Bounded Variation Implies Bounded (Trivial Case)

Question

Suppose $f$ is of bounded variation on $[a,b]$. The trivial case for when $x=a$ doesn't seem obvious to me. Why is $f$ bounded when $x=a$? I know the inequality $|f(x)|\leq |f(a)|+M$ holds, where $M$ is a bound for $\sum|\Delta f_{k}|$, but that would mean $M\geq 0$. Does this show $f$ is bounded? I feel like I am missing something....do I need a partition?

Answer 1

The statement that $f$ has bounded variation on $[a,b]$ requires that $f:[a,b] \to \mathbf{R}$ is a function with $f(a) \in \mathbf{R}$ and there exists a positive number $M$ such that

$$\sum_{k=1}^n|f(x_k) - f(x_{k-1})| \leqslant M,$$ for $any$ partition $P = (x_0,x_1,\ldots, x_n)$ where $a= x_0 < x_1 < \ldots, x_n = b$.

Given any point $x \in (a,b]$ we can select a partition where $x = x_j > a$ and see using the reverse and forward triangle inequalitites that that

$$|f(x) - |f(a)| \leqslant |f(x) - f(a)| \\\leqslant |f(x) - f(x_{j-1})| + \ldots +|f(x_1)-f(a)|,$$

whence, it follows that for all $x \in (a,b],$

$$|f(x)| \leqslant |f(a)| + \sum_{k=1}^j|f(x_k) - f(x_{k-1})| \leqslant |f(a)| +M$$

If $x = a$ we have the trivial bound $|f(x)| \leqslant |f(a)|$.