Let $X$ be a topological vector space and let $f:[a,b]\to X$. We say that $f$ is of bounded semi-variation in $[a,b]$ if the set $SV(f,[a,b])$ consisting of all the elements of the form $$\sum_{i=1}^n a_i[f(t_i)-f(t_{i-1})]$$ where $a=t_0<t_1<\dots<t_n=b$ and $|a_i|\le 1$, $a_i$ being complex numbers, is a bounded subset in $X.$
Question. With the above definition, is the set $$f[a,b]=\{f(x):x\in [a,b]\}$$ a bounded set in $X$ whenever $f$ is of bounded semi-variation?
If the answer is yes, then in this case, we say that $f$ is bounded and I would be thankful if you can give me some ideas or tips to resolved the question. Many thanks in advance.
Since $f$ has finite semivariation, then there exist bounded set $B\subset X$ such that $$ \sum_{i=1}^n a_i[f(t_i)-f(t_{i-1})]\in B $$ for all $a=t_0<t_1<\dots<t_n=b$ and $|a_i|\le 1$, $a_i$ being complex numbers.
Fix $x\in[a,b]$ and set $n=2$, $t_0=a$, $t_1=x$, $t_2=b$, $a_1=1$, $a_2=-1$, then $$ (f(x)-f(a))-(f(b)-f(x))\in B $$ this means that $$ f(x)\in 0.5(f(a)+f(b)+B) $$ Since $x\in[a,b]$ is arbitrary $$ f([a,b])\subset 0.5(f(a)+f(b)+B) $$ The set in the rifght hand side is clearly bounded, hence so does $f([a,b])$.