Almost-Everywhere Differentiability Of BV Functions

From Monotone Functions to Bounded Variation

A nondecreasing function can have jumps and need not be differentiable everywhere, but Lebesgue's Differentiability Theorem for Monotone Functions says it is differentiable almost everywhere. A function of bounded variation need not itself be monotone. The Jordan decomposition supplies the bridge: it writes such a function as a constant plus the difference of two nondecreasing functions. Where both components have finite derivatives, their difference does too.

Throughout, let \(a<b\) be finite real numbers, and use the finite, complete measure space \(([a,b],\mathcal{L}_{[a,b]},m)\), where \(\mathcal{L}_{[a,b]}\) is the Lebesgue sigma-algebra restricted to \([a,b]\), and \(m\) is Lebesgue measure. Functions in this discussion are real-valued and finite everywhere. Differentiability means existence of a finite two-sided derivative at a point \(x\in(a,b)\). Endpoints are excluded; the two endpoints form a null set, so this does not affect an almost-everywhere conclusion.

Definition: For a real-valued function \(f\) on \([a,b]\), its variation on \([a,x]\) is \(V_a^x(f)=\sup\sum_{i=1}^{n}|f(x_i)-f(x_{i-1})|\), where the supremum is over all finite partitions \(a=x_0<x_1<\cdots<x_n=x\). The function \(f\) is of bounded variation, written \(f\in BV[a,b]\), if \(V_a^b(f)<\infty\).

We use the Jordan decomposition theorem from the earlier material on functions of bounded variation. In its normalized form, if \(f\in BV[a,b]\), there are nondecreasing, finite functions \(g,h:[a,b]\to\mathbb{R}\) such that \(f(x)=f(a)+g(x)-h(x)\), with \(g(a)=h(a)=0\). In particular, the constant \(f(a)\) has no effect on differentiability. The normalized components can be expressed using the variation function, as shown below.

Jordan Components from the Variation Function

Write \(V(x)=V_a^x(f)\). The variation is additive on adjacent intervals, as established in the earlier material, and for \(a\le x\le y\le b\) it satisfies \(V(y)-V(x)\ge |f(y)-f(x)|\). The normalized Jordan components are

g(x)=\frac{V(x)+f(x)-f(a)}{2}, \qquad h(x)=\frac{V(x)-f(x)+f(a)}{2}.

Verification. At \(x=a\), \(V(a)=0\), so \(g(a)=h(a)=0\). If \(x\le y\), then

g(y)-g(x)=\frac{V(y)-V(x)+f(y)-f(x)}{2}\ge0,

because \(V(y)-V(x)\ge |f(y)-f(x)|\ge -(f(y)-f(x))\). Likewise,

h(y)-h(x)=\frac{V(y)-V(x)-f(y)+f(x)}{2}\ge0,

because \(V(y)-V(x)\ge |f(y)-f(x)|\ge f(y)-f(x)\). Thus both components are nondecreasing. They are finite because \(V(x)\le V_a^b(f)<\infty\) and \(f\) is real-valued. Finally, subtracting the formulas gives \(g(x)-h(x)=f(x)-f(a)\), or \(f(x)=f(a)+g(x)-h(x)\), as required.

Worked Example: The Absolute-Value Function

Let \(f(x)=|x|\) on \([-1,1]\). Its variation from \(-1\) to \(x\) is \(V(x)=1+x\): for \(-1\le x\le0\), the function decreases from \(1\) to \(|x|=-x\), giving variation \(1-(-x)=1+x\); for \(0\le x\le1\), it decreases from \(1\) to \(0\) and then increases to \(x\), giving variation \(1+x\). In particular, \(V(1)=2\).

Since \(f(-1)=1\), the normalized formulas give, for \(-1\le x\le0\), \(g(x)=(1+x-x-1)/2=0\) and \(h(x)=(1+x+x+1)/2=1+x\). For \(0\le x\le1\), they give \(g(x)=(1+x+x-1)/2=x\) and \(h(x)=(1+x-x+1)/2=1\). Thus \(g(x)=\max(x,0)\), \(h(x)=\min(x+1,1)\), and \(f(x)=1+g(x)-h(x)\). For \(-1<x<0\), \(f'(x)=-1\); for \(0<x<1\), \(f'(x)=1\). At \(x=0\), the right difference quotient is \(|t|/t=1\) for \(t>0\), while the left difference quotient is \(|t|/t=-1\) for \(t<0\), so no derivative exists. Check: the total variation is \(2\), the sum of the decrease from \(1\) to \(0\) and the increase from \(0\) to \(1\); the derivative exists except at the single point \(0\).

Almost-Everywhere Differentiability

Theorem: If \(f:[a,b]\to\mathbb{R}\) is of bounded variation, then \(f\) is differentiable at \(m\)-almost every \(x\in(a,b)\). The derivative is finite at every point in the full-measure set obtained in the proof.

Proof. By the Jordan decomposition, write \(f=f(a)+g-h\) with \(g,h:[a,b]\to\mathbb{R}\) nondecreasing and finite. Apply Lebesgue's Differentiability Theorem for Monotone Functions to \(g\) and to \(h\), on the stated finite interval with Lebesgue measure. There are Lebesgue-measurable null sets \(N_g,N_h\subseteq(a,b)\) such that \(g\) has a finite derivative at every \(x\in(a,b)\setminus N_g\) and \(h\) has a finite derivative at every \(x\in(a,b)\setminus N_h\). The theorem for monotone functions is the earlier result being used here.

The union \(N=N_g\cup N_h\) is null: by subadditivity, \(m(N)\le m(N_g)+m(N_h)=0\). Fix any \(x\in(a,b)\setminus N\). Both derivatives exist and are finite there. For every nonzero \(t\) sufficiently close to \(0\) with \(x+t\in[a,b]\),

\frac{f(x+t)-f(x)}{t} = \frac{g(x+t)-g(x)}{t} - \frac{h(x+t)-h(x)}{t}.

As \(t\to0\), the first quotient tends to \(g'(x)\) and the second tends to \(h'(x)\). The limit of their difference is therefore \(g'(x)-h'(x)\), a finite real number. The constant \(f(a)\) cancels in the quotient. Hence \(f'(x)\) exists and equals \(g'(x)-h'(x)\). This holds at every \(x\in(a,b)\setminus N\), whose complement in \((a,b)\) has measure zero. Thus \(f\) is differentiable \(m\)-almost everywhere. \(\square\)

The proof uses a precise set argument: the points where the conclusion is guaranteed are those where both Jordan components are differentiable. The exceptional set for \(f\) is contained in the union of the two exceptional sets, and a finite union of null sets is null. No assertion is made that every point where one component fails to be differentiable must also be a point where \(f\) fails to be differentiable; cancellation can occur.

Worked Example: A Step Function with a Jump

On \([0,1]\), define \(f(x)=0\) for \(0\le x<\tfrac12\), and \(f(x)=1\) for \(\tfrac12\le x\le1\). This function is nondecreasing. Its variation is \(0\) on \([0,\tfrac12)\) and \(1\) on \([0,1]\), so \(V_0^1(f)=1\). Because \(f(0)=0\), the normalized components are \(g=f\) and \(h=0\), giving \(f=0+g-h\).

If \(0<x<\tfrac12\), then every sufficiently small increment stays in the region where \(f=0\), so the difference quotient is \(0\). If \(\tfrac12<x<1\), every sufficiently small increment stays where \(f=1\), and again the quotient is \(0\). At \(x=\tfrac12\), \(f(x)=1\). For \(t>0\) sufficiently small, \((f(x+t)-f(x))/t=0\). For \(t<0\) sufficiently small, \(f(x+t)=0\), so the quotient is \((0-1)/t=-1/t\), which has no finite limit as \(t\uparrow0\). Thus the derivative exists and equals \(0\) except at \(\tfrac12\). Check: there is exactly one jump of size \(1\), matching the total variation \(1\), and the exceptional set \(\{\tfrac12\}\) has Lebesgue measure zero.

Worked Example: A Piecewise-Linear Tent

Define \(f:[0,2]\to\mathbb{R}\) by \(f(x)=x\) for \(0\le x\le1\), and \(f(x)=2-x\) for \(1\le x\le2\). On the first segment it increases by \(1\), and on the second it decreases by \(1\), so \(V_0^2(f)=2\). More generally, \(V_0^x(f)=x\) for \(0\le x\le2\): before \(1\) the variation is \(x\), and after \(1\) it is \(1+(x-1)=x\).

Here \(f(0)=0\). For \(0\le x\le1\), the Jordan formulas give \(g(x)=(x+x)/2=x\) and \(h(x)=(x-x)/2=0\). For \(1\le x\le2\), they give \(g(x)=(x+2-x)/2=1\) and \(h(x)=(x-(2-x))/2=x-1\). Both components are nondecreasing, and \(f=g-h\). Directly from its two linear pieces, \(f'(x)=1\) for \(0<x<1\) and \(f'(x)=-1\) for \(1<x<2\). At \(x=1\), the right difference quotient is \(-1\), while the left difference quotient is \(1\), so \(f'\) does not exist there. Check: the component derivatives give \(g'-h'=1-0=1\) on \((0,1)\) and \(g'-h'=0-1=-1\) on \((1,2)\), agreeing with the direct calculation.

What the Conclusion Does—and Does Not—Say

The conclusion applies to every real-valued \(f\in BV[a,b]\), without requiring continuity. The step-function example shows why: a function can have a jump and still be differentiable almost everywhere. The theorem does not say that \(f\) is differentiable at every point, or even that it is continuous at every point. It also does not say that the derivative alone records all the variation: for the step function, the derivative is zero almost everywhere, although its total variation is \(1\).

Common mistake: dropping the additive constant. The normalized formulas above yield \(g-h=f-f(a)\), not \(f\). For example, let \(f(x)=3+x\) on \([0,1]\). Its variation function is \(V(x)=x\), so the formulas give \(g(x)=(x+(3+x)-3)/2=x\) and \(h(x)=(x-(3+x)+3)/2=0\). The incorrect claim \(f=g-h\) would give \(3+x=x\), which fails (for example, at \(x=0\), it gives \(3=0\)). The correct identity is \(f(x)=f(a)+g(x)-h(x)=3+x\). The missing constant does not affect difference quotients, but it is necessary for the function identity.

The central mechanism is therefore simple but exact: bounded variation gives a difference of two monotone functions; each monotone component has a finite derivative off a null set; and outside the union of those null sets, subtracting the two difference-quotient limits gives the derivative of the original function.

Check Your Understanding

For each question, specify the interval and the exceptional points or null sets involved.

For a function \(f\in BV[a,b]\), explain why the normalized Jordan components satisfy \(g-h=f-f(a)\), rather than \(g-h=f\).
Why is the set where both Jordan components are differentiable a full-measure set?
For the step function in the example, compute the left and right difference quotients at \(x=\tfrac12\). Which one-sided limit prevents differentiability?
For the tent function, use \(f=g-h\) to compute \(f'\) separately on \((0,1)\) and \((1,2)\).
Does almost-everywhere differentiability imply that a bounded-variation function is continuous everywhere? Use the step-function example to justify your answer.

Almost-Everywhere Differentiability Of BV Functions

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