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I am trying to understand a step in the following proof of completeness of $L^p$ in Stein-Shakarchi's Functional Analysis. (See the proof on page 5 of the link or at the end of this post.)

At the beginning of the proof, it is said that

Let $\{f_n\}_{n=1}^\infty$ be a Cauchy sequence in $L^p$, and consider a subsequence $\{f_{n_k}\}_{k=1}^\infty$ of $\{f_n\}_{n=1}^\infty$ with the following property $\|f_{n_{k+1}}-f_{n_k}\|\le 2^{-k}$ for all $k\geq 1$.

Question: Why can the sequence be considered as it is?

On a YouTube video, it explains about a similar subsequence.

I still don't understand why for $n,m>n_k$ $\Vert f_{n}-f_{m}\Vert_p\implies \Vert f_{n_k}-f_{n_{k+1}}\Vert_p$, thus an increasing subsequence. Why is it justified to make $n$ to depend on $k$, $n_k$?


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  • $\begingroup$ What happened to all the $\sum$s? $\endgroup$ – Lord Shark the Unknown May 19 at 18:22
  • $\begingroup$ I don't know, I don't know why the author omitted them all; $\endgroup$ – galleta May 19 at 18:23
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    $\begingroup$ Basically the trick here is to replace the Cauchy sequence by a subsequence that converges really quickly. $\endgroup$ – Lord Shark the Unknown May 19 at 18:23
  • $\begingroup$ @LordSharktheUnknown What kind of convergence? $\endgroup$ – galleta May 19 at 18:55
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The authors mention at the beginning of the proof that

The argument is essentially the same as for $L^1$ (or $L^2$); see Section 2, Chapter 2 and Section 1, Chapter 4 in Book III.

It is said clearly there (see also a snapshot at the end) that

The existence of such subsequence is guaranteed by the fact that $\|f_{n}-f_{m}\|\leq \epsilon$ whenever $n,m\geq N(\epsilon)$, so that it suffices to take $n_k=N(2^{-k})$.

[Added for elaboration.]

There exists an integer $N(2^{-1})>0$ such that for all $n,m\geq N(2^{-1})$, $$ \|f_{n}-f_{m}\|\leq 2^{-1}\tag{1}. $$ There exists an integer $N(2^{-2})>N(2^{-1})$ such that for all $n,m\geq N(2^{-2})$, $$ \|f_{n}-f_{m}\|\leq 2^{-2}\tag{2}. $$ There exists an integer $N(2^{-3})>N(2^{-2})$ such that for all $n,m\geq N(2^{-3})$, $$ \|f_{n}-f_{m}\|\leq 2^{-3}\tag{2}. $$ ... so on and so forth.

Now, let $n_1=N(2^{-1})$, $n_2=N(2^{-2})$, $n_3=N(2^{-3})$, $\cdots$.

Since $n_1,n_2\geq N(2^{-1})$, we have by (1) $$ \|f_{n_2}-f_{n_1}\|\leq 2^{-1}. $$

Since $n_2,n_3\geq N(2^{-2})$, we have by (2) $$ \|f_{n_3}-f_{n_2}\|\leq 2^{-2}. $$

Since $n_3,n_4\geq N(2^{-3})$, we have by (3) $$ \|f_{n_4}-f_{n_3}\|\leq 2^{-3}. $$

... so on and so forth.


The following is a snapshot of the beginning of the proof for completeness of $L^1$ in Stein-Shakarchi's Book III (page 70 Theorem 2.2).

enter image description here

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  • $\begingroup$ Why can we take the sequence as an increasing sequence? $\endgroup$ – galleta Jun 8 at 16:40
  • $\begingroup$ What do you mean by "increasing sequence"? Are you asking why $n_1<n_2<n_3<\cdots$? $\endgroup$ – Jack Jun 8 at 17:51
  • $\begingroup$ See my edit. Not that $\{f_{n_k}\}_{k=1}^\infty$ is NOT necessarily an increasing sequence. In fact one may talk about a sequence of complex-valued functions and thus no such thing as "increasing sequence" here. One can say that the subscripts $\{n_k\}_{k=1}^\infty$ form an increasing sequence though. $\endgroup$ – Jack Jun 8 at 18:07
  • $\begingroup$ Got it Jack thanks. Do we take this W.L.O.G $N(2^{-2})>N(2^{-1})$? Because $2^{-2}<2^{-1}$ actually $\endgroup$ – galleta Jun 8 at 19:44
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    $\begingroup$ Well, I would not use "WLOG" there. Since $2^{-2}<2^{-1}$, we are taking a smaller $\epsilon$. On the other hand, we want to pick one positive integer $N(2^{-2})$ such that (i) $N(2^{-2})>N(2^{-1})$ (ii) for all $n,m\geq N(2^{-2})$, the inequality (2) holds. There are of course lots of other choices of $N(2^{-2})$ that would work, but we only need one such $N(2^{-2})$. $\endgroup$ – Jack Jun 8 at 20:00
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If $\{f_n\}_n$ is Cauchy, for each $k\in\Bbb N$ we can find an $N_k$ (depending on the $k$ we just chose) such that

$$ \|f_n - f_m\| < 2^{-k} \quad\text{ whenever } n,m\geq N_k.$$

For $k+1$, we likewise get

$$ \|f_n - f_m\| < 2^{-k-1} \quad\text{ whenever } n,m\geq N_{k+1}.$$

Now, both inequalities will hold for $n,m \geq \max\{N_k,N_{k+1}\}$, so we can choose $N_{k+1} > N_k$, and similarly we can choose an $N_{k+2} > N_{k+1} > N_{k}$ for $2^{-k-2}$ so that

$$\|f_{N_k} - f_{N_{k+1}}\| < 2^{-k}\quad\text{and}\quad \|f_{N_{k+1}} - f_{N_{k+2}}\| < 2^{-k-1}$$

by choosing $n = N_k$ and $m = N_{k+1}$, etc in the first inequalities above.

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  • $\begingroup$ Thank you. Does the subsequence $f_{N_k}$ converge pointwise? $\endgroup$ – galleta May 19 at 23:24
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    $\begingroup$ You're welcome :) And no, $L^p$ convergence does not imply pointwise convergence in general. $\endgroup$ – ryan221b May 19 at 23:26
  • $\begingroup$ it does not hold in general then there exist the possibility of $f_{N_k}$ to converge pointwise? $\endgroup$ – galleta May 19 at 23:40
  • $\begingroup$ Yes, if you take all the functions to be equal, for example. Or, for continuous functions in $L^1([0,1])$, uniform convergence implies both pointwise and $L^1$ convergence, so both can occur simultaneously. $\endgroup$ – ryan221b May 20 at 0:00
  • $\begingroup$ On this $\|f_{N_k} - f_{N_{k+1}}\| < 2^{-k}\quad\text{and}\quad \|f_{N_{k+1}} - f_{N_{k+2}}\| < 2^{-k-1}$ what is the whenever condition? $\endgroup$ – galleta Jun 6 at 19:47

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