Ex2-2

Solution to Chapter 2, Exercise 2
Theorem 2.1 (i)
If fis of bounded variation on [a,b], then f is bounded on [a,b].
Proof
It can be proved by taking contrapositive. We need to consider two cases. Case 1: Both |f(a)|< \infty and |f(b)|< \infty; Case 2: |f(a)|=\infty or |f(b)|=\infty. 
Case1 : Both |f(a)|< \infty and |f(b)|< \infty.
Assume that f is unbounded on [a,b], then there exists a sequence \{c_{j}\}\subset[a,b] such that |f(c_j)|>j. Now consider a partition \Gamma_j=\{a,c_j,b\}. Then we approximate the variation by\begin{align}
S_{\Gamma_j}&=\sum\limits_{k=1}^2\vert f(x_k)-f(x_{k-1})\vert\\
&=\vert f(c_j)-f(a)\vert+\vert f(b)-f(c_j)\vert\\
&=\vert f(c_j)-f(a)\vert+\vert f(c_j)-f(b)\vert\\
&\ge\vert f(c_j)\vert-\vert f(a)\vert+\vert f(c_j)\vert-\vert f(b)\vert\\
&\ge2\vert f(c_j)\vert-\vert f(a)\vert-\vert f(b)\vert\\
&\ge2j-\vert f(a)\vert-\vert f(b)\vert.
\end{align}
Next we take the sup to find the variation
\begin{align}
V(f,[a,b])&=\sup_{\Gamma}S_\Gamma\\
&\ge\sup_{j}S_{\Gamma_j}\\
&\ge\sup_{j}\{2j-\vert f(a)\vert-\vert f(b)\vert\}.
\end{align}
Since \displaystyle j can be arbitrarily large, so \sup_{j}\{2j-\vert f(a)\vert-\vert f(b)\vert\} is unbounded. Consequently, V(f,[a,b]) is unbounded. Under our assumption, f is not of bounded variation.
Case 2: |f(a)|=\infty or |f(b)|=\infty.
Similarly, there exists a sequence \{c_{j}\}\subset[a,b] such that |f(c_j)|>j since f is unbounded. We choose a partition \Gamma=\{a,c_j,d,b\}, where d \in (a,b) satisfies that |f(d)| < \infty. 
Then, we calculate the variation by\begin{align}
S_{\Gamma_j}&=\sum\limits_{k=1}^3\vert f(x_k)-f(x_{k-1})\vert\\
&=\vert f(c_j)-f(a)\vert+\vert f(d)-f(c_j)\vert+\vert f(b)-f(d)\vert\\
&\ge\vert f(d)-f(c_j)\vert\\
&\ge\vert f(c_j)\vert-\vert f(d)\vert
\end{align}Next we take the sup to find the variation\begin{align}
V(f,[a,b])&=\sup_{\Gamma}S_\Gamma\\
&\ge\sup_{j}S_{\Gamma_j}\\
&\ge\sup_j (\vert f(c_j)\vert-\vert f(d)\vert).
\end{align}Since \displaystyle j can be arbitrarily large, V(f,[a,b]) is unbounded in this case, which means f is not of bounded variation.
Another proof of Theorem 2.1 (i)
Let V(f;[a,b])=\sup_\Gamma S_\Gamma\leq M. Let x\in(a,b).
Consider the partition \Gamma=\{a,x,b\}.
Then |f(x)-f(a)|\leq|f(x)-f(a)|+|f(b)-f(x)|=S_\Gamma\leq\sup_\Gamma S_\Gamma\leq M.
This implies f(a)-M\leq f(x)\leq f(a)+M, so f is bounded.
Proof of Theorem 2.1 (ii)
Suppose fand gare 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]<+\inftyas well.
Since fand gare 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 Nfor 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/gis of bounded variation. Then the result will follow since f/gis the product of fand 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
.