User:Caliburn/s/10

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

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

Let $V_f$ be the total variation of $f$.

Then:


 * $\ds V_f = \int_a^b \size {\map {f'} x} \rd x$

Proof
We prove that:


 * $\ds V_f \le \int_a^b \size {\map {f'} x} \rd x$

and:


 * $\ds \int_a^b \size {\map {f'} x} \rd x \le V_f$

For each finite subdivision $P$ of $\closedint a b$, write:


 * $P = \set {x_0, x_1, \ldots, x_n }$

with:


 * $a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$

From the Fundamental Theorem of Calculus: Second Part, we have:


 * $\ds \map f {x_i} - \map f {x_{i - 1} } = \int_{x_{i - 1} }^{x_i} \map {f'} x \rd x$

So, we have:

We then have, for each finite subdivision $P$:

By the definition of total variation, we then have:


 * $\ds V_f \le \int_a^b \size {\map {f'} x} \rd x$

For the other inequality, we note that we have:


 * $\ds \map L {\size {f'}, P} = \sum_{k \mathop = 1}^n \inf_{x \in \closedint {x_{k - 1} } {x_k} } \size {f'} \paren {x_k - x_{k - 1} }$

where $\map L {\size {f'}, P}$ is the lower sum of $\size {f'}$ with respect to $P$.

By the Mean Value Theorem, for each $k \in \N$ with $k \le N$ there exists $c_k \in \closedint {x_{k - 1} } {x_k}$ such that:


 * $\ds \map {f'} {c_k} = \frac {\map f {x_k} - \map f {x_{k - 1} } } {x_k - x_{k - 1} }$

By the definition of infimum, we have:


 * $\ds \inf_{x \in \closedint {x_{k - 1} } {x_k} } \size {f'} \le \size {\map {f'} {c_k} }$

so:

We therefore have:


 * $\ds \map L {\size {f'}, P} \le \sum_{k \mathop = 1}^n \size {\map f {x_k} - \map f {x_{k - 1} } } = \map {V_f} P$

for every finite subdivision $P$.