Function of Bounded Variation is Bounded

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Theorem

Let $a, b$ be real numbers with $a < b$.

Let $f : \closedint a b \to \R$ be a function of bounded variation.


Then $f$ is bounded.


Proof

We use the notation from the definition of bounded variation.

Since $f$ is of bounded variation, there exists $M \ge 0$ such that:

$\map {V_f} {P ; \closedint a b} \le M$

for all finite subdivisions $P$ of $\closedint a b$.

Let $x$ be a real number with:

$a < x < b$

Then $\set {a, x, b}$ is a finite subdivision of $\closedint a b$.

We have:

$\map {V_f} {\set {a, x, b} ; \closedint a b} = \size {\map f x - \map f a} + \size {\map f b - \map f x}$

Since $x \in \openint a b$ was arbitrary, we therefore have:

$\size {\map f x - \map f a} + \size {\map f b - \map f x} \le M$

for all $x \in \openint a b$.

We have:

\(\ds \size {\map f x - \map f a} + \size {\map f b - \map f x}\) \(\ge\) \(\ds \size {\map f x - \map f a}\)
\(\ds \) \(\ge\) \(\ds \size {\size {\map f x} - \size {\map f a} }\) Reverse Triangle Inequality: Real and Complex Fields
\(\ds \) \(\ge\) \(\ds \size {\map f x} - \size {\map f a}\)

So for all $x \in \openint a b$, we have:

$\size {\map f x} \le \size {\map f a} + M$

Since $M \ge 0$, this inequality is also satisfied for $x = a$.

We therefore have:

$\size {\map f x} \le \map \max {\size {\map f a} + M, \size {\map f b} }$

for all $x \in \closedint a b$.

So $f$ is is bounded.

$\blacksquare$


Sources

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