Sum of subspaces

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.

Proposition

Let E be a Banach space and F and G two closed subspaces of E. The two following statements are equivalent.

  1.     F+G is a closed subspace of E.
  2.     There exists a constant C such that for each z \in F+G, there exist x \in F and y \in G with \|x\|\le C\|z\| and \|y\|\le C\|z\| satisfying z=x+y.

Proof

Let us first prove that 1 implies 2. Let (z_n) be a sequence of vectors in F+G such that z_n \to z. The sequence (z_n) 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 1

There exists a subsequence (z_{n_k}) of (z_n) such that  \|z_{n_{k+1}}-z_{n_k}\|\le 2^{-k}.

Proof of Lemma 1

We prove it recursively. Since (z_n) is a Cauchy sequence, there exists an integer N such that for all m and n greater than N, \|z_n-z_m\|\le 1/2. We set n_1=N. Now there is an integer M, that we can choose strictly greater than $latex  n_1$, such that for all m and n greater than M\|z_n-z_m\|\le 1/4. We set n_2=M. Since n_2>n_1, we have, by the initial choice of n_1, \|z_{n_2}-z_{n_1}\|\le 1/2. We can now pick n_3>n_2 such that  \|z_n-z_{n_3}\|\le 1/8 for all n\ge n_3, and it will automatically satisfy \|z_{n_3}-z_{n_2}\|\le 1/4. If we keep on, we obtain the desired subsequence z_{n_1}, z_{n_2},z_{n_3},z_{n_4},\dots.

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 (z_n). We therefore assume in the rest of the proof that \|z_{n+1}-z_{n}\|\le 2^{-n}.

Lemma 2

There are sequences (x_n) and $(y_n)$ of vectors in F and G respectively such that x_n+y_n=z_n, \|x_{n+1}-x_n\| \le C\|z_{n+1}-z_{n}\| and  \|y_{n+1}-y_n\| \le C\|z_{n+1}-z_{n}\|.

Proof of Lemma 2

We first pick any x_1 \in F and y_1 \in G such that x_1+y_1=z_1. Now, according to 1, there exist \tilde{x}\in F and \tilde{y} \in G such that \tilde{x}+\tilde{y}=z_2-z_1, \|\tilde{x}\| \le C\|z_2-z_1\| and \|\tilde{y}\| \le C\|z_2-z_1\|. We then set x_2=x_1+\tilde{x} and x_2=x_1+\tilde{x}. The other terms of the sequences are built recursively in the same manner.

The series \sum(x_{n+1}-x_n) is absolutely convergent and therefore convergent. The sequence (x_n) is therefore convergent. Let us call x its limit. Since F is closed, x belongs to F. In the same way, (y_n) converges to some y in G. By passing to the limit in x_n+y_n=z_n, we get x+y=z and therefore z\in F+G. We have proved that F+G is closed (This took longer than I thought it would).

Now for the converse. If 1 holds, the spaces F, G, and F+G are closed subspaces of E and therefore Banach spaces. We consider the Cartesian product F\times G, equipped with the norm \|(x,y)\|=\max(\|x\|,\|y\|). It is also a Banach space. We consider the mapping S:(x,y)\to x+y from  F\times G to F+G.  It is clearly linear, bounded, and surjective. Then, we bring out the big gun, namely the Banach-Schauder theorem.

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 S, we obtain that there is some \varepsilon>0 such that B(0,\varepsilon)\cap (F+G) \subset S((B(0,1)\cap F)\times (B(0,1)\cap G)), where $latex  B(x,R)$ denotes the open ball in E of radius R and center x. Now, let us pick any z \in F+G, z \neq 0. The vector z'=\frac{\varepsilon}{2\|z\|}z belongs to B(0,\varepsilon)\cap (F+G), therefore there exist x' \in B(0,1)\cap F and  y'\in B(0,1)\cap G such that z'=x'+y'. By setting x=\frac{2\|z\|}{\varepsilon}x' and  y=\frac{2\|z\|}{\varepsilon}y', we see that we have proved 2 with C=\frac{2}{\varepsilon}.

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.

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