Question about positive variations of function Given a function $f:\mathbb{R} \to \mathbb{C}$, Folland in his book "Real Analysis" defines the total variation of $f$ at $x \in \mathbb{R}$ as
$$
T_f(x) = \sup \left\{\sum_{1}^n |f(x_j)-f(x_{j-1})|: -\infty < x_0 <x_1 < \dots x_n = x , n \in \mathbb{N} \right \}
$$
and the space $BV(R)$ as the set of those functions with finite $T_f(x)$ for $x \in \mathbb{R}$. Then, for $f \in BV(R)$, he defines the positive variation of $f$ denoted by  $v(f)^+$
$$
v(f)^+ := \frac{1}{2} (T_f+f)(x) = \sup \left\{\sum_{1}^n [f(x_j)-f(x_{j-1})]^+: -\infty < x_0 <x_1 < \dots x_n = x , n \in \mathbb{N} \right \} + f(-\infty)$$
but I do not understand where the term $f(-\infty)$ comes from.
What I was able to get is
$$ 
v(f)^+  =  \sup \left\{\sum_{1}^n [f(x_j)-f(x_{j-1})]^+ + \frac{1}{2}f(x_0): -\infty < x_0 <x_1 < \dots x_n = x , n \in \mathbb{N} \right \}
$$
What am I missing?
 A: I agree that the formula needs the term $1/2$ as suggested by Kavi Rama Murthy.
First of all, recall that given $x \in \mathbb{R}$, $2x^+=|x|+x$. Take $x \in \mathbb{R}$, $n \in \mathbb{N}$ and $\left\{ x_j: j=0,...,n \right\}$ such that $-\infty<x_0<...<x_n=x$. Take a strictly decreasing sequence $\left(y_k\right)_{k \in \mathbb{N}}$ such that $\lim_{k \to \infty} y_k = -\infty$ and $y_1<x_0$. Given $k \in \mathbb{N}$, note that
\begin{align*}
\sum_{j=1}^n \left[ f(x_j)-f(x_{j-1}) \right]^+
&\leq
\left[ f(x_0)-f(y_k) \right]^++\sum_{j=1}^n \left[ f(x_j)-f(x_{j-1}) \right]^+
\\
&\leq
\frac{1}{2}\left\{ \left| f(x_0)-f(y_k) \right|+\left[ f(x_0)-f(y_k) \right]+\left[ f(x)-f(x_0)\right]+\sum_{j=1}^n \left| f(x_j)-f(x_{j-1}) \right|\right\}
\\
&\leq
\frac{1}{2}\left\{ \left| f(x_0)-f(y_k) \right|+\left[ f(x)-f(y_k) \right]+\sum_{j=1}^n \left| f(x_j)-f(x_{j-1}) \right|\right\}
\\
&\leq
\frac{1}{2}\left[ T_f(x)+f(x)-f(y_k)\right]
\end{align*}
Passing to the limit, we obtain
$$
\sum_{j=1}^n \left[ f(x_j)-f(x_{j-1}) \right]^+
\leq
\frac{1}{2}\left[ T_f(x)+f(x)-f(-\infty)\right]
$$
Therefore, we obtain the inequality $\leq$ where the left-hand side is obtained as the supremum of those sums over the considered partitions.
We are left with the question of whether the right-hand side is reached by such supremum.
It indeed is: take a sequence of partitions $\left( P_k \right)_{k \in \mathbb{N}}$ where given $k \in \mathbb{N}$, $n_k \in \mathbb{N}$ and $P_k=\left\{ x^k_i : i=0,...,n_k\right\}$ with $x^k_i < x^k_{i+1}$ for any $i=0,...,n_k$ such that $\lim_{k \to \infty}\sum_{i=0}^{n_k} \left[ f(x_i)-f(x_{i-1}) \right]^+$ is the supremum of those sums over the set of finite partitions.
Let $\left( y_k \right)_{k \in \mathbb{N}}$ be a sequence such that $\lim_{k \to \infty} y_k=-\infty$ and given $k \in \mathbb{N}$, $y_k<x^k_0$. Consider the new sequence of partitions $\left( P_k \cup \{ y_k \} \right)_{k \in \mathbb{N}}$. Each element of such sequence gives us an even greater sum, so it still converges to the supremum, and we got the new term we needed to obtain the $f(-\infty)$.
