A function of bounded variation is bounded

Question

It seems that a function of bounded variation is bounded. But how do we prove it?

Maybe to show this, we should prove that if a function is not bounded, then it is not of bounded variation.

So let a function $f$ be unbounded. Now how can I show that $\sup$$t_f$ is not finite?

where $t_f$ =$\sum_{j=1}^n\lvert f(x_j)-f(x_{j-1})\rvert$ where {$a=x_0 \lt x_1 \lt...\lt x_n=b$} is any partition of [$a,b$].

Another one: https://math.stackexchange.com/q/389234. – Martin R Jun 11 '20 at 08:53 — Martin R, Jun 11 '20 at 08:53

score 2 · Answer 1 · answered Jun 11 '20 at 08:45

2

For any $x\in [a,b]$ consider the trivial partition $\{a,x,b\}$, then, $|f(x)|-|f(a)|+|f(x)|-|f(b)|\leq |f(x)-f(a)|+|f(b)-f(x)|\leq M<\infty$, where $M$ is the total variation.

answered Jun 11 '20 at 08:45

Sumanta

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A function of bounded variation is bounded

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