Cauchy sequences - can we control the rate at which elements "get closer"?

The key to this observation in Reed and Simon is that we are choosing a subsequence by a particular process. And we can force this subsequence to obey a certain "rate of closeness." Here's how this works in general.

Let $(X,d)$ be an arbitrary metric space. Suppose that $\{x_n\}_{n=1}^\infty$ is a Cauchy sequence in $X$. First, choose an integer $n_1 \in \mathbb{N}$ so that $n, m \ge n_1$ implies $d(x_n, x_m) < \frac{1}{2}$. We have the ability to do this because $\{x_n\}$ is Cauchy.

Next, choose $n_2 \in \mathbb{N}$ so that $n, m \ge n_2$ implies $d(x_n, x_m) < \frac{1}{4} = \frac{1}{2^2}$ (again, we can do this because $\{x_n\}$ is Cauchy). In fact, we can go ahead and assume $n_2 > n_1$, since any integer $\tilde{n} > n_2$ also satisfies the property:

$$n,m \ge \tilde{n} \implies d(x_n, x_m) < \frac{1}{2}.$$

Now, we finish the argument by induction. Suppose we have chosen $n_1 < n_2 < \cdots < n_k$ with the property that

$$n,m \ge n_k \implies d(x_n,x_m) < \frac{1}{2^k}.$$

Notice in particular that this means $d(x_{n_i}, x_{n_{i+1}}) < \frac{1}{2^i}$ for $i = 1, \dots, k-1.$

Then, just as we did for the $n_2$ case, we pick $n_{k+1} > n_k$ so that $$n,m \ge n_{k+1} \implies d(x_n, x_m) < \frac{1}{2^{k+1}}.$$

And this implies $d(x_{n_k}, x_{n_{k+1}}) < \frac{1}{2^k}$. So in this way we construct a subsequence $\{x_{n_k}\}_{k=1}^\infty$ that satisfies the "rate of closeness" we desire.

This construction works for the example you give if we set $X = L^1$, $d(f, g) = \|f-g\|_1$. The potentially confusing part in Reed and Simon is the relabeling of the subsequence. Basically, Reed and Simon throw away the subscripts on the subsequence $\{x_{n_k}\}_{k=1}^\infty$ and just write the subsequence as $\{x_k\}_{k=1}^\infty$. And this is confusing because it makes the subsequence look just like the original sequence. But in fact they are now just using the subsequence.

We agree that the question is whether, for any strictly positive sequence $(\epsilon(n))$, there exists a subsequence $(f_{\varphi(n)})$ of $(f_n)$ such that $d(f_{\varphi(n)},f_{\varphi(n+1)})\leq \epsilon(n)$ for every $n$.

The answer is "yes". As you recall:

The definition of a Cauchy sequence says that for each $\epsilon>0$ we can choose and $N$ such that $n,m>N$ implies $d(f_n,f_m)\leq\epsilon$.

For each $n$, use the definition of a Cauchy sequence with $\epsilon=\epsilon(n)$, call $N(n)$ an integer $N$ such that the condition in the definition holds and let $\varphi(n)=\max\{N(k);k\leqslant n\}$. Then indeed, for every $n$, $d(f_{\varphi(n)},f_{\varphi(n+1)})\leq \epsilon(n)$ as desired.

Let $\varepsilon_1 = 2^{-2}$. Let $N_1$ be such that $n,m > N_1$ implies that $\Vert f_n - f_m \Vert_1 \le \varepsilon_1$.

Now, take some $k_1 > N_1$, and let $f_{k_1}$ be the first element in the subsequence.

Now, we can recursively define the sequence as follows:

Let $N_n, \varepsilon_n, k_n$ be given for some $n \ge 1$.

Define $\varepsilon_{n + 1} = \frac{\varepsilon_n}{2}$. (Alternately, take $\varepsilon_n = 2^{-n - 1}$ for all $n \ge 1$.)

Let ${\hat N}_{n + 1}$ be such that if $m,\ell > \hat N_{n + 1}$, then $\Vert f_m - f_\ell \Vert_1 < \varepsilon_{n + 1}$.

Define $N_{n + 1} = \max(k_n, \hat N_{n + 1})$.

Then, if $m,\ell > N_{n + 1}$, then $\Vert f_m - f_\ell \Vert_1 < \varepsilon_{n + 1}$, as $N_{n + 1} \ge \hat N_{n + 1}$. Moreover, both $m$ and $\ell$ are greater than $k_n$.

Now, we can take some $k_{n + 1} > N_{n + 1}$, and as both $k_{n + 1}$ and $k_n$ are greater than $N_{n}$, we have that $\Vert f_{k_n} - f_{k_{n + 1}} \Vert_1 < \varepsilon_n = 2^{-n - 1}$, and that $k_{n + 1} > N_{n + 1} \ge k_n$.

This therefore constructs a subsequence as desired.