Total variation of a function ($V_a^b(f)$)

The total variation of a function $f:[a,b]\to \mathbb{R}$ measures its accumulated rises and falls, rather than just its change between the endpoints. For each partition of an interval $\Pi =(a=t_0<\cdots <t_n=b)$, add the absolute values of its sampled increments. Then take the supremum over all such partitions of an interval:The function has bounded variation when $V_a^b(f)<\infty$. This definition uses its value at every point of the interval, including both endpoints.^[ref]

For a nondecreasing function, all increments are nonnegative and the sum telescopes to $f(b)-f(a)$. More generally, the triangle inequality gives $|f(b)-f(a)|\le V_a^b(f)$. Equality can fail dramatically when the function changes direction: an object can travel a long distance and finish near its starting position. Here we measure movement of the function's value; arc length of its function graph also includes horizontal movement.

On the vector space of continuous functions $f:[0,1]\to \mathbb{R}$ of bounded variation with $f(0)=0$, the total variation of a function is a norm. Its triangle inequality and absolute homogeneity of a norm follow by applying the corresponding absolute value identities to every partition of an interval, then taking the supremum. The partition of an interval $(0,t,1)$ gives $|f(t)|\le V_0^1(f)$, so zero total variation of a function forces $f=0$. The same inequality gives $\Vert f{\Vert }_{\infty }\le \Vert f{\Vert }_V$, proving that convergence for this norm implies uniform convergence. Without $f(0)=0$, every constant function would have zero total variation of a function.

Define a piecewise linear function by joining the five points $(0,0),(1,2),(2,-1),(3,3),(4,1)$. Its net change is $1$, but the increments at these knots are $2,-3,4,-2$. Their absolute values giveThis is its exact total variation, not merely one lower bound. Given any partition of an interval, insert all five knots. The triangle inequality ensures that this partition refinement cannot decrease the sum. On each resulting monotone function segment, the absolute increments telescope, so every refined sum is $11$. The knot partition of an interval itself attains $11$.

Sampling only at $0,4$ gives $1$. Sampling at $0,1,3,4$ gives $2+1+2=5$. Inserting the remaining knot at $2$ reveals the descent and rise that were hidden inside $[1,3]$, giving $11$. These are nested partitions of an interval: their sums increase because each is a partition refinement of the preceding one. Arbitrarily chosen unrelated partitions of an interval need not produce an increasing sequence of sums.

For each positive integer $n$, put $f_n(t)=\sin (2\pi nt)/n$ on $[0,1]$. Its supremum norm is exactly $1/n$, so this sequence of functions converges to zero by uniform convergence. Nevertheless,Each of its $n$ complete sine oscillations contributes $4/n$: rise from zero to $1/n$, fall through zero to $-1/n$, then return to zero. Taking a partition of an interval through all extrema attains $4$. Every further partition refinement merely subdivides monotone function segments, so the same reasoning as in partition example of total variation proves the upper bound too.

For a continuously differentiable function on a closed bounded interval, total variation also satisfiesApplied here, the derivative is $2\pi \cos (2\pi nt)$, and integrating its absolute value over $n$ periods gives $4$. The differentiability hypothesis matters: the partition definition applies much more widely than this derivative formula.^[ref] Small pointwise size and small accumulated movement are different approximation goals.

Let $f$ have bounded variation, and let $g,h$ be continuous functions on $[a,b]$. In any tagged sum for the Riemann-Stieltjes integral,Taking the limit proves the displayed error bound. Thus approximating $g$ uniformly to accuracy $\epsilon$ controls this integration error by $\epsilon V_a^b(f)$, even when the integrator is not differentiable. For the piecewise linear function in partition example of total variation, accuracy $0.001$ gives an error at most $0.011$. Total variation supplies exactly the movement budget that an endpoint difference would miss.

Camille Jordan developed bounded variation as part of his work in real analysis; his treatment of functions and arc length was part of the influence of his analysis textbooks.^[ref] The Jordan decomposition of a function of bounded variation expresses the underlying idea especially well: complicated changes of direction can be separated into two functions that only increase.

For $p\ge 1$, the displayed quantity measures the accumulated oscillation of a path in a metric space. For $p=1$ it is total variation. A Hölder continuous path of exponent $\alpha$ has finite $p$-variation on a compact interval if $p\alpha \ge 1$. Applying this construction to the Carnot-Carathéodory distance gives the defining regularity of a weak geometric p-rough path.

For a càdlàg function, replace the last sampled value at $2^{-n}\lceil 2^{n}t\rceil$ by $X_t$. Right continuity makes this change tend to zero. The corrected sum is bounded by the total variation of a function. Conversely every finite partition of an interval can be approximated by dyadic points from the right, retaining the endpoint $t$. The triangle inequality bounds that partition's limiting sum by the dyadic sums. Taking the supremum proves convergence, including the value $+\infty$.

If continuously differentiable $F_j,F$ converge uniformly on a compact interval and have uniformly bounded total variation of a function, then $\int b{F}_j^{\prime }\to \int b{F}^{\prime }$ for every continuous $b$. For continuously differentiable test functions this is integration by parts. Uniform approximation by polynomials extends the conclusion to continuous tests using the variation bound.

Every real function on $[a,b]$ has bounded variation if and only if it is a difference of two nondecreasing functions. For a normalized version, let $v(t)=V_a^t(f)$ and $v(a)=0$. Its accumulated positive and negative variations areBoth are nondecreasing functions, $P(a)=N(a)=0$, andThis is the Jordan decomposition of a function of bounded variation.^[ref] Equivalently one may write $f=f_{+}-f_{-}$ with $f_{+}=(v+f)/2$ and $f_{-}=(v-f)/2$; those functions need not vanish at $a$.

For the piecewise linear function in partition example of total variation, the successive values of $P$ are $0,2,2,6,6$, while those of $N$ are $0,0,3,3,5$. At $t=4$, the total rise is $6$ and the total fall is $5$: their difference gives the net change $1$, and their sum gives the total variation $11$. A declining segment increases $N$ while $P$ stays constant; an ascending segment does the opposite.

For a short proof outline, total variation is additive over adjacent intervals, because any two partitions of an interval have a common partition refinement. Thus $v(t)-v(s)=V_s^t(f)\ge |f(t)-f(s)|$ for $s<t$. Substituting this inequality into the formulas for $P$ and $N$ shows their increments are nonnegative. Conversely, if $f=g-h$ for two nondecreasing functions, the triangle inequality bounds each partition sum by $g(b)-g(a)+h(b)-h(a)$. This proves finite total variation without restricting the number of changes of direction.