Let $BV[a,b]$ be collection of all bounded variation real function on $[a,b]$. If $f,g\in BV[a,b]$ and $\alpha,\beta\in\mathbb{R}$ prove $$\alpha f+\beta g\in BV[a,b].$$
I try as below.
Given $f,g\in BV[a,b]$ then $$ V(f,[a,b])=\sup \sum\limits_{k=1}^n\vert f(x_k)-f(x_{k-1})\vert\leq M_1 $$ and $$ V(g,[a,b])=\sup \sum\limits_{k=1}^n\vert g(x_k)-g(x_{k-1})\vert\leq M_2. $$ I want to prove $$\alpha f+\beta g\in BV[a,b],$$ i.e. $$ V(\alpha f+\beta g,[a,b])=\sup \sum\limits_{k=1}^n\vert (\alpha f+\beta g)(x_k)-(\alpha f+\beta g)(x_{k-1})\vert\leq M. $$ Is it right the prove as follows? \begin{align} V(\alpha f+\beta g,[a,b])&=\sup \sum\limits_{k=1}^n\vert (\alpha f+\beta g)(x_k)-(\alpha f+\beta g)(x_{k-1})\vert\\ &= \sup \sum\limits_{k=1}^n\vert \alpha f(x_k)+\beta g(x_k)-\alpha f(x_{k-1})-\beta g(x_{k-1})\vert\\ &= \sup \sum\limits_{k=1}^n\vert \alpha (f(x_k)-f(x_{k-1}))+\beta (g(x_k)-g(x_{k-1}))\vert\\ &\leq \alpha\sup \sum\limits_{k=1}^n\vert (f(x_k)-f(x_{k-1}))\vert+\beta\sup \sum\limits_{k=1}^n \vert(g(x_k)-g(x_{k-1}))\vert\\ &= \alpha M_1+\beta M_2\\ &= M \end{align}
EDIT \begin{align} V(\alpha f+\beta g,[a,b])&=\sup \sum\limits_{k=1}^n\vert (\alpha f+\beta g)(x_k)-(\alpha f+\beta g)(x_{k-1})\vert\\ &= \sup \sum\limits_{k=1}^n\vert \alpha f(x_k)+\beta g(x_k)-\alpha f(x_{k-1})-\beta g(x_{k-1})\vert\\ &= \sup \sum\limits_{k=1}^n\vert \alpha (f(x_k)-f(x_{k-1}))+\beta (g(x_k)-g(x_{k-1}))\vert\\ &\leq \vert\alpha\vert\sup \sum\limits_{k=1}^n\vert (f(x_k)-f(x_{k-1}))\vert+\vert\beta\vert\sup \sum\limits_{k=1}^n \vert(g(x_k)-g(x_{k-1}))\vert\\ &= \vert\alpha\vert M_1+\vert\beta\vert M_2\\ &= M \end{align}