Prove variation of a differentiable function

[stripes]

Homework Statement

Prove that if a function f is once-differentiable on the interval [a, b], then

$Vf = \int ^{b}_{a} | f'(x) | dx,$

where $Vf = sup_{P} \sum^{i=n-1}_{0} | f(x_{i+1}) - f(x_{i}) |$ where the supremum is taken over all partitions $P = \left\{ a = x_{0} < x_{1} < ... < x_{n} = b .\right\}$

Use Taylor's Theorem (or the mean value theorem) and the definition of the integral.

Hint, adding an extra point to a partition cannot decrease $\sum^{i=n-1}_{0} | f(x_{i+1}) - f(x_{i}) |$

In our class, we use this definition for a Riemann integral:

Homework Equations

The Attempt at a Solution

Well I've gotten as far as writing down the definition of the integral and the definition of the total variation. Not sure how to apply the mean value theorem here. I did apply it on a later question but we were given a function to work with. Not sure how to start this one.

 

[mighty2000]

Thank you for your post. This is an interesting problem that requires some knowledge of calculus and the definition of the Riemann integral. Let's start by writing down the definitions of the integral and the total variation:

The definition of the Riemann integral is:

\int ^{b}_{a} f(x) dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_{i}) \Delta x_{i}

where the limit is taken as the norm of the partition \Delta x_{i} goes to 0.

The definition of the total variation is:

Vf = sup_{P} \sum^{i=n-1}_{0} | f(x_{i+1}) - f(x_{i}) |

where the supremum is taken over all partitions P = \left\{ a = x_{0} < x_{1} < ... < x_{n} = b .\right\}

Now, let's use Taylor's Theorem to prove the given statement:

By Taylor's Theorem, we have:

f(x_{i+1}) = f(x_{i}) + f'(c_{i})(x_{i+1}-x_{i})

where c_{i} is some point between x_{i} and x_{i+1}.

Substituting this into the definition of the total variation, we get:

Vf = sup_{P} \sum^{i=n-1}_{0} | f'(c_{i}) | \Delta x_{i}

Now, since f is once-differentiable on the interval [a, b], we know that f' is continuous on [a, b]. This means that f' is bounded on [a, b]. Let's say | f'(x) | \leq M for all x in [a, b].

Therefore, we can write:

Vf = sup_{P} \sum^{i=n-1}_{0} | f'(c_{i}) | \Delta x_{i} \leq M \sup_{P} \sum^{i=n-1}_{0} \Delta x_{i} = M (b-a)

Now, let's look at the definition of the Riemann integral. We know that the integral is equal to the limit of the sum as the norm of the partition goes to 0. This means