In Simon & Reed's book Methods of Modern Mathematical Physics, it is proven in chapter 1 (Theorem 1.12) that $L^1$ is complete (Riesz-Fisher theorem). The proof starts off as follows:

Let $f_n$ be a Cauchy sequence of functions in $L^1$. It is enough to show that some subsequence converges (this has been shown earlier) so we pass to a subsequence (labeled in the same way) with $\left|\left|f_n-f_{n+1}\right|\right|_1\leq 2^{-n}$.

This arouses my suspicion (although surely I will turn out to be wrong): Can we pick a subsequence such that $\left|\left|f_n-f_{n+1}\right|\right|_1\leq 2^{-n}$? This seems strange to me because it seems to say something about the rate of at which elements "get closer" under this norm (I should probably specify that $\left|\left| f\right|\right|_1=\int \left| f \right| dx$), rather than just saying that they do get arbitrarily close at some point. In particular, it seems to say that each progressive element is "twice as close".

The definition of a Cauchy sequence says that for each $\epsilon>0$ we can choose and $N$ such that $n,m>N$ implies $\left|\left| f_n-f_m\right|\right|_1=d(f_n,f_m)\leq\epsilon$, where $d(\cdot,\cdot)$ is the metric induced by the norm. This, to me, does not seem equivalent to saying that for any strictly positive $\epsilon(n)$ there is a subsequence such that $d(f_n,f_{n+1})\leq \epsilon(n)$. Am I mistaken?


4 Answers 4


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.


This can be done. Let's start out with the original sequence $(f_n)$. Since we are assuming it is Cauchy, we know that for each $\epsilon >0$ there is a number $N$ for which $k,l>N$ yields $||f_k - f_l||_1 \leq \epsilon$. What we can do now is choose $\epsilon = 2^{-n}$ for any $n$ we wish. What we are guaranteed then is a cutoff $N_n$ for which $||f_k-f_l||<2^{-n}$ for any $k,l>N_n$. In particular, this works for $l=k+1$ if $k$ is chosen to be greater than $N_n$.

Given this, the subsequence you are speaking of is generated for each $n$ by referring to the definition of a Cauchy sequence and choosing a particular $k$ and $l$.


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