Ex2-2

Solution to Chapter 2, Exercise 2

Proof of Theorem 2.1 (ii)

Suppose $$f$$and $$g$$are of bounded variation on $$[a,b]$$and $$\Gamma=\{x_0,x_1,...,x_m\}$$is a partition of $$[a,b]$$. Since$$ \begin{align} S_\Gamma[cf;a,b]&=\sum_{i=1}^m\left|cf(x_i)-cf(x_{i-1})\right|\\ &=\left| c \right|\sum_{i=1}^m\left|f(x_i)-f(x_{i-1})\right|\\ &=\left| c \right|S_\Gamma[f;a,b], \end{align} $$we have$$ \begin{align} V[cf;a,b]&=\sup_\Gamma{S_\Gamma[cf;a,b]}\\ &=\left| c \right|V[f;a,b]. \end{align} $$Therefore, if $$V[f;a,b]<+\infty$$, then $$V[cf;a,b]<+\infty$$.

Observe that$$ \begin{align} S_\Gamma[f+g;a,b]&=\sum_{i=1}^m\left|f(x_i)+g(x_i)-f(x_{i-1})-g(x_{i-1})\right|\\ &\leq\sum_{i=1}^m\left| f(x_i)-f(x_{i-1}) \right|+\sum_{i=1}^m\left| g(x_i)-g(x_{i-1}) \right|\\ &=S_\Gamma[f;a,b]+S_\Gamma[g;a,b]\\ &\leq V[f;a,b]+V[g;a,b]. \end{align} $$Thus, $$V[f+g;a,b]=\sup_\Gamma{S_\Gamma[f+g;a,b]}\leq V[f;a,b]+V[g;a,b].$$Therefore, $$V[f+g;a,b]<+\infty$$as well.

Since $$f$$and $$g$$are of bounded variation, by Theorem 2.1(i) they are bounded as well. For some $$M,N>0$$, $$|f(x)|\leq M$$ and $$|g(x)|\leq N$$for all $$x$$. Then$$ \begin{align} S_\Gamma[fg;a,b]&=\sum_{i=1}^m |f(x_i)g(x_i)-f(x_{i-1})g(x_{i-1})|\\ &=\sum_{i=1}^m |f(x_i)g(x_i)-f(x_i)g(x_{i-1})+f(x_i)g(x_{i-1})-f(x_{i-1})g(x_{i-1})|\\ &\leq\sum_{i=1}^m |f(x_i)g(x_i)-f(x_i)g(x_{i-1})|+\sum_{i=1}^m|f(x_i)g(x_{i-1})-f(x_{i-1})g(x_{i-1})|\\ &=\sum_{i=1}^m |f(x_i)||g(x_i)-g(x_{i-1})|+\sum_{i=1}^m|g(x_{i-1})||f(x_i)-f(x_{i-1})|\\ &\leq M\sum_{i=1}^m |g(x_i)-g(x_{i-1})|+N\sum_{i=1}^m|f(x_i)-f(x_{i-1})|\\ &=MS_\Gamma[g;a,b]+NS_\Gamma[f;a,b]\\ &\leq MV[g;a,b]+NV[f;a,b] \end{align} $$Therefore, $$V[fg;a,b]\leq MV[g;a,b]+NV[f;a,b]<+\infty$$.

Suppose $$\exists\epsilon>0$$ such that $$|g(x)|\geq\epsilon$$ for all $$x\in[a,b]$$. Let $$M=1/\epsilon$$. Then$$\left| \frac{1}{g(x)} \right|\leq \frac{1}{\epsilon}=M.$$It suffices to show that $$1/g$$is of bounded variation. Then the result will follow since $$f/g$$is the product of $$f$$and $$1/g$$. Notice that$$ \begin{align} S_\Gamma[1/g;a,b]&=\sum_{i=1}^m\left| \frac{1}{g(x_i)}-\frac{1}{g(x_{i-1})} \right|\\ &=\sum_{i=1}^m \left| \frac{g(x_{i-1})-g(x_i)}{g(x_i)g(x_{i-1})} \right|\\ &\leq \sum_{i=1}^m M^2|g(x_i)-g(x_{i-1})|\\ &=M^2 S_\Gamma[g;a,b]\\ &\leq M^2 V[g;a,b]. \end{align} $$Therefore, $$ V[1/g;a,b]\leq M^2 V[g;a,b]<+\infty $$.