Two Different Ways to Compose a Function
Total variation measures the accumulated change of a function along an interval. Composition can alter that change in two distinct ways. A monotone change of parameter changes how the domain interval is traversed; a Lipschitz outer function changes the values taken along that traversal. The first operation cannot increase variation, and it preserves variation when the parameter map covers the entire original interval. The second operation multiplies variation by at most the Lipschitz constant.
All functions in this tutorial are real-valued, and all intervals are closed and bounded intervals in \(\mathbb{R}\). No measure space or integral is involved. For \(u:[a,b]\to\mathbb{R}\), a partition is a finite sequence \(a=x_0<x_1<\cdots<x_n=b\), and its variation sum is \(\sum_{j=1}^n|u(x_j)-u(x_{j-1})|\). The total variation is the supremum of these sums. We say \(u\in BV[a,b]\) if this supremum is finite. On a degenerate interval \(a=b\), we set \(V_a^a(u)=0\).
We use the elementary partition fact that inserting extra points into a partition cannot decrease its variation sum: if \(r\le s\le t\), then \(|u(t)-u(r)|\le |u(s)-u(r)|+|u(t)-u(s)|\), by the triangle inequality. This also lets us compare an increasing list of points that omits one or both endpoints with the full-interval variation: add the endpoints, and, if necessary, remove repeated points.
Monotone Changes of Parameter
Proof. If \(c=d\), the left side is zero, so the inequality holds. Suppose \(c<d\), and take any partition \(c=t_0<\cdots<t_n=d\). If \(\phi\) is nondecreasing, then \(\phi(t_0)\le\cdots\le\phi(t_n)\). The sum \(\sum_{j=1}^n|f(\phi(t_j))-f(\phi(t_{j-1}))|\) is bounded by \(V_a^b(f)\): remove repeated image points, add \(a\) and \(b\) if they are missing, and use the triangle inequality to see that the resulting full-interval partition has a sum at least as large. If \(\phi\) is nonincreasing, reverse the order of its image points and apply the same argument; reversing a list does not change the sum of absolute successive differences. Thus every partition sum for \(f\circ\phi\) is at most \(V_a^b(f)\). Taking their supremum proves the inequality and shows \(f\circ\phi\in BV[c,d]\).
Now suppose \(\phi\) is onto. If \(a=b\), both variations are zero. Otherwise, surjectivity and monotonicity imply that either \(\phi(c)=a,\ \phi(d)=b\), with \(\phi\) nondecreasing, or \(\phi(c)=b,\ \phi(d)=a\), with \(\phi\) nonincreasing. Take any partition \(a=x_0<\cdots<x_n=b\). In the nondecreasing case, choose \(t_j\in[c,d]\) with \(\phi(t_j)=x_j\), choosing \(t_0=c\) and \(t_n=d\). Since \(x_i<x_j\) and \(\phi\) is nondecreasing, every such choice satisfies \(t_i<t_j\); hence these points form a partition of \([c,d]\). Its sum for \(f\circ\phi\) is exactly \(\sum_{j=1}^n|f(x_j)-f(x_{j-1})|\). In the nonincreasing case the corresponding preimages occur in reverse order, and reversing them gives a partition with the same sum. Thus every partition sum for \(f\) is at most \(V_c^d(f\circ\phi)\). Taking the supremum gives the reverse inequality, and therefore equality. \(\square\)
Worked Example: A Full and a Partial Reparameterization
Let \(f(x)=x\) on \([0,1]\). First take \(\phi(t)=t^2\) on \([0,1]\). This map is nondecreasing and onto \([0,1]\), so the theorem predicts that \(f\circ\phi\) has the same variation as \(f\). Directly, \(f\circ\phi(t)=t^2\), which is increasing. For any partition \(0=t_0<\cdots<t_n=1\), its variation sum telescopes: $$ \sum_{j=1}^n(t_j^2-t_{j-1}^2)=t_n^2-t_0^2=1. $$ Thus \(V_0^1(f\circ\phi)=1\). The same telescoping computation gives \(V_0^1(f)=1\), checking equality.
Next take \(\psi(t)=t/2\) on \([0,1]\). It is nondecreasing but its image is only \([0,1/2]\), so it is not onto \([0,1]\). Now \(f\circ\psi(t)=t/2\), and every partition sum is $$ \sum_{j=1}^n\left(\frac{t_j}{2}-\frac{t_{j-1}}{2}\right)=\frac12. $$ Therefore \(V_0^1(f\circ\psi)=1/2\le1=V_0^1(f)\). The inequality is strict because this parameter map traverses only half the range of \(f\). Check: the endpoint increment of \(f\circ\psi\) is \(|1/2-0|=1/2\), agreeing with the computed variation.
The theorem does not require a strictly monotone parameter map. Flat portions merely repeat values, contributing zero increments. Nor does it say that every monotone map preserves variation: covering the whole interval is the extra condition that gives equality.
Lipschitz Maps of the Values
Proof. If \(a=b\), the variation on the left is zero, so the assertion holds. For \(a<b\), take any partition \(a=x_0<\cdots<x_n=b\). Since each \(f(x_j)\) lies in \(J\), the Lipschitz condition gives $$ \sum_{j=1}^n|g(f(x_j))-g(f(x_{j-1}))| \le L\sum_{j=1}^n|f(x_j)-f(x_{j-1})| \le L\,V_a^b(f). $$ Taking the supremum over all partitions proves the stated estimate. The right side is finite because \(f\in BV[a,b]\), so \(g\circ f\) also has bounded variation. This argument includes \(L=0\): then \(g\) is constant on \(J\), and the composition has variation zero. \(\square\)
Worked Example: A Lipschitz Map Can Reduce Variation
On \([-1,1]\), set \(f(x)=x\), and on \(\mathbb{R}\) set \(g(t)=\max(t,0)\). The map \(g\) is \(1\)-Lipschitz: if \(s,t\) have the same sign, the inequality \(|g(s)-g(t)|\le|s-t|\) follows directly; if \(s\le0\le t\), then \(|g(s)-g(t)|=t\le t-s=|t-s|\). The identity \(f\) is increasing, so every partition sum telescopes to \(1-(-1)=2\), giving \(V_{-1}^1(f)=2\).
The composition is \(g(f(x))=0\) for \(-1\le x\le0\) and \(g(f(x))=x\) for \(0\le x\le1\). For any partition, the increments on the nonpositive part vanish, and those on the nonnegative part telescope to at most \(1\). The partition \(-1<0<1\) attains \(1\), so \(V_{-1}^1(g\circ f)=1\). The theorem gives \(1\le1\cdot2\), and the exact value confirms the bound. Check: the sum on the displayed partition is \(|0-0|+|1-0|=1\).
A Lipschitz condition controls every pair of values, which is why it can be applied to each increment of each partition. Without such control, a bounded-variation input can acquire infinite variation after composition.
Worked Example: A Non-Lipschitz Outer Map Can Create Infinite Variation
Let \(f(x)=x\) on \([0,1]\), so the telescoping partition calculation gives \(V_0^1(f)=1\). Define \(g:[0,1]\to\mathbb{R}\) by \(g(0)=0\) and \(g(t)=t\sin(1/t^2)\) for \(t>0\). This function is continuous at zero because \(|g(t)|\le t\), but it is not Lipschitz there.
For \(n\ge0\), put \(q_n=\pi/2+n\pi\) and \(t_n=q_n^{-1/2}\). Then \(0<t_{n+1}<t_n<1\), and direct substitution gives \(g(t_n)=(-1)^n t_n\). Use the partition \(0<t_N<t_{N-1}<\cdots<t_0<1\), including \(1\) as the final endpoint. Its variation sum is at least the contribution from the listed points: $$ |g(t_N)-g(0)|+\sum_{n=0}^{N-1}|g(t_n)-g(t_{n+1})| =t_N+\sum_{n=0}^{N-1}(t_n+t_{n+1}). $$ In particular this is at least \(\sum_{n=0}^{N-1}t_n\). For \(n\ge1\), \(q_n=\pi(n+1/2)\le(3\pi/2)n\), so \(t_n\ge ((3\pi/2)n)^{-1/2}\). The partial sums of \(n^{-1/2}\) diverge, hence these partition sums are unbounded as \(N\) increases. Thus \(V_0^1(g\circ f)=V_0^1(g)=\infty\), even though \(V_0^1(f)=1\).
To verify the missing hypothesis, the Lipschitz ratio at successive points is $$ \frac{|g(t_n)-g(t_{n+1})|}{|t_n-t_{n+1}|} =\frac{t_n+t_{n+1}}{t_n-t_{n+1}}. $$ Here \(t_{n+1}/t_n=\sqrt{(n+1/2)/(n+3/2)}\to1\), so this ratio is unbounded. Thus \(g\) is not Lipschitz. Check: each adjacent pair contributes the sum of the two positive amplitudes because their signs alternate; the unbounded variation is not inferred merely from the oscillation of the formula.
How to Use the Estimates
The two results answer different questions. For \(f\circ\phi\), the outer function \(f\) is unchanged and the parameter \(\phi\) moves through its domain in one direction. Its image points stay ordered, so no new sequence of oscillations is introduced. If \(\phi\) is onto, every partition of the original interval can be recovered through ordered preimages, giving exact preservation.
For \(g\circ f\), the parameter is unchanged but each pair of values is transformed. The Lipschitz inequality bounds each transformed increment by \(L\) times its original increment, and summing gives the variation estimate. The constant \(L\) need not be the best possible bound, and the estimate need not be an equality. In particular, these results do not assert that arbitrary composition preserves bounded variation: a nonmonotone parameter map or an outer map without a suitable Lipschitz bound requires a separate argument.
Common mistake. It is incorrect to claim that every monotone reparameterization preserves variation. For \(f(x)=x\) on \([0,1]\) and \(\psi(t)=t/2\), the wrong conclusion would be \(V_0^1(f\circ\psi)=V_0^1(f)=1\). The correction is the direct calculation \(f\circ\psi(t)=t/2\), whose partition sums telescope to \(1/2\); the general theorem gives only \(V_0^1(f\circ\psi)\le V_0^1(f)\) unless the parameter map is onto. Likewise, the Lipschitz estimate is an upper bound, not an equality: the positive-part example has variations \(1\) and \(2\), respectively.
Check Your Understanding
For each question, identify which composition result applies and check all interval and mapping hypotheses.
- If \(f(x)=3x-2\) on \([0,1]\) and \(\phi(t)=1-t\) on \([0,1]\), compute \(V_0^1(f)\) and \(V_0^1(f\circ\phi)\). Why does equality hold?
- Give a nondecreasing map from \([0,1]\) into \([0,1]\) that is not onto, and explain why the theorem gives an inequality rather than equality.
- If \(f\in BV[a,b]\) and \(g\) is \(L\)-Lipschitz on an interval containing \(f([a,b])\), what bound follows for \(V_a^b(g\circ f)\)? Explain the case \(L=0\).
- For \(f(x)=x\) on \([-1,1]\) and \(g(t)=\max(t,0)\), compute both variations and compare them with the Lipschitz estimate.
- In the oscillatory example, why do the values at successive \(t_n\) force the variation sums to become unbounded?