- #1
SqueeSpleen
- 141
- 5
Given [itex][a,b][/itex] a bounded interval, and [itex]f \in L^{p} ([a,b]) 1 < p < \infty[/itex], we define:
[itex]F(x) = \displaystyle \int_{a}^{x} f(t) dt[/itex], [itex]x \in [a,b][/itex]
Prove that exists [itex]K \in R[/itex] such that for every partition:
[itex]a_{0} = x_{0} < x_{1} < ... < x_{n} = b[/itex] :
[itex]\displaystyle \sum_{i=0}^{n-1} \frac{| F(x_{i+1}) - F(x_{i}) |^{p}}{|x_{i+1}-x_{i} |^{p-1}} \leq K[/itex]
I know that [itex]F(x)[/itex] is absolutely continuous and of bounded variation.
[itex]\frac{|F(x_{i+1}) - F(x_{i})|}{|x_{i+1} - x_{i}|} = F(\xi)[/itex] for some [itex]\xi \in [x_{i}, x_{i+1}][/itex] (Lagrange Intermediate Value Theorem)
Then
[itex]\frac{|F(x_{i+1}) - F(x_{i})|^{p}}{|x_{i+1} - x_{i}|^{p-1}} = f(\xi)^{p} (x_{i+1}-x_{i})[/itex]
As [itex]f \in L^{p}[/itex] when [itex]x_{i+1} \to x_{i}[/itex] [itex]f(x_{i+1}-x_{i})[/itex] when grows slower in some neighborhood of [itex]x_{i}[/itex] than [itex]\frac{1}{x_{i+1}-x_{i}}[/itex] because that function doesn't isn't [itex]L^{1}[/itex].
I was trying prove that the closenes of the [itex]x_{n}[/itex] 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 [itex]\displaystyle \int_{a}^{b} | f (t) |^{p} dt[/itex]
And when it doesn't I think that [itex](2M)^{p} (b-a)[/itex] where [itex]M[/itex] is the maximum of [itex]F(x)[/itex] with [itex]x \in [a,b][/itex] (which exists because [itex]F(x)[/itex] is continuous in that closed set).
So I guess [itex]K = (2M)^{p} (b-a) + \displaystyle \int_{a}^{b} | f (t) |^{p} dt [/itex] works
But I'm not sure how to prove it.
[itex]F(x) = \displaystyle \int_{a}^{x} f(t) dt[/itex], [itex]x \in [a,b][/itex]
Prove that exists [itex]K \in R[/itex] such that for every partition:
[itex]a_{0} = x_{0} < x_{1} < ... < x_{n} = b[/itex] :
[itex]\displaystyle \sum_{i=0}^{n-1} \frac{| F(x_{i+1}) - F(x_{i}) |^{p}}{|x_{i+1}-x_{i} |^{p-1}} \leq K[/itex]
I know that [itex]F(x)[/itex] is absolutely continuous and of bounded variation.
[itex]\frac{|F(x_{i+1}) - F(x_{i})|}{|x_{i+1} - x_{i}|} = F(\xi)[/itex] for some [itex]\xi \in [x_{i}, x_{i+1}][/itex] (Lagrange Intermediate Value Theorem)
Then
[itex]\frac{|F(x_{i+1}) - F(x_{i})|^{p}}{|x_{i+1} - x_{i}|^{p-1}} = f(\xi)^{p} (x_{i+1}-x_{i})[/itex]
As [itex]f \in L^{p}[/itex] when [itex]x_{i+1} \to x_{i}[/itex] [itex]f(x_{i+1}-x_{i})[/itex] when grows slower in some neighborhood of [itex]x_{i}[/itex] than [itex]\frac{1}{x_{i+1}-x_{i}}[/itex] because that function doesn't isn't [itex]L^{1}[/itex].
I was trying prove that the closenes of the [itex]x_{n}[/itex] 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 [itex]\displaystyle \int_{a}^{b} | f (t) |^{p} dt[/itex]
And when it doesn't I think that [itex](2M)^{p} (b-a)[/itex] where [itex]M[/itex] is the maximum of [itex]F(x)[/itex] with [itex]x \in [a,b][/itex] (which exists because [itex]F(x)[/itex] is continuous in that closed set).
So I guess [itex]K = (2M)^{p} (b-a) + \displaystyle \int_{a}^{b} | f (t) |^{p} dt [/itex] works
But I'm not sure how to prove it.