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By struggling with the proof that $\ell^p$ is complete, I looked up different proofs by different authors, and I ended up focusing on the one given by Kreyszig in his classic book on functional analysis, because I found it the most perspicuous for my level. Still, there are some point that are not completely clear, thus I will write down here the whole proof and I will add my remarks and doubts.

[I]

Theorem: The space $\ell^p$ is complete; here $p$ is fixed and $1 \leq p < \infty$.

Proof: Let $(x_n)$ be any Cauchy sequence in $\ell^p$, where $x_m = ( \xi^{(m)}_1, \xi^{(m)}_2, \dots )$. Then, for every $\varepsilon > 0 $ there is an $N$ such that for all $m, n >N$, \begin{equation} d ( x_m, x_n ) = \Bigg( \sum_{j=1}^\infty |\xi^{(m)}_j - \xi^{(n)}_j |^p \Bigg)^{\frac{1}{p}} < \varepsilon \hspace{2cm} \text{(1)} . \end{equation} It follows that for every $j = 1,2, \dots$ we have for $m,n >N$ $$ |\xi^{(m)}_j - \xi^{(n)}_j | < \varepsilon \hspace{2cm} \text{(2)}. $$ We choose a fixed $j$. From (2) we see that $( \xi^{(1)}_j, \xi^{(2)}_j, \dots )$ is a Cauchy sequence of numbers. It converges since $\Re$ is complete, say $\xi^{(m)}_j \to \xi_j$ as $m \to \infty$. Using these limits, we define $x = (\xi_1, \xi_2, \dots)$ and show that $x \in \ell^p$ and $x_m \to x$.

This looks fine,

[II]

From (1) we have for all $m,n >N$ $$ \sum_{j=1}^k |\xi^{(m)}_j - \xi^{(n)}_j |^p < \varepsilon^p \hspace{2cm} (k=1,2,\dots). $$

First problems!

I see where the $\varepsilon^p$ comes from, but what puzzles me is the $k$ on the top of $\sum$. I see it comes from (1) (indeed, if it works for $j \to \infty$, then it has to work necessarily for a finite $k$. Still, I don't see why we actually need to do it.
Why don't we simply stick to (1)?

[III]

Letting $n \to \infty$, we obtain for $m>N$ $$ \sum_{j=1}^k |\xi^{(m)}_j - \xi_j |^p \leq \varepsilon^p \hspace{2cm} (k=1,2,\dots). $$

Same problem as before, plus the mysterious $\leq$ between LHS and RHS. Indeed, I think the idea should be that the sequence converge and it is Cauchy, thus the $\varepsilon > 0$ is the same that we use in the definition of Cauchy sequence, and in the standard limit definition. But then it should be still $<$ and not $\leq$.

[IV]

We may now let $k \to \infty$; then for $m >N$ $$ \sum_{j=1}^\infty |\xi^{(m)}_j - \xi_j |^p \leq \varepsilon^p \hspace{2cm} \text{(3)}. $$

Bypassing the previous related problems, this is fine.

[V]

This shows that $x_m - x = ( \xi^{(m)}_j - \xi_j ) \in \ell^p$.

Not completely sure I see why it is actually the case.

[VI]

Since $x_m \in \ell^p$, it follows by means of the Minkowsky inequality, that $$ x = x_m + (x - x_m) \in \ell^p. $$

The reference to the Minkowsky inequality is really mysterious.

[VII]

Furthermore, the series in (3) represents $[d(x_m,x_)]^p$, so that (3) implies that $x_m \to x$. Since $(x_m)$ was an arbitrary Cauchy sequence in $\ell^p$, this proves the completeness of $\ell^p$, where $1 \leq p < \infty$. QED

Sorry for the vivisection of this rather straightforward proof, but I have the feeling that by properly catching each step here, I could improve dramatically my overall understanding of real analysis tools and procedures.

Thank you for your time and patience. As always, any feedback is most welcome!


Edit for BOUNTY:

I am editing this question because I am putting a bounty. True enough, I received very helpful comments, but still they were only comments, and I would really love to have an answer, because I do feel a lot of the things that look here mysterious have to be fairly important.

I also have the feeling that some of the problems I showed in these questions, e.g. the one that is below under (1), can give the feeling that it is impossible I can actually try to read something sort of advanced, without having a solid grasps of other things, and maybe this could put a potential answerer in the akward position of feeling “Come on, man, are you kidding me? Don’t let me waste my time. I cannot go back to teach you 2+2!”. However, that’s how things are, and this is mostly due to the fact that I am self-taught, which put me in the position to choose my own topics. But, actually, exactly those very naive questions are the ones that would set me on the right path to keep on studying properly.

Hence, I renumbered the parts in which I divided the proof of the theorem in order to easily refer the questions to each part.

1) Is the change from $\infty$ to $k$ in [II] related to the fact that we are dealing with a series, and thus we standardly see how a series behave with its finite terms, before letting the limit goes to $\infty$ (which is what happen in [III]?

2) Is in step [V] implicitly assumed that it is the case due to the fact that it is for all $j$?

3) How does step [VI] come from the Minkowsky inequality?

Of course, any other addition or explanation is most welcome.

Thanks for your time!

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    $\begingroup$ To the moderators: It is the first time that I use the tag self-learning, and I know it is not completely correct, according to the description of the tag. But, before doing that, I read quite some posts on the MetaMath concerning this tag, and apparently it is commonly used as a signal that somebody is self-learning something. Considering this is my case (I can hardly imagine a more extreme case of self-learning) I decided to use this tag. In case this is problematic, I would be happy to point out this issue again on the Meta level. Thanks! $\endgroup$ – Kolmin Jan 8 '15 at 10:01
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    $\begingroup$ The first problem is just the rewriting of Cauchy limit, and don't forget that when your are passing to a limit: $<$ becomes $\le$. Notice that you are taking $\Bbb{R}$ with the usual distance: It's a complete space. $\endgroup$ – user169373 Jan 8 '15 at 10:16
  • $\begingroup$ 1. Needed to use the fact that Cauchy sequence converges on $\Bbb R$ (under the usual metric). $\endgroup$ – IAmNoOne Jan 8 '15 at 10:21
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    $\begingroup$ @Kolmin just think about simple example, as $u_n=2+\frac{1}{n}$ and $v_n=2-\frac{1}{n}$ :-). You proved that $\sum_{k=1}^{\infty}\Vert x_m^k-x_k\Vert^p\le \varepsilon^p$ it implies directly that $x_m-x\in l^p$ by definition. $x_m-x\in l^p$ $\endgroup$ – user169373 Jan 8 '15 at 10:33
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    $\begingroup$ @Kolmin, what if I tell you that for each integer $n < 0, 1/n < 0$. If we pass the limit, do we get a weaker inequality? $\endgroup$ – IAmNoOne Jan 8 '15 at 10:34
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I will use your post to add the explanations inline.

[I]

Theorem: The space $\ell^p$ is complete; here $p$ is fixed and $1 \leq p < \infty$.

Proof: Let $(x_n)$ be any Cauchy sequence in $\ell^p$, where $x_m = ( \xi^{(m)}_1, \xi^{(m)}_2, \dots )$. Then, for every $\varepsilon > 0 $ there is an $N$ such that for all $m, n >N$, \begin{equation} d ( x_m, x_n ) = \Bigg( \sum_{j=1}^\infty |\xi^{(m)}_j - \xi^{(n)}_j |^p \Bigg)^{\frac{1}{p}} < \varepsilon \hspace{2cm} \text{(1)} . \end{equation} It follows that for every $j = 1,2, \dots$ we have for $m,n >N$ $$ |\xi^{(m)}_j - \xi^{(n)}_j | < \varepsilon \hspace{2cm} \text{(2)}. $$ We choose a fixed $j$. From (2) we see that $( \xi^{(1)}_j, \xi^{(2)}_j, \dots )$ is a Cauchy sequence of numbers. It converges since $\Re$ is complete, say $\xi^{(m)}_j \to \xi_j$ as $m \to \infty$. Using these limits, we define $X = (\xi_1, \xi_2, \dots)$ and show that $x \in \ell^p$ and $x_m \to x$.

This looks fine,

Answer: OK.

[II]

From (1) we have for all $m,n >N$ $$ \sum_{j=1}^k |\xi^{(m)}_j - \xi^{(n)}_j |^p < \varepsilon^p \hspace{2cm} (k=1,2,\dots). $$

First problems!

I see where the $\varepsilon^p$ comes from, but what puzzles me is the $k$ on the top of $\sum$. I see it comes from (1) (indeed, if it works for $j \to \infty$, then it has to work necessarily for a finite $k$. Still, I don't see why we actually need to do it.

Why don't we simply stick to (1)?

Answer: The reason is the next step. They want to take limit as $n\to\infty$. But for the series it is not clear (we don't know if) we can interchange the limit with the summation (which is kind of what we need to prove).

On the other hand, the truncated sum is a finite sum. More than that, it is a continuous function $f(y_1,y_2,...,y_k):=\sum_{j=1}^{k}|\xi_j^{m}-y_j|^p$. So, taking limit of $f(\xi_j^{(n)},\xi_j^{(n)},...,\xi_j^{(n)})$ as $n\to\infty$ for the finite sum is the same as evaluating at the limit point $\xi_j$ to get $f(\xi_1,\xi_2,...,\xi_k)=\sum_{j=1}^{k}|\xi_j^{(m)}-\xi_j|^p$.

Expanded answer to address the comment:

Why continuity? What we want to to take $\lim_{n\to\infty}$ on both sides of (1) in such a way that we can take the limit inside as in

$$\lim_{n\to\infty}\sum_{j=1}^{\infty}|\xi_j^{(m)}-\xi_j^{(n)}|^p=\sum_{j=1}^{\infty}|\xi_j^{(m)}-\lim_{n\to\infty}\xi_j^{(n)}|^p$$

But it is not clear that if is true. On the other hand, the finite sum is a finite sum of continuous functions. It is therefore continuous and we can do $$\lim_{n\to\infty}\sum_{j=1}^{k}|\xi_j^{(m)}-\xi_j^{(n)}|^p=\sum_{j=1}^{k}|\xi_j^{(m)}-\lim_{n\to\infty}\xi_j^{(n)}|^p=\sum_{j=1}^{k}|\xi_j^{(m)}-\xi_j|^p\leq \lim_{n\to\infty}\epsilon^p=\epsilon^p$$ giving us the next step.

Remember, a function is continuous if always $\lim_{n\to\infty}a_n=a$ implies $$\lim_{n\to\infty}f(a_n)=f(\lim_{n\to\infty}a_n)$$

[III]

Letting $n \to \infty$, we obtain for $m>N$ $$ \sum_{j=1}^k |\xi^{(m)}_j - \xi_j |^p \leq \varepsilon^p \hspace{2cm} (k=1,2,\dots). $$

Same problem as before, plus the mysterious $\leq$ between LHS and RHS. Indeed, I think the idea should be that the sequence converge and it is Cauchy, thus the $\varepsilon > 0$ is the same that we use in the definition of Cauchy sequence, and in the standard limit definition. But then it should be still $<$ and not $\leq$.

Answer: This is standard when working with inequalities and limits. If $a_n<b_n$ and $\lim a_n=a$ and $\lim b_n=b$ then the most we can say is $a\leq b$. For example take $a_n=0$ and $b_n:=1/n$. We know that $0<1/n$, but the limits of both sides are equal.

[IV]

We may now let $k \to \infty$; then for $m >N$ $$ \sum_{j=1}^\infty |\xi^{(m)}_j - \xi_j |^p \leq \varepsilon^p \hspace{2cm} \text{(3)}. $$

Bypassing the previous related problems, this is fine.

Answer: OK.

[V]

This shows that $x_m - x = ( \xi^{(m)}_j - \xi_j ) \in \ell^p$.

Not completely sure I see why it is actually the case.

Answer: The previous inequality computed, inequality (3), is $||x_m-x||_p\leq\epsilon<\infty$. This is the definition of being in $\ell^p$.

[VI]

Since $x_m \in \ell^p$, it follows by means of the Minkowsky inequality, that $$ x = x_m + (x - x_m) \in \ell^p. $$

The reference to the Minkowsky inequality is really mysterious.

Answer: First recall that Minkowsky's inequality is $$||x+y||_p\leq ||x||_p+||y||_p$$ for all $x,y\in\ell^p$. This is just the triangle inequality for the norm $||\cdot||_p$.

We have that $$||x||_p=||x_m+(x-x_m)||_p\leq||x_m||_p+||x-x_m||_p$$ The inequality above is Minskowsky's inequality (which is just the triangle inequality for the $||\cdot||_p$ norm).

Now let us finish bounding the right hand side to show that it is finite.

The last term, $||x-x_m||_p$, we had it already bounded by $\epsilon$ for all $m>N$.

The term $||x_m||_p$ is finite because $x_m\in\ell^p$.

Therefore $||x||_p$ is smaller than the finite number $||x_m||_p+||x-x_m||_p$ and it is therefore finite. Then, by definition, $x\in\ell^p$.

[VII]

Furthermore, the series in (3) represents $[d(x_m,x)]^p$, so that (3) implies that $x_m \to x$. Since $(x_m)$ was an arbitrary Cauchy sequence in $\ell^p$, this proves the completeness of $\ell^p$, where $1 \leq p < \infty$. QED

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    $\begingroup$ @Pp..: First of all, thank you for your detailed answer! I do have two minor questions. The first concerns point [II] and your reference to the continuous function $f(y_1, \dots, f_k)$: honestly I found this statement a bit mysterious. :) $\endgroup$ – Kolmin Jan 14 '15 at 21:22
  • $\begingroup$ @Pp..: The second question concerns point [VI]. Thus, correct me if I am wrong, the idea that you are conveying is that given the formula $||x||_p=||x_m+(x-x_m)||_p\leq||x_m||_p+||x-x_m||_p$, what you are saying is that $|| x ||_p \in \ell^p$, due to the fact that – thanks to the Minkowsky inequality – we know that the LHS is equal or smaller than the RHS, that is bounded and... by being bounded is necessarily in $\ell^p$$! $\endgroup$ – Kolmin Jan 14 '15 at 21:26
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    $\begingroup$ @Kolmin I added further explanation about [II]. About [VI]: The definition of $x=\{x_n\}$ being in $\ell^p$ is that $||x||_p$ is a finite number. The inequalities you mention show that $||x||_p$ is smaller than the sum of two norms, both of which are finite. $\endgroup$ – Pp.. Jan 14 '15 at 21:45
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    $\begingroup$ @Pp..:Thanks a lot for your detailed answer! $\endgroup$ – Kolmin Jan 14 '15 at 22:46

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