Proof.

(a)

If \(a< x < b\), the partition \(\{a, x, b\}\) gives \[|f(x) - f(a)| \leq |f(x) - f(a)| + |f(b) - f(x)| \leq V(f) .\] This also holds for \(x = a\) and \(x=b\). Thus \(|f(x)| \leq |f(a)| + V(f)\) for every \(x\in [a,b]\).

(b)

A typical term in \(V(fg, P)\) satisfies \begin {align*} &|f(x_i) g(x_i) - f(x_{i - 1}) g(x_{i - 1})| \\ &\qquad = \bigl |f(x_i)\bigl (g(x_i) - g(x_{i-1})\bigr ) + g(x_{i-1})\bigl (f(x_i) - f(x_{i - 1})\bigr )\bigr |\\ &\qquad \leq M |g(x_i) - g(x_{i-1})| + N |f(x_i) - f(x_{i - 1})| . \end {align*}

Summing over \(i\), \[V(fg,P) \leq M V(g,P) + N V(f,P) \leq M V(g) + N V(f)\] for every partition \(P\). Taking the supremum over \(P\) gives the result.

(c)

Since \(|f(x)|\leq |x|\), \(f(x)\longrightarrow 0 = f(0)\) as \(x\longrightarrow 0^+\); and \(f\) is continuous on \((0,b]\) as a product and composite of continuous functions. So \(f\) is continuous on \([0,b]\).

Let \(y_k = \frac {2}{k\pi }\), \(k\in \mathbb {N}\). Then \(f(y_k) = \frac {2}{k\pi }\sin \frac {k\pi }{2}\), so \(f(y_k) = 0\) for even \(k\) and \(|f(y_k)| = \frac {2}{k\pi }\) for odd \(k\). Choose \(k_0\) with \(y_{k_0}\leq b\). For \(n > k_0\), let \(P_n\) be the partition of \([0,b]\) consisting of \(0\), \(b\) and the points \(y_k\) with \(k_0\leq k\leq 2n\). Each pair of consecutive points \(y_{2j}<y_{2j-1}\) in \(P_n\) contributes \(|f(y_{2j-1}) - f(y_{2j})| = \frac {2}{(2j-1)\pi }\), so \[V(f,P_n) \geq \frac {2}{\pi }\sum _{j=k_0}^{n} \frac {1}{2j-1} \geq \frac {2}{\pi }\sum _{j=k_0}^{n} \frac {1}{2j} = \frac {1}{\pi }\sum _{j=k_0}^{n} \frac {1}{j} .\] The harmonic series diverges, so \(V(f,P_n)\longrightarrow \infty \) as \(n\longrightarrow \infty \), and \(V(f) = \infty \).

(d)

If \(P\) is any partition of \([a,c]\) and \(Q\) is any partition of \([c,b]\), then \(P\cup Q\) is a partition of \([a,b]\) and \(V(f,P) + V(f,Q) = V(f, P\cup Q) \leq V(f: [a,b])\). Taking suprema over \(P\) and \(Q\), \[V(f: [a,c]) + V(f: [c,b]) \leq V(f: [a,b]) .\] Conversely, let \(P_0\) be any partition of \([a,b]\), and let \(P' = P_0\cup \{c\}\). Adding a point can only increase the sum, by the triangle inequality, and \(P'\) splits into a partition \(P_1\) of \([a,c]\) and a partition \(Q_1\) of \([c,b]\). Thus \begin {align*} V(f,P_0) &\leq V(f,P') = V(f,P_1) + V(f, Q_1)\\ &\leq V(f: [a,c]) + V(f: [c,b]) . \end {align*}

Taking the supremum over \(P_0\) gives the reverse inequality.

Proof. The function \(F\) “follows” \(f\), reflecting decreasing portions of the graph so that they become increasing. Let \(a\leq x < y\leq b\). By part (d) of Theorem 3.1.2, \[F(y) - F(x) = V(f:[x,y]) \geq 0,\] so \(F\) is increasing. Moreover \begin {align*} G(y) - G(x) &= V(f: [x,y]) - \bigl (f(y) - f(x)\bigr )\\ &\geq V(f: [x, y]) - |f(y) - f(x)| \geq 0, \end {align*}

since the partition \(\{x,y\}\) of \([x,y]\) shows that \(|f(y) - f(x)|\leq V(f:[x,y])\). So \(G\) is increasing. □

Proof. Let \(P: a = x_0<x_1<\dots <x_n = b\). By the Mean Value Theorem (Section 4.2), applied on each \([x_{k-1},x_k]\), there are points \(t_k \in (x_{k-1}, x_k)\) with \(f(x_k) - f(x_{k-1}) = f'(t_k) (x_k - x_{k-1})\). Hence \[V(f,P) = \sum ^n_{k = 1}|f'(t_k)| (x_k - x_{k - 1}) \leq M\sum ^n_{k = 1} (x_k - x_{k - 1}) = M (b - a) .\] Since \(P\) was arbitrary, \(V(f)\leq M(b-a)\). □

Solution. At \(0\), \(\frac {f(x) - f(0)}{x} = x\cos \frac {1}{x}\longrightarrow 0\), so \(f'(0) = 0\). For \(x\neq 0\), \[f'(x) = 2x \cos \frac {1}{x} + \sin \frac {1}{x},\] so \(|f'(x)| \leq 2 + 1 = 3\) for all \(x\in (0,1)\). Also \(f\) is continuous on \([0,1]\). By Theorem 3.1.4, \(f\) is of bounded variation, with \(V(f)\leq 3\).

Solution. The function decreases on \([-1,0]\) and increases on \([0,2]\). By part (d) of Theorem 3.1.2 and the note on monotone functions, \[V(f:[-1,2]) = V(f:[-1,0]) + V(f:[0,2]) = |f(0) - f(-1)| + |f(2) - f(0)| = 1 + 4 = 5 .\] Note that \(|f(2) - f(-1)| = 3\) is smaller: the variation counts the descent from \(1\) to \(0\) as well as the ascent from \(0\) to \(4\).

In general, if \([a,b]\) splits into finitely many intervals on each of which \(f\) is monotone, then \(V(f)\) is the sum of \(|f(\text {right end}) - f(\text {left end})|\) over these intervals.

Solution. The sine function increases on \([0,\frac {\pi }{2}]\), decreases on \([\frac {\pi }{2},\frac {3\pi }{2}]\) and increases on \([\frac {3\pi }{2}, 2\pi ]\). Hence \[V(\sin : [0,2\pi ]) = (1 - 0) + \bigl (1 - (-1)\bigr ) + \bigl (0 - (-1)\bigr ) = 4 .\] The same reasoning on \([0,x]\) gives \[F(x) = \begin {cases} \sin x, & 0\leq x\leq \frac {\pi }{2},\\ 2 - \sin x, & \frac {\pi }{2}\leq x\leq \frac {3\pi }{2},\\ 4 + \sin x, & \frac {3\pi }{2}\leq x\leq 2\pi . \end {cases}\] Then \(G(x) = F(x) - \sin x\) equals \(0\), \(2 - 2\sin x\) and \(4\) on the three intervals. Both \(F\) and \(G\) are increasing and continuous, and \(\sin x = F(x) - G(x)\).

Solution. Here \(f'(x) = 3(x^2 - 1)\), so \(f\) increases on \([-2,-1]\), decreases on \([-1,1]\) and increases on \([1,2]\). The values are \(f(-2) = -2\), \(f(-1) = 2\), \(f(1) = -2\), \(f(2) = 2\). Hence \[V(f:[-2,2]) = |2-(-2)| + |-2 - 2| + |2 - (-2)| = 4 + 4 + 4 = 12 .\] For the decomposition, \(F(x) = V(f:[-2,x])\) is \[F(x) = \begin {cases} f(x) + 2, & -2\leq x\leq -1,\\ 6 - f(x), & -1\leq x\leq 1,\\ f(x) + 10, & 1\leq x\leq 2, \end {cases}\] and \(G(x) = F(x) - f(x)\) equals \(2\) on \([-2,-1]\), \(6 - 2f(x) = 6 - 2x^3 + 6x\) on \([-1,1]\), and \(10\) on \([1,2]\). One checks that \(G\) increases from \(2\) to \(10\) on \([-1,1]\), since \(G'(x) = -2f'(x)\geq 0\) there.

Solution. For \(n\in \mathbb {N}\) choose a partition \(P: 0 = x_0 < x_1 < \dots < x_{2n} = 1\) in which \(x_i\) is rational for even \(i\) and irrational for odd \(i\); this is possible because every interval contains both rational and irrational numbers. Then every term of \(V(f,P)\) equals \(|f(x_i) - f(x_{i-1})| = 1\), so \(V(f,P) = 2n\). Since \(n\) is arbitrary, \(V(f) = \infty \). (Note that \(f\) is bounded, so boundedness does not imply bounded variation.)

Solution. At \(0\), \(\left |\frac {f(x)-f(0)}{x}\right | = \left |x\sin \frac {1}{x^2}\right |\leq |x|\longrightarrow 0\), so \(f'(0) = 0\); elsewhere \(f\) is differentiable by the chain rule. Now let \[x_k = \frac {1}{\sqrt {\left (k+\frac {1}{2}\right )\pi }},\qquad k\in \mathbb {N},\] so that \(\sin \frac {1}{x_k^2} = \sin \left (k+\frac {1}{2}\right )\pi = (-1)^k\) and \(f(x_k) = \frac {(-1)^k}{(k+\frac {1}{2})\pi }\). Consecutive values have opposite signs, so \[|f(x_k) - f(x_{k+1})| = \frac {1}{(k+\frac {1}{2})\pi } + \frac {1}{(k+\frac {3}{2})\pi } > \frac {1}{(k+\frac {1}{2})\pi }.\] Using the partition \(0 < x_n < x_{n-1}<\dots <x_1 < 1\), \[V(f) \geq \sum _{k=1}^{n-1} \frac {1}{(k+\frac {1}{2})\pi } \geq \frac {1}{\pi }\sum _{k=1}^{n-1}\frac {1}{k+1}\longrightarrow \infty .\] So \(f\notin BV[0,1]\). This does not contradict Theorem 3.1.4: here \(f'(x) = 2x\sin \frac {1}{x^2} - \frac {2}{x}\cos \frac {1}{x^2}\) is unbounded near \(0\). Compare the first example, where the derivative was bounded.