Direct sum of two closed subspaces of Banach space is not closed

Question

I'm looking for an example of two closed subspace of a Banach space (or even a Hilbert space) whose sum is not closed.

We have $l^2$ as Banach space and $A$ and $B$ are closed subspaces of $l^2$ : $A=\{a\in l^2$| $a^{2n}=0$ }$ $ B={$a\in l^2| a^{2n}=a^{2n-1}/2n \} $.

I have to prove $A+B$ is not closed, then I can use this example as a counter example.

What is $n$? Do you mean $A={a\in \ell^2\mid a^{2n}=0 \text{ for some } n}$? And in $B$ for all $n$ or for some $n$? — This Is Me, Dec 28 '15 at 10:07
@ThisIsMe it means for all n , $a_2n$ is zero , for B is like that too. — Albert, Dec 28 '15 at 12:54

score 4 · Answer 1 · answered Mar 06 '16 at 07:41

There are two observations to make:

$A+B$ is dense in $l^2$, because it contains every sequence that is eventually zero.
The sequence $(1/n)_{n\ge 1}$ is not in $A+B$. Indeed, if it's written as $a+b$ with $a\in A$, $b\in B$, then $b_{2n}=1/(2n)$, hence $b_{2n-1}=1$ for all $n$.

If a subspace is dense and is not all of $l^2$, it is not closed.

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