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Let $V$ and $I$ be two closed subsets of a Banach space $A$.

The set $V$ is a convex cone, and $I$ is a linear subspace of $A$. I also know that $V\cap I=\{0\}$.

I would like to know whether $I+V$ is closed. I've seen that there is a criterion of Dieudonné which I can't use here because I know that neither $V$ nor $I$ is locally compact.

So my question is : are there any other criteria that I could try to use ?

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Maybe I should add that by "criterion" I mean a sufficient condition. –  Fabien Besnard Feb 14 '12 at 15:51

3 Answers 3

up vote 4 down vote accepted

When $V$ is also a subspace, the standard equivalence to the sum being closed is that the unit spheres of $V$ and $I$ are a positive distance apart. I bet this is true when $V$ is just a convex cone but don't have time right now to think about it (the given condition is clearly sufficient for closedness of the sum; necessity is the direction that requires thought).

EDIT: It will come to no surprise to those who know me that I lose my bet. Let $A$ be $\ell_p$, $1\le p < \infty$ ($c_0$ is also OK with suitable notational changes) and $x^*$ the linear functional $\sum 2^{-n} e_n$, where $e_n$ is the unit vector basis for $A^* = \ell_q$, $1/p + 1/q = 1$. $I$ is the kernel of $x^*$ and $V$ are the non negative vectors in $A$. In Pietro's notation, $\pi(V)$ is closed because it is a cone which contains its base point in a one dimensional space and hence, as Pietro remarked, $V+I$ is closed in $A$. For $n>1$ let $x_n=e_n-2^{-n+1}e_1 \in I$ and observe that $e_n-x_n\to 0$.

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Ok, I managed to show that if the unit spheres of the subspaces $V-V$ and $I$ are a positive distance apart then $V+I$ is closed under the hypotheses I have given. The subspace $V-V$ need not be closed, only $V$ has to be. The sufficient condition was all I cared for, and this one works in the case I have at hand. So I thank you and accept your answer. –  Fabien Besnard Feb 20 '12 at 16:32

In general, there is no easy criterion. I recall the construction of two closed subspaces of a Banach space whose sum is not closed: Let $T:X\to Y$ be a linear map between Banach spaces with closed graph $G= \{ (x,T(x)): x\in X \}$. Then $L=\{(\xi,0): \xi\in X\}$ is another closed (even complemented) subspace such that $G+L= X\times T(X)$ which is closed if and only if $T(X)$ is closed in $Y$. Moreover, $G\cap L=0$ if $T$ is injective. A concrete example is obtained for the inclusion $\ell_1 \hookrightarrow \ell_2$ (in this case the sum is dense).

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I'd like to mention the following, even though it is just a reformulation: For $V\subset X $ and $I\subset X$ a closed linear subspace, the sum $V+I$ is closed in $X$ if and only if $\pi(V)$ is closed in the quotient $X/I\\ $, $\pi:X\to X/I$ being the quotient map.

(Reason: Indeed, if $\pi(V)$ is closed, so is $V+I=\pi^{-1}\pi(V)$. On the other hand, if $V+I$ is closed and $\xi\in \overline{ \pi(V) }$, then $\xi$ is limit of a sequence $\xi_n\in \pi(V) $ with $\| \xi_n - \xi_{n+1}\|_{X/I}\le 2^{-n}$ and by definition of the quotient norm there exists an inductively defined sequence $w_n\in V+I$ such that $\xi_n=\pi w_n$ and $\|w_n -w_{n+1}\| _ X < 2^{-n}$; therefore $(w_n)_n $ is a Cauchy sequence in the closed set $V + I$ and converges to an element $w\in V + I$ such that $\pi w=\xi$, which proves that $\pi(V)$ is closed in $X/I$).

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