Functions of Bounded Variation

Show that every monotone function $f$ on $[a, b]$ is of bounded variation with $V_a^b(f) = |f(b) - f(a)|$ .

V_a^b(f) := \sup_P \sum_{i=1}^{m} \bigl|f(x_i) - f(x_{i-1})\bigr|

\sum_{i=1}^{m} \bigl|f(x_i) - f(x_{i-1})\bigr| = \sum_{i=1}^{m} \bigl(f(x_i) - f(x_{i-1})\bigr) = f(b) - f(a) \quad (f \text{ increasing})

n = 40

:

f(x) = \sin(2\pi x)

is not monotone, yet

V_{40} \approx 3.9968

— when

f

is monotone, the sum telescopes on every partition, fixing the variation at

|f(b) - f(a)|

.

Show proof

Let $f$ be increasing on $[a, b]$ and take any partition $P = \{x_0 < x_1 < \dots < x_m\}$ ( $[x_0, x_m] = [a, b]$ ). Monotonicity gives $f(x_i) - f(x_{i-1}) \ge 0$ on each subinterval, so the absolute values unwrap: $\sum_{i=1}^{m}|f(x_i) - f(x_{i-1})| = \sum_{i=1}^{m}(f(x_i) - f(x_{i-1}))$ , which telescopes to $f(b) - f(a)$ .
The decreasing case is the same: $f(x_i) - f(x_{i-1}) \le 0$ , so $|f(x_i) - f(x_{i-1})| = f(x_{i-1}) - f(x_i)$ and the sum is $f(a) - f(b) = |f(b) - f(a)|$ . Every partition of a monotone $f$ gives the same variation $|f(b) - f(a)|$ .
Hence the supremum over partitions is $|f(b) - f(a)| < \infty$ , i.e. $V_a^b(f) = |f(b) - f(a)|$ : monotone functions are of bounded variation. The Cantor function (see the limits chapter) is increasing on $[0, 1]$ with $c(0) = 0$ , $c(1) = 1$ , so $V_0^1(c) = 1$ .

Jordan decomposition: $f$ is of bounded variation on $[a, b]$ if and only if $f = g - h$ with $g, h$ increasing on $[a, b]$ . Hint: set $v(x) := V_a^x(f)$ .

v(x) := V_a^x(f), \qquad f = \frac{v + f}{2} - \frac{v - f}{2}

V_x^y(f) \ge \pm\bigl(f(y) - f(x)\bigr) \quad (a \le x < y \le b)

n = 80

:

V_{80} \approx 3.9992

— the cumulative variation

v(x) = V_0^x(\sin 2\pi\,\cdot)

is itself increasing, so

f = \frac{v+f}{2} - \frac{v-f}{2}

splits

f

into a difference of two increasing functions.

Show proof

Assume $f$ is of bounded variation on $[a, b]$ and set $v(x) := V_a^x(f)$ for $a \le x \le b$ . Variation is additive over adjacent intervals, $V_a^y(f) = V_a^x(f) + V_x^y(f)$ for $a \le x < y \le b$ , so $v(y) - v(x) = V_x^y(f) \ge 0$ : $v$ is increasing.
The key inequality: the trivial partition $\{x, y\}$ of $[x, y]$ is one of the partitions in the supremum, so $V_x^y(f) \ge |f(y) - f(x)| \ge \pm(f(y) - f(x))$ . Hence $v(y) - v(x) \ge f(y) - f(x)$ and also $v(y) - v(x) \ge -(f(y) - f(x))$ .
Set $g := \frac{v + f}{2}$ and $h := \frac{v - f}{2}$ . Then $g(y) - g(x) = \frac{1}{2}\bigl((v(y) - v(x)) + (f(y) - f(x))\bigr) \ge 0$ and $h(y) - h(x) = \frac{1}{2}\bigl((v(y) - v(x)) - (f(y) - f(x))\bigr) \ge 0$ , so $g$ and $h$ are both increasing, and $g - h = f$ .
Conversely, if $f = g - h$ with $g, h$ increasing, then by problem 1 both are of bounded variation, and the triangle inequality gives $|f(x_i) - f(x_{i-1})| \le |g\text{-diff}| + |h\text{-diff}|$ , so $V_P(f) \le V_P(g) + V_P(h) = (g(b) - g(a)) + (h(b) - h(a)) < \infty$ : $f$ is of bounded variation. So $f$ is BV $\iff$ $f = g - h$ (unique up to an additive constant).

Show that $f$ with $f(0) = 0$ and $f(x) = x\sin(1/x)$ for $x > 0$ is NOT of bounded variation on $[0, 1]$ .

u_k = \frac{1}{2\pi k + \pi/2},\; f(u_k) = u_k, \qquad v_k = \frac{1}{2\pi k + 3\pi/2},\; f(v_k) = -v_k

V(f) \ge \sum_{k=1}^{N}(u_k + v_k) \ge \sum_{k=1}^{N}\frac{1}{2\pi k} = \frac{H_N}{2\pi} \longrightarrow \infty

n = 200

:

V_{200} \approx 2.1388

— already past the proof's lower bound

\sum_{k=1}^{200}(u_k + v_k) \approx 1.6849 > 1

, and since

H_N/(2\pi) \to \infty

the variation admits no finite upper bound.

Show proof

Take $u_k = \frac{1}{2\pi k + \pi/2}$ where $\sin(1/x) = 1$ (so $f(u_k) = u_k$ ) and $v_k = \frac{1}{2\pi k + 3\pi/2}$ where $\sin(1/x) = -1$ (so $f(v_k) = -v_k$ ). Comparing denominators, $0 < v_k < u_k < 1$ , and $u_{k+1} = \frac{1}{2\pi k + 5\pi/2} < v_k$ , so the points line up as $0 < v_N < u_N < \dots < v_1 < u_1 < 1$ .
Take the partition $P_N = \{0, v_N, u_N, \dots, v_1, u_1, 1\}$ . The contributions from $[0, v_N]$ and $[u_1, 1]$ are nonnegative, so dropping them only lowers the sum; on each pair $(v_k, u_k)$ we get $|f(u_k) - f(v_k)| = |u_k - (-v_k)| = u_k + v_k$ . Hence $V_{P_N}(f) \ge \sum_{k=1}^{N}(u_k + v_k)$ .
Now $u_k + v_k \ge 2v_k = \frac{2}{2\pi k + 3\pi/2} \ge \frac{1}{2\pi k}$ : the last inequality is equivalent to $4\pi k \ge 2\pi k + 3\pi/2$ , i.e. $4k \ge 3$ for $k \ge 1$ . Therefore $V_{P_N}(f) \ge \sum_{k=1}^{N} \frac{1}{2\pi k} = \frac{H_N}{2\pi} \to \infty$ as $N \to \infty$ .
Since the partition variations diverge, $\sup_P V_P(f) = \infty$ : $f$ is not of bounded variation on $[0, 1]$ . Numerical check: $u_1 = \frac{1}{2.5\pi} \approx 0.12732$ , $v_1 = \frac{1}{3.5\pi} \approx 0.09095$ , $\sum_{k=1}^{200}(u_k + v_k) \approx 1.6849 > 1$ — the scene's $V_{200} \approx 2.1388$ already exceeds this lower bound. (Contrast: $x^2\sin(1/x)$ IS of bounded variation, with $V \le 3$ , since $\int_0^1 |f'| \le 2$ .)

Functions of Bounded Variation

Part of these learning paths

Examples & Exercises