)=5. $$ So no partition sum exceeds $5$. Now choose the partition $(-2,0,3)$, which includes the turning point. Its function values are $2,0,3$, giving $$ S_P(f)=|0-2|+|3-0|=2+3=5. $$ A partition sum equals the upper bound, so the supremum is exactly $V_{-2}^{3}(f)=5$. Check: the bound applies to every partition, while the displayed partition attains it; together these give both inequalities needed for equality.

What Finite Variation Controls

Proposition: If $f:[a,b]\to\mathbb{R}$ has $V_a^b(f)<\infty$, then $f$ is bounded on $[a,b]$.

Proof. If $a=b$, the domain has one point, so $f$ is bounded there. Suppose $a<b$, and set $V=V_a^b(f)<\infty$. For any $x\in(a,b)$, the list $(a,x,b)$ is a partition, and its sum is at most $V$. Its first term is $|f(x)-f(a)|$, so $|f(x)-f(a)|\le V$. For $x=b$, use the partition $(a,b)$ to get $|f(b)-f(a)|\le V$, and for $x=a$ the difference is zero. Thus for every $x\in[a,b]$, $$ |f(x)|\le |f(a)|+|f(x)-f(a)|\le |f(a)|+V. $$ The right-hand side is finite and independent of $x$, proving that $f$ is bounded. $\square$

The proposition gives a useful necessary consequence of finite variation. It does not say that every bounded function has finite variation: boundedness controls the size of individual values, whereas variation adds changes across arbitrarily many consecutive intervals. The supremum over all partitions is what detects repeated oscillation.

Common Pitfall: Using Signed Increments

A tempting but incorrect calculation is to replace each absolute increment in the definition by the signed increment $f(x_i)-f(x_{i-1})$. The signed sum telescopes to $f(b)-f(a)$, regardless of the interior values, and can therefore conceal changes in both directions. For $f(x)=|x|$ on $[-2,3]$, that mistaken calculation gives $f(3)-f(-2)=3-2=1$ and would report variation $1$. The correction is to keep the absolute value in every term: the partition $(-2,0,3)$ gives $2+3=5$, and the bound proved above shows that no partition gives more. Thus the correct variation is $5$, not $1$.

For an affine function, the endpoint calculation happens to work because all increments have the same sign (or are all zero). In the absolute-value example, the sign changes at zero, so the net change no longer records the full amount of movement. When computing from a partition, check that every increment is treated with the absolute value required by the definition.

Check Your Understanding

For each question, start from the partition definition and make clear whether your calculation concerns one partition sum or the supremum over all partitions.

Find $V_0^4(f)$ for $f(x)=7$, and explain why every partition gives the same sum.
Compute $V_{-2}^{1}(f)$ for $f(x)=2x+5$. Verify the value using the endpoint partition.
For $f(x)=x^2$ on $[-1,1]$, compare the endpoint change with the partition sum from $(-1,0,1)$. What does this show about using signed endpoint change?
Prove directly from the definition that $V_0^2(f)=4$ for $f(x)=x^2$ on $[0,2]$.
Explain why finite total variation implies boundedness, and identify the partition used to control $|f(x)-f(a)|$ when $a<x<b$.