Ex2-27

If f is an even function on [-1,1], verify the formulas:
(i) V[f;-1,1]=2V_P[f;-1,1];
(ii) V[f;0,1]=V_P[f;-1,1];
(iii) V_P[f;0,1]=V_N[f;-1,0].
f is an even function on [-1,1] implies that it is symmetric to the y-axis. By the definition of even functions, 
f(x)=f(-x)~\text{for all}~x\in[0,1].
This implies that f(-1)=f(1).
Consider the case when V is finite. Then, by Theorem 2.6, V_P and V_N is finite. Moreover, V_P+V_N=V and V_P-V_N=f(1)-f(-1)=0.
(i)\begin{align}
V_P[f;-1,1]&=\frac{1}{2}[V[f;-1,1]+f(1)-f(-1)]\\
&=\frac{1}{2}V[f;-1,1]\\
\implies V[f;-1,1] &=2V_P[f;-1,1].
\end{align}
(ii) Since the graph remains unchanged after reflection about the y-axis, we have V[f;-1,0]=V[f;0,1], thus
\begin{align}
V[f;-1,1]&=V[f;-1,0]+V[f;0,1]\\
\implies V[f;-1,1] &=2V[f;-1,0]=2V[f;0,1].
\end{align}
Combining with the result of (i),
\begin{align}
2V_P[f;-1,1]=V[f:-1,1]=2V[f;0,1]\\
\implies V[f;0,1]=V_P[f;-1,1].
\end{align}(iii)
 The idea of this proof is more or less right, but it is too vague. Someone should edit it to make the calculations more explicit. 
Given f an even function. Consider a partition \Gamma=\{-1=-x_m, -x_{m-1}, \ldots -x_1,x_0=0,x_1,\ldots, x_{m-1}, x_m=1\}. Choose x_i, x_{i+1}\in \Gamma, where i\in [0,m-1]. If f is increasing (decreasing) in the interval [-x_{i+1},-x_{i}], where \{-x_{i+1},-x_{i}\}\in[-1,0], then f is decreasing (increasing) in the interval [x_{i+1},x_{i}], where \{x_{i+1},x_i\}\in [0,1]. Moreover, they are with the same variation. Thus, by reviewing the definition of V_P and V_N,
V_P[f;0,1]=V_N[f;-1,0].