# Absolute continuity, function of partition bound

1. Nov 20, 2013

### SqueeSpleen

Given $[a,b]$ a bounded interval, and $f \in L^{p} ([a,b]) 1 < p < \infty$, we define:
$F(x) = \displaystyle \int_{a}^{x} f(t) dt$, $x \in [a,b]$
Prove that exists $K \in R$ such that for every partition:
$a_{0} = x_{0} < x_{1} < ... < x_{n} = b$ :
$\displaystyle \sum_{i=0}^{n-1} \frac{| F(x_{i+1}) - F(x_{i}) |^{p}}{|x_{i+1}-x_{i} |^{p-1}} \leq K$

I know that $F(x)$ is absolutely continous and of bounded variation.

$\frac{|F(x_{i+1}) - F(x_{i})|}{|x_{i+1} - x_{i}|} = F(\xi)$ for some $\xi \in [x_{i}, x_{i+1}]$ (Lagrange Intermediate Value Theorem)
Then
$\frac{|F(x_{i+1}) - F(x_{i})|^{p}}{|x_{i+1} - x_{i}|^{p-1}} = f(\xi)^{p} (x_{i+1}-x_{i})$
As $f \in L^{p}$ when $x_{i+1} \to x_{i}$ $f(x_{i+1}-x_{i})$ when grows slower in some neighborhood of $x_{i}$ than $\frac{1}{x_{i+1}-x_{i}}$ because that function doesn't isn't $L^{1}$.
I was trying prove that the closenes of the $x_{n}$ can't make the sum diverges, and that neither the number of terms but I don't know how to approach all the issues at the same time.
When the partition norm tends to zero we got $\displaystyle \int_{a}^{b} | f (t) |^{p} dt$
And when it doesn't I think that $(2M)^{p} (b-a)$ where $M$ is the maximum of $F(x)$ with $x \in [a,b]$ (which exists because $F(x)$ is continous in that closed set).
So I guess $K = (2M)^{p} (b-a) + \displaystyle \int_{a}^{b} | f (t) |^{p} dt$ works
But I'm not sure how to prove it.

2. Nov 20, 2013

### brmath

I got hung up when you wrote:

$\frac {F(x_{i+1} - F(x_i)|}{|x_{i+1} - x_i} = F(\zeta)$... surely you mean $F'(\zeta)$ which would be $f(\zeta)$.. From the next line I see you did mean that.

Sum up your $f^p(\zeta)(x_{i+1} - x_i).$ As the width of the partition goes to 0 you will have the Riemann integral of $f^p$.

I think you have to show (or state) that if it is Riemann integrable then it is Lebesgue integrable.
Can you take it from there?