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.
Let 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 .
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).
There 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 .
There 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.
A bounded linear mapping between Banach space that is surjective is an open 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.