Two closed subspace whose sum is not closed?

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An example of two closed subspaces in a Hilbert space where their sum is not closed can be found in the space \(\mathcal{H}=\ell^2(\mathbb{N})\). The operator \(T\) defined by \(T(en)=(1/n)*en\) creates a bounded linear operator on \(\mathcal{H}\). The subspaces \(A\) and \(B\) are defined as \(A={(x,0) : x \in \mathcal{H}}\) and \(B={(x,Tx) : x \in \mathcal{H}}\) in the direct sum \(\mathcal{H} \oplus_2 \mathcal{H}\). The sum \(A+B\) is closed if and only if the range of \(T\) is closed, but since the range of \(T\) is a proper dense subspace, \(A+B\) cannot be closed. This topic is also covered in literature, such as Halmos's "A Hilbert Space Problem Book."
quasar987
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What would be an example of two closed subspaces of a normed (or Banach) space whose sum A+B = {a+b: a in A, b in B} is not closed?

I suppose we would have to look in infinite dimensional space to find our example, because this is hard to imagine in R^n!
 
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How about this: Consider the Hilbert space \mathcal{H}=\ell^2(\mathbb{N}) of square-summable sequences of reals. Let {en} be the standard o.n. basis for \mathcal{H}, and define T on \mathcal{H} by letting T(en)=(1/n)*en and extending linearly. This is a bounded linear operator on \mathcal{H}. Next, consider the space \mathcal{H} \oplus_2 \mathcal{H}, which is simply the direct sum of two copies of \mathcal{H} given the 2-norm coordinate wise. (This is still a Hilbert space.) Let A={(x,0) : x in \mathcal{H}} and B={(x,Tx) : x in \mathcal{H}}. Then A and B are subspaces of \mathcal{H} \oplus_2 \mathcal{H}, and A+B is closed there iff {Tx : x in \mathcal{H}} is closed in \mathcal{H}. But the range of T is a proper dense subspace of \mathcal{H}. Thus, A+B cannot be closed.
 
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I was googling to see if there's a better example, and I found the following http://www.hindawi.com/GetArticle.aspx?doi=10.1155/S0161171201005324. You might find it interesting.

Also, apparently this problem is discussed in the books A Hilbert Space Problem Book by Halmos and Elements of Operator Theory by Kubrusly. Try to see if your library has a copy of either.
 
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Very nice! and congratulations on the fruitful google search ;)
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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