If $[a',b']$ is a subinterval of $[a,b]$ show that $P[a',b'] \leq P[a,b]$
where $P$ is the positive variation defined by $P=sup \sum_{i=1}^m [f(x_i)-f(x_{i-1})]^+$ where $x^+$ is defined by $x$ if $x>0$ and $0$ if $x\leq 0$
I know this to be true for total variation - i.e $V[a',b']\leq V[a,b]$, and I think I know the general outline of that proof, but I don't believe I can follow it to obtain a proof for the positive variation (or negative variation).
I know that $P= \frac{1}{2} (V+f(b)-f(a))$ but I wasn't able to get a proof by simply manipulating things, can anyone show me the method or provide the hint/trick I am missing?
Thank you
Let $(x_1, x_2, \ldots, x_m)$ be any partition of $[a', b']$. Then $(a, x_1, x_2, \ldots, x_2, b)$ is a partition of $[a, b]$. It follows that $$ \sum_{i=1}^m [f(x_i)-f(x_{i-1})]^+ \le [f(x_1)-f(a)]^+ + \sum_{i=1}^m [f(x_i)-f(x_{i-1})]^+ + [f(b)-f(x_m)]^+ \\ \le P[a, b] \, . $$ So we have $$ \sum_{i=1}^m [f(x_i)-f(x_{i-1})]^+ \le P[a, b] $$ for all partitions $(x_1, x_2, \ldots, x_m)$ of $[a', b']$. Then the same inequality holds for the supremum over all partitions of $[a', b']$, i.e. $P[a', b'] \le P[a, b] $.