In connection with my previous post, I just read about a criterion that determines when the sum of two closed subspaces of a Banach space is closed. This is explained for instance in Haïm Brezis’s textbook on Functional Analysis, but I assume it can be found in a lot of other places.

PropositionLet be a Banach space and and two closed subspaces of . The two following statements are equivalent.

- is a closed subspace of .
- There exists a constant such that for each , there exist and with and satisfying .

**Proof**

Let us first prove that 1 implies 2. Let be a sequence of vectors in such that . The sequence is a Cauchy sequence. The following trick is often useful, since it allows us to use absolutely convergent series (which, in a Banach space, are convergent).

Lemma 1There exists a subsequence of such that .

**Proof of Lemma 1**

We prove it recursively. Since is a Cauchy sequence, there exists an integer such that for all and greater than , . We set . Now there is an integer , *that we can choose strictly greater than* $latex n_1$, such that for all and greater than , . We set . Since , we have, by the initial choice of , . We can now pick such that for all , and it will automatically satisfy . If we keep on, we obtain the desired subsequence .

In the rest of the proof, we assume that such a subsequence has been chosen. We will make an abuse of notation by denoting this subsequence by . We therefore assume in the rest of the proof that .

Lemma 2There are sequences and $(y_n)$ of vectors in and respectively such that , and .

**Proof of Lemma 2**

We first pick any and such that . Now, according to 1, there exist and such that , and . We then set and . The other terms of the sequences are built recursively in the same manner.

The series is absolutely convergent and therefore convergent. The *sequence* is therefore convergent. Let us call its limit. Since is closed, belongs to . In the same way, converges to some in . By passing to the limit in , we get and therefore . We have proved that is closed (This took longer than I thought it would).

Now for the converse. If 1 holds, the spaces , , and are closed subspaces of and therefore Banach spaces. We consider the Cartesian product , equipped with the norm . It is also a Banach space. We consider the mapping from to . It is clearly linear, bounded, and surjective. Then, we bring out the big gun, namely the Banach-Schauder theorem.

TheoremA bounded linear mapping between Banach space that is

surjectiveis anopen mapping(the direct image of any open set is an open set).

Applying this theorem to , we obtain that there is some such that , where $latex B(x,R)$ denotes the open ball in of radius and center . Now, let us pick any , . The vector belongs to , therefore there exist and such that . By setting and , we see that we have proved 2 with .

Of course, this criterion cries out for a reformulation in terms of quotient topology. I will do it at some point, but I have procrastinated enough this week. Time to go to bed.