Total Variation is Non-Negative

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

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

Let $V_f$ be the total variation of $f$ on $\closedint a b$.

Then:


 * $V_f \ge 0$

with equality $f$ is constant.

Proof
We use the notation from the definition of bounded variation.

Note that by the definition of absolute value, we have:


 * $\size x \ge 0$

for all $x \in \R$.

Let $P$ be a finite subdivision of $\closedint a b$.

Then:


 * $\displaystyle \map {V_f} P = \sum_{i \mathop = 1}^{\size P - 1} \size {\map f {x_i} - \map f {x_{i - 1} } } \ge 0$

So, by the definition of supremum, we have:


 * $\displaystyle V_f = \sup_P \paren {\map {V_f} P} \ge 0$

Let $f$ be constant.

Then:


 * $\size {\map f {x_i} - \map f {x_{i - 1} } } = 0$

for all $x_{i - 1}, x_i \in \closedint a b$.

So, for any finite subdivision $P$ we have:

We then have:

So if $f$ is constant, then $f$ is of bounded variation with:


 * $V_f = 0$

It remains to show that if $V_f = 0$, then $f$ is constant.

It suffices to show that if $f$ is not constant then $V_f > 0$.

Since $f$ is non-constant, we can pick $x \in \openint a b$ such that either:


 * $\map f x \ne \map f a$

or:


 * $\map f x \ne \map f b$

So that either:


 * $\size {\map f x - \map f a} > 0$

or:


 * $\size {\map f b - \map f x} > 0$

Note that we then have:

We must then have:


 * $V_f \ge \map {V_f} {\set {a, x, b} } > 0$

Hence the claim.