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Homework Help: Complex conjugate & lp space

  1. Nov 16, 2008 #1
    I have two questions

    1. The problem statement, all variables and given/known data

    Show that the function defined by complex conjugation is not differentiable.


    2. Relevant equations

    If z = x + iy then the complex conjugate of z is x - iy


    3. The attempt at a solution

    f'(z) = lim z -> z1 [tex]\frac{f(z) - f(z1)}{z - z1}[/tex]
    So I need to show that this limit is not defined. So I think the method to do this is to show that the limit as x approaches x1, y approaches y1 of [tex]\frac{(x-x1) + i(y1 - y)}{(x-x1) + i(y - y1)}[/tex] is undefined. I thought that it is pretty obvious that this does not work but I don't know how to actually show it.

    1. The problem statement, all variables and given/known data

    Show that the lp spaces are complete for p = 1 to infinite

    2. Relevant equations

    Each lp space is the vector space of infinite sequences that are p summable. The norm for these spaces is the p sum, whole raised to the exponent of 1/p. These spaces are metric spaces since they satisfy Minkowski's inequality.


    3. The attempt at a solution

    I think that l-infinite is obviously complete. It consists of all bounded sequences so the limit of a Cauchy sequence in the space is bounded and thus, contained in l-infinite.

    For p finite, let x_n be a cauchy sequence in an lp space. Then let x be the candidate of the limit for x_n as n tends to infinite. Then I think it is easily shown that x is bounded since ||x|| < ||x_n|| + c for some c > 0 (since x_n) is Cauchy. I know the goal is to show that x is p summable. However I do not know how to proceed after this. I know the proofs for showing things like B (X, Y) ( bounded linear operators from X to complete Y) is complete but am having a hard time figuring out the second step here.
     
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  3. Nov 16, 2008 #2

    morphism

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    Do you know what the Cauchy-Riemann equations are? They will trivialize your first problem.

    As for your second problem, I'm not too convinced by your proof that l^infty is complete. You have to show that every cauchy sequence converges - obviously the limit (if the sequence converges) will lie in l^infty. But the question is, why does the sequence converge in the first place? Same comment applies to your attempt at proving that l^p is complete. It's not enough to show that x is in l^p; you have to show that x exists first, and then show that it lies in l^p. Also keep in mind that a sequence in l^p is a sequence of sequences of scalars!

    By the way, there's an interesting remark to be made about your title. A very easy proof of the fact that the l^p spaces are complete follows from a 'conjugation' argument. Namely, if 1<p<infinity, then l^p is the conjugate space of l^q, where 1/p+1/q=1. And l^1 is the conjugate of c_0, and l^infty is the conjugate of l^1. Conjugate spaces are complete because they are of the form B(X,Y) with Y=base field=complete.
     
  4. Nov 16, 2008 #3
    For the complex conjugate question, no I do not know what those equations are since we did not do them in class and so I assume that we should not use them. I thought that was I had for the attempt at the solution was correct (though perhaps it could be phrased better).

    I was under the assumption that it is only necessary to show that a metric space is complete if every Cauchy sequence converges to a value in the space. I tried to do this by first showing that is it bounded, then showing that the limit of the sequences of sequences in the lp space is real valued (since R is complete). Then I wrote that the memebers of the infinite sequences of the limit are all p summable because the terms of the original sequences were all p summable. So my arguement looked clear to me, but you mentioned that I have to first show that the limit exists, which I did not do. Perhaps I am just going about it the wrong way.
     
  5. Nov 16, 2008 #4
    And I read over your proof at the end and looked up some of the terms that were used and I believe that I should not go that route (because we have not covered conjugate spaces) but thanks for the insight.

    I guess that I am more wrong than I thought I was..
     
  6. Nov 16, 2008 #5

    morphism

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    Your approach for proving that complex conjugation isn't differentiable is actually pretty sound. I just figured I'd point you towards the CR-eqns approach in case you knew what they were. Here's a hint to help you finish your argument: if w is a nonzero complex number, then 1/w=w*/|w|^2, where w* is the complex conjugate of w.


    OK, moving on to your second problem. A metric space X is complete if every cauchy sequence {x_n} in X converges in X. So to prove that l^p is complete, you have to take an arbitrary cauchy sequence in l^p, prove that it converges, and show that the limit is in l^p.

    And don't worry about the end of my post. I just figured it was kind of funny how your title related to what is probably the least tedious proof of the completeness of the l^p spaces!
     
  7. Nov 16, 2008 #6
    Okay well I'm sure I'll be able to understand your proof in a couple of years from now..

    For the lp space problem, I am pretty sure that I have bits of it right (as mentioned a few posts above) but there are some parts that I do not understand. For example, to 'prove that it converges', is it okay to say that this is satisfied by R being a complete metric space?

    And to show that the limit is in the lp space, I made the statement that since the limit is a sequence of sequences and each sequence in that sequence is the limit of a p summable sequence, so it itself must be p summable and so the limit is in the lp space. I think that this is a correct statement but I am unsure if this satisfies proving that the limit is in the lp space. Thank you
     
  8. Nov 16, 2008 #7
    And I think your hint for the complex conjugate arguement confused me even more but I guess that's a good thing!
     
  9. Nov 16, 2008 #8

    morphism

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    First let's focus on proving that the sequence converges. And let's stick to the case where p=1.

    So we have a cauchy sequence {x_n} in l^1. Each element in this sequence is itself a sequence of real numbers. You noted that R is complete. This alone doesn't guarantee that {x_n} converges, but it will sure help. If {x_n} were to converge to some x, can you guess what x could be?
     
  10. Nov 16, 2008 #9

    morphism

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    Heh, alright. Let me try again:

    [tex]\frac{f(z) - f(z_1)}{z - z_1} = \frac{\overline{z} - \overline{z_1}}{z - z_1} = \overline{z - z_1} \cdot \frac{1}{z - z_1}.[/tex]

    Can you see how to apply my hint?
     
  11. Nov 16, 2008 #10
    Oh I thought that in a metric space, Cauchy sequences were convergent and so the only thing I had to do was to show that the sequences converges to a value in the lp space.

    But for your case of l1, l1 is composed of absolutely convergent series so the limit of a sequence would be absolutely convergent.
     
  12. Nov 17, 2008 #11

    HallsofIvy

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    No, it is not true that Cauchy sequences in a metric space converge. For example, the set of rational numbers with d(x,y) defined as |x- y| is a metric space but Cauchy sequences do not converge.

    However, in looking at the limit of
    [tex]\frac{(x-x1) + i(y1 - y)}{(x-x1) + i(y - y1)}[/tex]
    as x goes to x1 and y goes to y2, probably the simplest thing to do is to look at the limit along the x and y directions separately. If they are different, the limit itself does not exist. If y= y1, then we have
    [tex]\lim_{x\rightarrow x1}\frac{x-x1}{x-x1}= 1[/tex]

    If x= x1, then we have
    [tex]\lim_{y\rightarrow y1}\frac{y1- y}{y- y1}= -1[/tex]

    I have no idea what that has to do with lp spaces being complete. What that a separate problem?
     
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