Why can we consider this subsequence in $L^p$?

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?

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 $$k$$, $$n_k$$?

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

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 $$n_k=N(2^{-k})$$.

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). • Why can we take the sequence as an increasing sequence? – galleta Jun 8 at 16:40
• What do you mean by "increasing sequence"? Are you asking why $n_1<n_2<n_3<\cdots$? – Jack Jun 8 at 17:51
• 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. – Jack Jun 8 at 18:07
• Got it Jack thanks. Do we take this W.L.O.G $N(2^{-2})>N(2^{-1})$? Because $2^{-2}<2^{-1}$ actually – galleta Jun 8 at 19:44
• 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 $N(2^{-2})$. – Jack Jun 8 at 20:00

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.

• Thank you. Does the subsequence $f_{N_k}$ converge pointwise? – galleta May 19 at 23:24
• You're welcome :) And no, $L^p$ convergence does not imply pointwise convergence in general. – ryan221b May 19 at 23:26
• it does not hold in general then there exist the possibility of $f_{N_k}$ to converge pointwise? – galleta May 19 at 23:40
• 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. – ryan221b May 20 at 0:00
• 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? – galleta Jun 6 at 19:47