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I recently encountered the theorem that the sum of a closed (linear) subspace with a finite dimensional subspace is closed subspace of the Banach space in which it is contained. However, this came with the caveat that the statement does not hold for two arbitrary closed subspaces.

So, here's what I'm looking for:

Find a Banach space $X$ and closed subspaces $M,N$ such that $$ M+N=\{m+n\mid m\in M, n\in N\} $$ Is not closed in $X$.

Any references, hints, or answers are appreciated!

Ben Grossmann
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  • The statement mentioned, with some proofs – Ben Grossmann May 15 '16 at 21:26
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    From http://www.sciencedirect.com/science/article/pii/1385725885900113: If $M,N$ are two linearly independent closed linear subspaces of a Banach space $X$, then $M+N$ is closed if and only if there exists a constant $A>0$ such that for all $x,y \in X$ with $x \in M$ and $y \in N$, we have $|x| \leq A|x+y|$.

    This should put anyone on the way to finding a counterexample, I guess.

     – sTertooy May 15 '16 at 21:43
  • Can you prove , a vector normed space is infinite dimension iff it contains two closed subspace , whose sum is not closed? – Red shoes Jul 13 '17 at 16:35
  • @Ashkan first of all, why are you commenting on this question which is more than a year old. Second, if you have a new question, post your own question. – Ben Grossmann Jul 13 '17 at 16:51
  • I came cross to this question as a similar question which I answered here https://math.stackexchange.com/a/2357241/219176 I just wanted let you know (give you a hint) how common these two subspaces are! As they appear in any infinite dimension banach space. Your hostile tone reveals you even can't tolerate a simple comment helping you to get better understanding of your own question !

    BTW, I don't need ask any question since I know the its answer.

     – Red shoes Jul 13 '17 at 17:35
  • @Ashkan I'll grant you that I was a bit hostile; sorry for that, and thank you for the information. However, you should know that from my perspective, your comment was a very strange, and it wasn't clear what your intention was. I would have been more receptive if it was clearer that you were "giving me a hint", even if it was a hint towards a question that has long since been answered to my satisfaction. Also, if your goal was to let me know about the result you linked, it would have been preferable if you just posted that link in your first comment. – Ben Grossmann Jul 13 '17 at 20:00
  • @Ashkan You should also know that it's not unusual for users to solicit math help in the comments by posting on an old question; this is what I assumed you were doing, and I find this practice to be very unpleasant. – Ben Grossmann Jul 13 '17 at 20:03

1 Answers1

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Let $X$ be a Hilbert space with orthonormal basis $\{e_n\}_{n\in\mathbb{N}}$. Let $x_n=e_{2n}$, and let $y_n=e_{2n}+\frac{e_{2n+1}}{n+1}$. Take $M$ to be the closed span of the $x_n$ and $N$ to be the closed span of the $y_n$. Note that $M+N$ contains $e_n$ for all $n$, so the closure of $M+N$ is all of $X$.

However, I claim that $M+N$ does not contain the vector $z=\sum \frac{e_{2n+1}}{n+1}$ and hence is not all of $X$. Indeed, if you could write $z=x+y$ for $x\in M$ and $y\in N$, it is clear that $y$ would have to be $\sum y_n$, since the only way to get a nonzero inner product with $e_{2n+1}$ when building an element of $M$ or $N$ is to use $y_n$. Since the sum $\sum y_n$ does not converge, there are no such $x$ and $y$.

Eric Wofsey
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  • I know this question is old, but I found it interesting enough to comment. Could you explain how to "build" elements from $M$ and $N$ to conclude your last point? – user23793 Dec 11 '17 at 15:14
  • By "build" I just mean taking linear combinations of the basis element. So if $z=x+y$ for $x\in M$ and $y\in N$, then $z=\sum a_mx_m+\sum b_my_m$ for some scalars $a_m$ and $b_m$. Computing $\langle z,e_{2n+1}\rangle$, we get just $b_n/(n+1)$, since all the $x_m$ and $y_m$ are orthogonal to $e_{2n+1}$ except for $y_n$ which has inner product $1/(n+1)$ with it. But we know $\langle z,e_{2n+1}\rangle=1/(n+1)$, so $b_n=1$ for all $n$. – Eric Wofsey Dec 11 '17 at 17:23
  • How to conclude that $e_1 \in M+N$ ? – Lucas Linhares Jun 19 '22 at 19:09
  • @ThiagoGuimarãesMelo: $e_1=-x_0+y_0$ – Eric Wofsey Jun 19 '22 at 19:10
  • But in your orthonormal basis is included the vector $e_{0} = 0_{X}$ ? In a orthonormal basis we must have $||e_n|| = 1$, for all $n$. – Lucas Linhares Jun 19 '22 at 19:16
  • @ThiagoGuimarãesMelo: I don't understand why you write $e_0=0_X$. – Eric Wofsey Jun 19 '22 at 19:19
  • Look, by your definition, $x_0 = e_{2\cdot0} = e_{0}$. But what means $e_0$ ? – Lucas Linhares Jun 19 '22 at 19:21
  • $e_0$ is just the one of the elements of the original orthonormal basis. – Eric Wofsey Jun 19 '22 at 19:21
  • Thank you Eric, now I unsdertood. – Lucas Linhares Jun 19 '22 at 19:41
  • I think I know why Thiago was confused. In principle, the vector $e_0$ is not a part of the orthonormal basis as its indexation starts with $n=1$. The easiest is just to start the indexation with $n=0$, and hence, ${e_n}_{n\in \mathbb N_0}$. Otherwise, everything is correct. – Marko Nov 13 '23 at 11:22