# 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.