A criterion for the sum of two closed sets to be closed ? 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 ?  
 A: 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 $e_n-x_n\to 0$.
A: 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).
A: 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 $X/I$). 
A: In finite dimensional space, your result is true. Take a linear functional with $\text{ker } T = I$. It is not hard to prove, using $I\cap V={\boldsymbol 0}$, that $T(V)$ is closed by seeing that $$\text{inf }_{\begin{array}{c}\boldsymbol x \in V \\  \|\boldsymbol x\| = 1\end{array}} \|T(\boldsymbol x)\| > 0,$$ with the minimum achieved. Thus, by Lemma H1 of the book Geometric Functional Analysis and its Applications, $I+V$ is closed.
A: Please, have a look at Sections 22 and 35 in G.J.O. Jameson, "Topology and Normed Spaces", Chapman and Hall, 1974.
