Complex conjugate & lp space

In summary: Do you see how to use the hint I gave you earlier?As for convergence, you need to actually show that the limit exists, not just that it exists and "is a sequence of sequences." And to show that the limit is in l^p, you have to show that it converges to an element of l^p!But in the real case, if a sequence is cauchy then it converges to a real number in R which is complete. So I was wondering if this was similar in this case (since it is a sequence of sequences, but I think I am confused, I have to show that the limit is a sequence of sequences that are p summable)..With the hint, I am
  • #1
VeeEight
615
0
I have two questions

Homework Statement



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


Homework Equations



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


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.

Homework Statement



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

Homework 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.


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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  • #2
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.
 
  • #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 argument 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.
 
  • #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..
 
  • #5
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!
 
  • #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
 
  • #7
And I think your hint for the complex conjugate argument confused me even more but I guess that's a good thing!
 
  • #8
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?
 
  • #9
VeeEight said:
And I think your hint for the complex conjugate argument confused me even more but I guess that's a good thing!
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?
 
  • #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.
 
  • #11
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?
 

1. What is a complex conjugate?

A complex conjugate is a pair of complex numbers that have the same real part but opposite signs for their imaginary parts. For example, the complex conjugate of 3 + 2i is 3 - 2i.

2. Why is the complex conjugate important in mathematics?

The complex conjugate is important in mathematics because it allows us to simplify complex calculations involving complex numbers. It also helps us find the roots of complex polynomials and understand the behavior of complex functions.

3. What is an lp space?

An lp space is a mathematical concept used to describe a collection of functions or sequences that share certain properties. The "l" stands for "Lebesgue", a type of mathematical measure, and the "p" represents a number that determines the specific properties of the space.

4. How is the lp space related to complex numbers?

The lp space is related to complex numbers because it allows us to define and analyze functions and sequences of complex numbers in a more general and rigorous way. It also allows us to study the convergence and divergence of complex series and integrals.

5. Can you give an example of a practical application of complex conjugates and lp space?

A practical application of complex conjugates and lp space is in signal processing, specifically in the field of digital filtering. The use of complex numbers and lp spaces allows for the efficient and accurate analysis and manipulation of signals, which is essential in many technologies such as telecommunications and audio processing.

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