Ex2-28

Let f be Lipschitz continuous on an interval [a,b]. Show that V[f;a,b] is strictly less than the length of the graph of f over [a,b].
Proof: 
Recall that f is said to be Lipschitz continuous on [a,b], or to be a Lipschitz function [a,b], if there is a constant C\geq 0 such that
|f(x)-f(y)|\leq C|x-y|~\text{for all}~x,y\in [a,b].
Consider a partition \Gamma=\{a=x_0,x_1,\ldots,x_m=b\}. With each partition \Gamma, we have
\begin{align}
S_{\Gamma}[f;a,b]&=\sum_{i=1}^m |f(x_{i})-f(x_{i-1})|\\
&=|f(x_{1})-f(x_{0})|+|f(x_{2})-f(x_{1})|+\ldots+|f(x_{m})-f(x_{m-1})|\\
&\leq C|x_{1}-x_{0}|+C|x_{2}-x_{1}|+\ldots+C|x_{m}-x_{m-1}|\\
&=C(x_{1}-x_{0})+C(x_{2}-x_{1})+\ldots+C(x_{m}-x_{m-1})\\
&=C[x_{1}-x_{0}+x_{2}-x_{1}+\ldots+x_{m}-x_{m-1}]\\
&=C[x_{m}-x_{0}]\\
&=C[b-a].
\end{align}
Since this inequality holds for any \Gamma, thus
\begin{align}
V[f;a,b]&=\underset{\Gamma}{sup}~S_{\Gamma}[f;a,b]\\
&\leq C[b-a].
\end{align}
Also, the length of the graph of f over [a,b] is defined as \int_a^b \sqrt{1+f'(x)^2}~dx. From Corollary 2.10, 
\begin{align}
V&=\int_a^b |f'(x)|~dx\\
&=\int_a^b \sqrt{[f'(x)]^2}~dx\\
&<\int_a^b \sqrt{1+f'(x)^2}~dx.
\end{align}