Theorem: Formula for Total Variation of Continuously Differentiable Function

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:
{{begin-eqn}}
{{eqn	| l = \size {\map f {x_{i + 1} } - \map f {x_i} }
	| r = \size {\int_{x_i}^{x_{i + 1} } \map {f'} x \rd x}
}}
{{eqn	| o = \le
	| r = \int_{x_i}^{x_{i + 1} } \size {\map {f'} x} \rd x
}}
{{end-eqn}}
We therefore have, for each finite subdivisions $P$:
{{begin-eqn}}
{{eqn	| l = \map {V_f} {P ; \closedint a b}
	| r = \sum_{i \mathop = 0}^n \size {\map f {x_{i + 1} } - \map f {x_i} }
}}
{{eqn	| o = \le
	| r = \sum_{i \mathop = 0}^n \int_{x_i}^{x_{i + 1} } \size {\map {f'} x} \rd x
}}
{{eqn	| r = \int_a^b \size {\map {f'} x} \rd x
}}
{{end-eqn}}
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: 
{{begin-eqn}}
{{eqn	| l = \sum_{i \mathop = 0}^n \paren {\inf_{\closedint {x_i} {x_{i + 1} } } \size {f'} } \paren {x_{i + 1} - x_i}
	| o = \le
	| r = \sum_{i \mathop = 0}^n \size {\map {f'} {c_i} } \paren {x_{i + 1} - x_i}
}}
{{eqn	| r = \sum_{i \mathop = 0}^n \size {\map f {x_{i + 1} } - \map f {x_i} } \paren {x_{i + 1} - x_i}
}}
{{eqn	| r = \map {V_f} {P ; \closedint a b}
}}
{{end-eqn}}
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.
{{qed}}
Category:Total Variation of Real Function
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