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 $\map {V_f} {\closedint a b}$ be the total variation of $f$ on $\closedint a b$.
Then:
- $\map {V_f} {\closedint a b} \ge 0$
with equality if and only if $f$ 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 $\closedint a b$.
Write:
- $P = \set {x_0, x_1, \ldots, x_n}$
with:
- $a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then:
- $\ds \map {V_f} {P ; \closedint a b} = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } } \ge 0$
So, by the definition of supremum, we have:
- $\ds \map {V_f} {\closedint a b} = \sup_P \paren {\map {V_f} {P ; \closedint a b} } \ge 0$
Now consider $f$ a constant function.
We show that $\map {V_f} {\closedint a b} = 0$.
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:
\(\ds \map {V_f} {P ; \closedint a b}\) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
We then have:
\(\ds \map {V_f} {\closedint a b}\) | \(=\) | \(\ds \sup_P \paren {\map {V_f} {P ; \closedint a b} }\) | Definition of Total Variation of Real Function on Closed Bounded Interval | |||||||||||
\(\ds \) | \(=\) | \(\ds \sup \set 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Definition of Supremum of Subset of Real Numbers |
So if $f$ is constant, then $f$ is of bounded variation with:
- $\map {V_f} {\closedint a b} = 0$
It remains to show that if $\map {V_f} {\closedint a b} = 0$, then $f$ is constant.
It suffices to show that if $f$ is not constant then $\map {V_f} {\closedint a b} > 0$.
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:
\(\ds \map {V_f} {\set {a, x, b} ; \closedint a b }\) | \(=\) | \(\ds \size {\map f x - \map f a} + \size {\map f b - \map f x}\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 0\) |
We must then have:
- $\map {V_f} {\closedint a b} \ge \map {V_f} {\set {a, x, b} ; \closedint a b} > 0$
Hence the claim.
$\blacksquare$
Sources
- 1973: Tom M. Apostol: Mathematical Analysis (2nd ed.) ... (previous) ... (next): $\S 6.4$: Total Variation