Formula for Total Variation of Continuously Differentiable Function

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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

Let $\map {V_} {\closedint a b}$ be the total variation of $f$ on $\closedint a b$.


Then:

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


Proof

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

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

with:

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

We first show that:

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

We then show that:

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

from which the claim follows.

From Differentiable Function is Continuous:

$f$ is continuous on $\closedint a b$.

From Restriction of Continuous Mapping is Continuous:

$f$ is continuous on $\closedint {x_i} {x_{i + 1} }$ for each $i$.

Since $f'$ is continuous, we can apply the Fundamental Theorem of Calculus to obtain:

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

Then:

\(\ds \size {\map f {x_{i + 1} } - \map f {x_i} }\) \(=\) \(\ds \size {\int_{x_i}^{x_{i + 1} } \map {f'} x \rd x}\)
\(\ds \) \(\le\) \(\ds \int_{x_i}^{x_{i + 1} } \size {\map {f'} x} \rd x\)

We therefore have, for each finite subdivisions $P$:

\(\ds \map {V_f} {P ; \closedint a b}\) \(=\) \(\ds \sum_{i \mathop = 0}^n \size {\map f {x_{i + 1} } - \map f {x_i} }\)
\(\ds \) \(\le\) \(\ds \sum_{i \mathop = 0}^n \int_{x_i}^{x_{i + 1} } \size {\map {f'} x} \rd x\)
\(\ds \) \(=\) \(\ds \int_a^b \size {\map {f'} x} \rd x\)

From the definition of total variation, it follows that:

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

We now turn attention to proving:

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

For each $i$, there exists $c_i \in \openint {x_i} {x_{i + 1} }$ such that:

$\ds \map f {x_{i + 1} } - \map f {x_i} = \map f {c_i} \paren {x_{i + 1} - x_i}$

by the Mean Value Theorem.

We then have:

$\ds \map L {\size {f'}, P} = \sum_{i \mathop = 0}^n \paren {\inf_{\closedint {x_i} {x_{i + 1} } } \size {f'} } \paren {x_{i + 1} - x_i}$

From the definition of infinum, we have:

$\ds \inf_{\closedint {x_i} {x_{i + 1} } } \size {f'} \le \size {\map {f'} {c_i} }$

for each $i$.

Then:

\(\ds \sum_{i \mathop = 0}^n \paren {\inf_{\closedint {x_i} {x_{i + 1} } } \size {f'} } \paren {x_{i + 1} - x_i}\) \(\le\) \(\ds \sum_{i \mathop = 0}^n \size {\map {f'} {c_i} } \paren {x_{i + 1} - x_i}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 0}^n \size {\map f {x_{i + 1} } - \map f {x_i} } \paren {x_{i + 1} - x_i}\)
\(\ds \) \(=\) \(\ds \map {V_f} {P ; \closedint a b}\)

That is:

$\ds \map L {\size {f'}, P} \le \map {V_f} {P ; \closedint a b}$

Taking the supremum over all finite subdivisions $P$ of $\closedint a b$, we have:

$\ds \map L {\size {f'} } \le \map {V_f} {\closedint a b}$

Note that $f$ is continuously differentiable, so:

$f'$ is continuous.

From Absolute Value of Continuous Real Function is Continuous:

$\size {f'}$ is continuous.

From Continuous Real Function is Darboux Integrable, we have:

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

So we obtain:

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

Hence the result.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Formula_for_Total_Variation_of_Continuously_Differentiable_Function&oldid=580949"