3.1 Algebraic Properties
Definition 3.1.1. Let \(f:[a,b]\longrightarrow \mathbb {R}\) be a function, and let \(\hspace {0.2cm} P: a = x_0 < x_1 < \cdots \cdots \cdots < x_n = b\hspace {0.3cm}\) be a partition of the interval \([a,b]\). Define \[ V_f[a,b] = V(f,P) = \sum ^n_{i = 1}\Big |f(x_i) - f(x_{i - 1})\Big |\] Then, the variation of \(f\) on \([a,b]\) is defined to be \[\boxed {V(f) = \sup _p(f,P)}\]
We say that \(f\) is of bounded variation on \([a,b]\) if \(V(f)< \infty \).
It follows from the definition that a monotone function is of bounded variation. To see this let \(f\) be say an increasing function on \([a,b]\) and let \(\hspace {0.2cm} P:\, a = x_0 < x_1 < x_2 < \cdots \cdots < x_n = b\hspace {0.2cm}\) be the partition. Then
\begin {align*} V(f,P) & = \sum ^n_{i=1} \begin {vmatrix} f(x_i) - f(x_{i -1})\\ \end {vmatrix}\\ & = \begin {vmatrix} f(x_1) - f(x_0)\\ \end {vmatrix} + \begin {vmatrix} f(x_2) - f(x_1)\\ \end {vmatrix} + \begin {vmatrix} f(x_3) - f(x_2)\\ \end {vmatrix} + \cdots \cdots + \begin {vmatrix} f(x_n) - f(x_{n - 1})\\ \end {vmatrix}\\ & = f(x_1) - f(x_0) + f(x_2) - f(x_1) + \cdots \cdots + f(x_n) - f(x_{n - 1})\\ & = f(x_n) - f(x_0)\\ & = f(b) - f(a)\\ \end {align*}
That if \(f\) is monotone, then the variation of \(f\) is given by \(\hspace {0.2cm}\displaystyle {V(f,P) = \big | f(b) - f(a)\big |}\hspace {0.2cm}\) which is finite.
Also, for arbitrary function \(f\), we have \(\hspace {0.2cm} V(-f) = V(f)\).
Suppose that \(f\) and \(g\) are functions defined on the same interval, then
\[ V(f + g, P) \leq V(f,P) + V(g,P)\]
by the triangle inequality. Hence \(\hspace {0.2cm}\boxed {V(f + g, P) \leq V(f) + V(g)}\hspace {0.2cm}\) for any \(P\). Taking the supremum over \(P\) we obtain
\[ V(f + g) \leq V(f) + V(g)\]
\(**\) Write the details for the proof of \(\hspace {0.2cm} V(f + g) \leq V(f) + V(g)\).
If \(f\) is of bounded variation we write \(f\in BV\).
- a).
- If \(f\in BV\), then \(f\) is bounded.
- b).
- If \(f,\, g \in BV\), with \(\begin {vmatrix} f\\ \end {vmatrix}\leq M\) and \(\begin {vmatrix} g\\ \end {vmatrix}\leq N\) on \([a,b]\), then \(\hspace {0.2cm} V(fg) \leq M V(g) + N V(f)\), hence \(fg\in BV\).
- c).
- If \(f(x) = x\,\sin (1/x),\, 0< x\leq b, \, f(0) = 0,\,\) then \(f\) is continuous but not of bounded variation.
- d).
- If \(\hspace {0.2cm} a< c< b\), then \begin {align*} V(f: [a,b]) & = V\big (f: [a,c]\big ) + V\big (f: [c,a]\big )\\\\ V_f\big (a,b\big ) & = V_f\big (a,c\big ) + V_f\big (c,b\big ) \end {align*}
Proof.
- a).
- If \(a< x < b,\,\) we have \begin {align*} \begin {vmatrix} f(x) - f(a)\\ \end {vmatrix} & \leq \begin {vmatrix} f(x) - f(a)\\ \end {vmatrix} + \begin {vmatrix} f(b) - f(x)\\ \end {vmatrix}\\ & \leq V(f) < \infty \end {align*}
Thus \(\, \begin {vmatrix} f(x)\\ \end {vmatrix} \leq \begin {vmatrix} f(a)\\ \end {vmatrix} + V(f) < \infty \).
- b).
- A typical term in the computation of \(V(fg, P)\) is
\[ \begin {vmatrix} f(x_i)\, g(x_i) - f(x_{i - 1})\, g(x_{i - 1})\\ \end {vmatrix} \leq \begin {vmatrix} f(x_i)\\ \end {vmatrix}\, \begin {vmatrix} g(x_i) - g(x_{i-1}) \end {vmatrix} + \begin {vmatrix} g(x_{i-1})\\ \end {vmatrix}\, \begin {vmatrix} f(x_i) - f(x_{i - 1})\\ \end {vmatrix}\]
Since \(\, \begin {vmatrix} f(x_i)\\ \end {vmatrix}\leq M\,\) and \(\, \begin {vmatrix} g(x_{i - 1})\\ \end {vmatrix} \leq N\), it follows that \begin {align*} V(fg,P) & \leq M V(g,P) + N V(f,P)\\ & \leq M\, V(g) + N\, V(f)\hspace {0.5cm} \text {for all}\hspace {0.2cm} P \end {align*}
Take the supremum \(P\) to get the desired result.
- c).
- Let \(\,Y_n = \dfrac {2}{n \pi }, \hspace {0.2cm} n = 1, 2, \cdots \cdots \) Then
\[f(Y_1) = \dfrac {2}{\pi } \, ,\hspace {0.3cm} f(Y_2) = 0\, ,\hspace {0.3cm} f(Y_3) = \dfrac {-2}{3\pi }\, , \hspace {0.2cm} f(Y_4) = 0\, , \hspace {0.3cm} f(Y_5) = \dfrac {2}{5\pi }\, \, \hspace {0.3cm}\cdots \cdots \]
\[f(Y_{2n}) = 0\hspace {0.4cm},\hspace {0.4cm} f(Y_{2n + 1}) = \pm \, \dfrac {2}{(2n + 1) \pi }\]
Let \(P_n\) be the partition formed by \(0, \hspace {0.2cm} Y_{2n + 1} \hspace {0.2cm},\hspace {0.2cm} Y_{2n}\hspace {0.2cm}, \cdots \cdots \). Then \begin {align*} V(f,P_n) & = \dfrac {2}{\pi }\Big (1 + \dfrac {1}{3} + \dfrac {1}{5} + \dfrac {1}{7} + \cdots \cdots \cdots + \dfrac {1}{2n + 1}\Big )\\\\ & > \dfrac {2}{\pi } \Big ( \dfrac {1}{4} + \dfrac {1}{6} + \dfrac {1}{8} + \cdots \cdots \cdots + \dfrac {1}{2n + 2}\Big )\\\\ & = \dfrac {1}{\pi }\Big (\dfrac {1}{2} + \dfrac {1}{3} + \dfrac {1}{4} + \dfrac {1}{5} + \cdots \cdots \cdots + \dfrac {1}{n + 1}\Big ) \longrightarrow \infty \hspace {0.2cm} \text {as}\hspace {0.2cm} n\longrightarrow \infty . \end {align*}
- d).
- If \(P\) is any partition of \([a,c]\) and \(Q\) is any partition of \([c,b]\) we have
\[ V(f,P) + V(f,Q) \leq V\big (f: [a,b]\big )\]
by definition so that
\[ V\big (f: [a,c]\big ) + V\big (f: [c,b]\big ) \leq V\big (f: [a,b]\big )\]
If \(P_0\) is any partition of \([a,b]\) we can refine \(P_0\) if necessary to obtain a partition \(P'\) of \([a,c]\) followed by a
partition \(Q'\) of \([c,b]\). The refining process can only increase the variation, by the triangle inequality.
Thus \begin {align*} V\big (f,P_0\big ) & \leq V\big (f,P'\big ) + V\big (f, Q'\big )\\ & \leq V\big (f: [a,c]\big ) + V\big (f: [c,b]\big ) \end {align*}
Take sup over \(P_0\) to finish the argument.
Theorem 3.1.3. If \(f\) is of bounded variation on \([a,b]\) define
\begin {align*} F(x) & = V\Big (f: [a,x]\Big )\hspace {0.5cm} a\leq x\leq b\\\\ G(x) & = F(x) - f(x) \end {align*}
Then \(F\) and \(G\) are increasing functions on \([a,b]\), so \(f\) is expressible as the difference of two increasing
functions.
Proof. Note that \(F\) ‘follows’ \(f\), reflecting decreasing portions about the horizontal axis to obtain increasing portions. It follows from the last Theorem (d) that \(F\) is increasing. To show that \(G\) is increasing, let \(x < y\). Then \begin {align*} G(y) - G(x) & = F(y) - F(x) - \big ( F(y) - f(y)\big )\\ & = V\Big (f: [a,y]\Big ) - V \Big (f: [a,x]\Big ) - \Big (f(y) - f(x)\Big )\\ & = V \Big ( f; [x,y]\Big ) - \Big (f(y) - f(x)\Big ) \hspace {0.3cm} \text {by (d) of the last Thm}\\ &\leq V\, \Big ( f: [x, y]\Big ) - \begin {vmatrix} f(y) - f(x)\\ \end {vmatrix}\geq 0\hspace {0.3cm} \text {by definition of variation} \end {align*}
If \(f\) is continuous on \([a,b]\) then so are \(F\) and \(G\).
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Theorem 3.1.4. If \(f\) is continuous on \([a,b]\) and if \(f'\) exists and is bounded in the interior \((a,b)\) i.e \(\hspace {0.2cm} \begin {vmatrix} f'(x)\\ \end {vmatrix}\leq M \, \) for all \(x\in (a,b), \, M> 0\, \)
then \(f\) is of bounded variation on \([a,b]\).
Proof. Apply the Mean Value Theorem, we have \(\hspace {0.3cm} f(x_k) - f(x_{k-1}) = f'(t_k)\, (x_k - x_{k-1})\hspace {0.2cm}\) where
\(t_k \in (x_{k-1} - x_k)\). This implies that \begin {align*} \sum ^n_{k = 1}\begin {vmatrix} f(x_k) - f(x_{k - 1})\\ \end {vmatrix} & = \sum ^n_{k = 1}\begin {vmatrix} f'(t_k)\\ \end {vmatrix} \, (x_k - x_{k - 1})\\ & \leq \sum ^n_{k = 1} M \, (x_k - x_{k - 1})\\ & = M \, \sum ^n_{k=1} (x_k - x_{k - 1} )\\ & = M \, (b - a)\\ \end {align*}
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Example 3.1.5. Let \(\, f(x) = \begin {cases} x^2\, \cos \dfrac {1}{x} & \text {if}\hspace {0.3cm} x\neq 0\\\\ 0 & \text {if}\hspace {0.3cm} x = 0\\ \end {cases}\)
Show that \(f\) is of bounded variation on the interval \([0,1]\).
Solution. Since \(\, f(0) = 0\,\) then \(\, f'(0) = 0\,\) and also if \(\, x\neq 0\, \) then \(\, f'(x) = \sin \dfrac {1}{x} + 2x\, \cos \dfrac {1}{x}\). Then \(\, \begin {vmatrix} f'(x)\\ \end {vmatrix} \leq 3\,\) for all \(\, x\in (0,1)\).
Thus \(f'\) is bounded on \([0,1]\) and by the previous theorem \(f\) is of bounded variation.
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