Hey everyone! Any input would be appreciated.
A note on notation, on means that is Riemann-Stieltjes integrable with respect to on the interval
Question 1: Let and define on by . Also, let be continuous at . Is on ? If so what is its value.
Answer: Using the following lemma:
Lemma: Suppose that is bounded on , has a finite number of discontinuities and is continuous at all those points, then on .
We can see that since has only one discontinuity, at , and is continuous there.
Next to find the value of this integral consider for any partitions. Now consider the interval created by . Since is a finite point set, it follows that has infinite points. Which indicates it contains values other than . So for any interval on we can see that . Now since we can see that for any partition . Now since it follows that
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Question 2: Let , this is known as Dirchlet's function. Show that for any
Answer: This one is simple I think. Consider any partition of . Now consider any interval formed by . For the same reason as above is infinite. Now since is dense in any subset of the reals it follows that there are value of in such that where it follows similarly to above that for any partition. Now following a similar argument noting that is dense in any subset of the reals we can see that for any partition . So for any partition . Where it follows that since it fails the criterion that given any we must have a partition such that
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Question 3: Does ?
Answer: No. Counterexample . The reason is that which is continuous thus integrable, but for the same reason as in question 2 is not integrable.
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Question 4: Suppose that , is continuous on and . Prove that .
Answer: It is obvious that if . So instead let us prove that if .
Since we can find a Partition of such that for some interval we have that , now since this interval is closed it follows that . Because of this . Now consider that since it follows that .
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Question 5: Prove that if on and that
Answer: Let be partitions of respectively. Let .
Let us note that . To prove this simply note that given an interval and that .
So since taking the infimum over both intervals gives .
Using the same argument except considering that we arrive at
Combining completes the proof.
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Question six: Prove Holder's inequality for integrals, i.e. if then
I can prove this, but it requires me first to prove that if the conditions of are met as above then and I am not sure how to do this.
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For 1-5 any criticism is appreciated ...for number six any help with the inequality would be great.
Thanks in advance
Thank you Kalagota. Let me rephrase this a little better
Overview: Before I do my proof let me explain what I attempt to do. I will first show that . I will then show that if at any point of then . This will show that the only function such that with is .
Part one: Here I will show that . Consider a partition of . Let be any interval formed by . It is clear that since that . So . So since is continuous it follows that
Part two: Let for some point of . We know use the following lemma.
Lemma: I will show that if is uniformly continuous and then there exists a neighborhood of such that .
Proof: Since is uniformly continuous we have that for every there exists a corresponding such that . Now let , this implies that . So assume that there does not exist a neighborhood of such that , this implies that . So picking we cannot a that satisfies which violates continuity.
So now noticing that being on continuous on implies that is uniformly continuous on since is compact, we can see that there exists a such that . So now pick a partition of such that . It is clear that there is an interval . So we can see that which implies that . So now since is continuous we can see that .
This proves that the only continuous function on such that is
These all look correct to me. One small point on Q.2:
The supremum of the function in any interval of the partition is 1, but the upper sum U for the partition is .
Not true. You need (or possibly ).
Assuming that and that u and v are ≥0, this is elementary calculus. Let . For a fixed value of v, f(u,v) is ≥0 at u=0 and as u→∞. The derivative is 0 when , at which point f has a minimum value of 0.
Yes, thank you. I had this correct down on my paperThe supremum of the function in any interval of the partition is 1, but the upper sum U for the partition is .
Pardon me, I transcribed the problem incorrectly.Not true. You need (or possibly ).[/tex]
I meant using purely non-calculus means, but I guess this works as wellAssuming that and that u and v are ≥0, this is elementary calculus. Let . For a fixed value of v, f(u,v) is ≥0 at u=0 and as u→∞. The derivative is 0 when , at which point f has a minimum value of 0.
Looks good to me.
Also good.---------------------------------------------------------------------
Question 2: Let , this is known as Dirchlet's function. Show that for any
Answer: This one is simple I think. Consider any partition of . Now consider any interval formed by . For the same reason as above is infinite. Now since is dense in any subset of the reals it follows that there are value of in such that where it follows similarly to above that for any partition. Now following a similar argument noting that is dense in any subset of the reals we can see that for any partition . So for any partition . Where it follows that since it fails the criterion that given any we must have a partition such that
Good.--------------------------------------------------------------------
Question 3: Does ?
Answer: No. Counterexample . The reason is that which is continuous thus integrable, but for the same reason as in question 2 is not integrable.
Yes, it is obvious. You were also not asked to prove it so there is no need to mention it! (This is NOT an "if and only if" statement.)------------------------------------------------------------------------
Question 4: Suppose that , is continuous on and . Prove that .
Answer: It is obvious that if .
Good.So instead let us prove that if .
Since we can find a Partition of such that for some interval we have that , now since this interval is closed it follows that . Because of this . Now consider that since it follows that .
I think you need to say a little more than this. Not every partition of [a, b] can be writtenas in this way. You need to assert that given any such partition, there exist a refinement that is of that form. If c is not itself a "boundary" of an interval in P, it lies in one of the intervals. Dividing that interval at c gives a refinement that can be written as .-------------------------------------------------------------------
Question 5: Prove that if on and that
Answer: Let be partitions of respectively. Let .
Let us note that . To prove this simply note that given an interval and that .
So since taking the infimum over both intervals gives .
Using the same argument except considering that we arrive at
Combining completes the proof.
That is called "Young's inequality" and a proof can be found here:--------------------------------------------------------------------
Question six: Prove Holder's inequality for integrals, i.e. if then
I can prove this, but it requires me first to prove that if the conditions of are met as above then and I am not sure how to do this.
Young's inequality - Wikipedia, the free encyclopedia
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For 1-5 any criticism is appreciated ...for number six any help with the inequality would be great.
Thanks in advance
Would this be a correct alternative solution to the above question? I did not using a lemma in my last one.
Answer: By definition we know that iff for every there exists a partition of such that . So if we can prove for this particular function then we have proved integrability.
So now consider as I stated in my first solution that since for any subinterval created by the partition there exists a an element of . This shows that which in turn implies that for any partition. Thus proving reduces to proving .
To do this first define a paritition of , we can see that . So now let represent the interval centered at , then it is apparent that . So now we can pick a parition that forces to be as small as we want, this in turn implies that we may pick a partition that makes as small as we want. This last fact is due to being continuous at . This proves that
So now for the value of the integral I stay with my previous response.
Im pretty dang sure that is correct...but with analysis I never know . Any criticism is very welcome!
Sorry to beat a dead horse everyone, I just coming up with things about his basic problem. To compute the value of couldnt we consider this? Note, let us slightly generliaze this to . It is clear that following the above arguments.
So now to find the value first consider that since that for any interval containing that where is just "excess" interval not containing . Now Consider that on any interval not containing , so that . Now note that repeated reasoning shows that then . But now consider that since that . So now then and since is continuous at this implies that
Just wanted to share.