Given a function $f$, prove: if $ f\in BV([a,b])$ then $f$ is bounded

Question

prove: if $f\in BV([a,b])$ then $f$ is a bounded function

i use the method of contrapositive i suppose f is not bounded such that f goes to ∞ i want to prove that f is not in BV ([a,b]) .

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Answer 1

A direct proof: we denote by $T$ the total variation of $f$. Let $x \in [a,b]$. Then $\{a,x,b\}$ is a partition of $[a,b]$, hence

$S(x):=|f(a)-f(x)|+|f(x)-f(b)| \le T$.

Therefore

$|f(x)| =|f(x)-f(a)+f(a)| \le |f(x)-f(a)|+|f(a)| \le S(x)+|f(a)| \le T+|f(a)|.$