A question on a function of bounded semivariation Let $(X,\tau)$ be a Hausdorff locally convex TVS and let $P(X)$ be a family of continuous seminorms on $X$ that generates the topology $\tau$. I got the following definition from one of the articles that I have downloaded. 
Definition. A function $f:[a,b]\to X$ is of bounded semi-variation on $[a,b]$ if for each $p\in P(X)$, there exists $K_p> 0$ such that
$$pSV(f,[a,b])=\sup\left\{p\left(\sum_{i=1}^n a_i[f(x_i)-f(x_{i-1})]\right)\right\}\le K_p$$
where the supremum is taken over all $a\le x_0\le x_1\le\dots\le x_n\le b$ and $|a_i|\le 1$. In this case, we say that $pSV(f,[a,b])$ is a $p$-variation of $f$ on $[a,b]$.
It should be noted that the definition above coincides with the usual definition of a function of bounded variation if $X=\mathbb{R}$.
Question. Is it possible that whenever we assume that $f:[a,b]\to X$ is of bounded semi-variation on $[a,b]$ and $c\in (a,b)$, we have
$$pSV(f,[a,b])=pSV(f,[a,c])+pSV(f,[c,b])?$$
I need some help and thanks in advance.
 A: The bounded semi-variation is sub-additive, i.e.
$$ \begin{align*} \DeclareMathOperator \psv{pSV} \psv(f,[a,b]) &\leq \psv(f,[a,c])+\psv(f,[c,b]), \qquad a <c<b. \tag{1} \end{align*}$$
In general, the left-hand side does not equal the right-hand side. (But equality holds e.g. for $X=\mathbb{R}$ endowed with the Euclidean norm.)
Proof of $(1)$:
Let $a<c<b$ and $a = x_0 \leq \ldots \leq x_n = b$, $|a_i| \leq 1$. There exists $j \in \{1,\ldots,n\}$ such that $$x_j>c \qquad \quad x_{j-1} \leq c.$$ By the triangle inequality, we have
$$\begin{align*} p \left( \sum_{i=1}^n a_i \cdot \big( f(x_i)-f(x_{i-1}) \big) \right) &\leq p \left( \sum_{i=1}^{j-1} a_i \big( f(x_i)-f(x_{i-1}) \big) \right) + p \left( \sum_{i=j}^{n} a_i \big( f(x_i)-f(x_{i-1}) \big) \right) \\ &\leq \psv(f,[a,c])+\psv(f,[c,b]) \end{align*}$$
Taking the supremum over all admissible partitions and coefficients, we find
$$\psv(f,[a,b]) \leq \psv(f,[a,c])+\psv(f,[c,b]).$$
Counterexample for super-additivity:
Consider $X=\mathbb{R}^2$ endowed with the maximal norm $$\|x\|_{\infty} := \max\{|x_1|,|x_2|\}, \qquad x \in \mathbb{R}^2.$$ Define $f:[0,1] \to \mathbb{R}^2$ by
$$f(t) := \begin{cases} (t,0) & t \in [0,1/2] \\ (0,t) & t \in (1/2,1] \end{cases}$$
Choose $c=1/2$. By definition of the norm and $f$,
$$\psv(f,[0,1/2]) = \sup \left( \left| \sum_{i=1}^n a_i (t_i-t_{i-1}) \right|; |a_i| \leq 1, 0=t_0<\ldots<t_n \leq 1/2 \right).$$
Therefore, we see that the bounded semivariation coincides with the bounded variation of $g(t) := t$ on the interval $[0,1/2]$. By the monotonicity, we conclude
$$\psv(f,[0,1(2]) = \frac{1}{2}.$$
Similarly,
$$\psv(f,[1/2,1]) = \sup \left( \left| \sum_{i=1}^n a_i (t_i-t_{i-1}) \right|; |a_i| \leq 1, 1/2=t_0<\ldots<t_n \leq 1 \right) = \frac{1}{2}.$$
On the other hand,
$$\begin{align*} \psv(f,[0,1]) &= \sup  \left\{ \left\| \sum_{i=1}^n a_i \cdot (f(t_i)-f(t_{i-1}) \right\|_{\infty}; |a_i| \leq 1, 0=t_0<\ldots<t_n=1 \right\} \\ &= \sup \bigg\{ \max \left( \sum_{i=1}^j a_i \cdot (t_i-t_{i-1}), \sum_{i=j+1}^n a_i \cdot (t_i-t_{i-1}) \right); \\&\quad |a_i| \leq 1, 0=t_0<\ldots<t_j<c<t_{j+1} \ldots<t_n=1 \bigg\}\\ &= \max(\psv(f,[0,1/2]),\psv(f,[1/2,1]) = \frac{1}{2} \\ &< \psv(f,[0,1/2])+\psv(f,[1/2,1]) = 1 \end{align*}$$
