Homework Help: Sum of two closed subspaces in a Banach space

1. Oct 31, 2013

mahler1

The problem statement, all variables and given/known data.

Let $E$ be a Banach space and let $S,T \subset E$ two closed subspaces. Prove that if dim$T< \infty$, then $S+T$ is also closed.

The attempt at a solution.

To prove that $S+T$ is closed I have to show that if $x$ is a limit point of $S+T$, then $x \in S+T$. Let $x$ be a limit point of $S+T$. Then, there is a sequence $\{x_n\}_{n \in \mathbb N} \subset S+T$ : $x_n→x$. For each $n \in \mathbb N$, $x_n$ is in $S+T$, this means that $x_n=s_n+t_n$ for some $s_n \in S$ and some $t_n \in T$. $\{s_n\}_{n \in \mathbb N} \subset S$ and $\{t_n\}_{n \in \mathbb n} \subset T$. Here I got stuck, I would like to show that these two sequences are convergent. I have to use all the facts I know (the completeness of $E$, that $S$ and $T$ are closed and that $T$ is finite dimensional), but I don't know how.

To be honest, I don't even understand intuitively why this statement is true. I mean, why one just requires for $T$ to be finite dimensional and not both of them? Why infinity can screw up closure with the sum if the two subspaces are closed?

Last edited: Oct 31, 2013
2. Oct 31, 2013

R136a1

Think about the quotient map $Q:X\rightarrow X/S$. Notice that $T+S = Q^{-1}(Q(T))$.

3. Oct 31, 2013

mahler1

I don't know what a quotient map is, but I'll look it up, if I understand the definition, I'll use your suggestion. Is there any other proof where I don't have to use that concept of abstract algebra?

4. Oct 31, 2013

brmath

You have $s_n \rightarrow s \in S;$ and $t_n \rightarrow t \in T$. So s + t is in S+T. It would be fun to say lim$_{n \rightarrow \infty}(s_n + t_n)$ = s + t. (fun because then you would be done.)

This is equivalent to saying that lim($s_n + t_n$) = lim$s_n$ + lim$t_n$. Now what do you know about splitting up limits like this? Can you always do it? Can you sometimes do it? If splitting it is subject to conditions, what would they be?

As for why T is finite dimensional that may have something to do with the above. As to why they can't both be finite dimensional, of course they could, but that is not what you are being asked.

By the way, what is X?

5. Nov 1, 2013

mahler1

Oops, I meant $E$, not $X$.

Wait, before trying to prove that lim($s_n + t_n$) = lim$s_n$ + lim$t_n$, wouldn't I have to prove that $\{s_n\}_{n \in \mathbb N}$ and $\{t_n\}_{n \in \mathbb N}$ are convergent sequences? I know that $E$ is a complete space, so any Cauchy sequence is convergent. But for me it's not obvious that $\{s_n\}_{n \in \mathbb N}$ and $\{t_n\}_{n \in \mathbb N}$ converge just from the fact that $\{x_n\}_{n \in \mathbb N}$ is convergent.

6. Nov 2, 2013

brmath

You are entirely right, we would have to show that $s_n$ and $t_n$ are Cauchy first in S + T which implies in S and T respectively. (and you said that the first time.). I think this is not true.

Let's let S and T just be closed intervals on the real line. Let $s_n + t_n \rightarrow c$. Let $t_n = c_n + k_n$ where $k_n$ = n mod2 and $c_n \rightarrow c$. Let $s_n = -k_n$. Then neither $s_n$ nor $t_n$ converges, even though $s_n + t_n$ converges to c.

I have looked around and I find that in general the sum of two closed subspaces of a Banach space need not be closed. https://www.physicsforums.com/showthread.php?t=214357. So it's all got to depend on T being finite dimensional.

I do not have an answer for you, and unfortunately not enough time to work one out. I hope some mentor will help here. If no one comes by, perhaps you can pm someone.

Here is how I would approach it: look at the case where both spaces are finite dimensional. Then every element of each space can be written in terms of a finite number of basis elements. If you can show that the sum of two closed spaces is closed, it will have to depend on the finiteness of the bases.

Well, in what way does it depend on that? If you can figure out that much, then you can figure out how the result remains without the finite basis for one of the spaces.

By the way, I looked into the quotient map approach and if it helps I can't see it.

7. Nov 2, 2013

brmath

Well, I kept thinking about it and worked out some of it last night. Let's start with a simple case E =$R^3$. Let S be the line generated by (1,0,0) T the line (0,1,0). Both are closed. Why? The standard definition is that a set is closed if it's complement C (all the points not in the set) is open. The complement of the x-axis is the set of point (a,b,c) such that a≠ 0.

Open means that for every point in the set there is a neighborhood (in $R^3$ we can think of it as a ball around the point), which is contained in C. This is nearly obvious in this case. You could write that out formally for $C_x$ the complement of the x-axis.

Then S + T is the xy-plane. It is closed because its complement is open -- same as above.

In fact, for every $R^n$ each subspace is closed; the sum of two subspaces is another subspace and is closed because its complement is open.

If E is an n-dimensional subspace which is not $R^n$ it is isomorphic to $R^n$ and the same conclusions apply.

In the case that E is infinite, let's look again at an example, the $R^{\infty}$ space. This is generated by the basis

\begin{align} &(1,0,0..[zeros]..)\\ &(0,1,0,....)\\ &(0,0,1,0,...)\\ &...\\ \end{align}

Let S be the subspace generated by the basis which is the alternate elements of the basis for E -- i.e. each element $s_n$ has a 1 in the 2n-1 spot and a zero elsewhere, so its elements look like (a,0,b,0,c,0,...). Again you can show this is closed. Let T = the line generated by (0,1). The S + T is the set of all vectors of the form (a,t,b,0,c,0,...) and again is closed.

The general case for $R^\infty$ is very similar to the example above. I am pretty sure any (infinite) space G which has an orthonormal basis is isomorphic to E. (The isomorphism would just map the basis elements of G into the basis elements of E).

What is left is Banach spaces which do not have an orthonormal basis. This I know very little about, but obvously the space in the example I linked you to must be one.

Hope this helps.