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Let $T:X\to Y$ be an operator between Banach spaces $X$ and $Y$. Assume that $X$ has a positive cone $C\subset X$, which generates $X$: every element of $X$ can be written as a difference of elements of the positive cone.

Assume now that we can show $$k\Vert x\Vert\le \Vert Tx\Vert$$ for some $k>0$ and every $x\in C$.

Question: Are there any natural, general conditions on $T$, $X$ or both that allow to conclude that the image of $X$ under $T$ is closed in $Y$ (the above inequality holds for all $x\in X$, possibly with a different $k$)?

The motivating example is the following. Let $G=(V,E)$ be an infinite (oriented) graph and take $X=\ell_1(V)$, $Y=\ell_1(E)$. Let $T$ be the discrete gradient, $Tf(x,y)=f(y)-f(x)$. Then it is enough to check the inequality only for positive $f$, then use the triangle inequality.

Update: Given Andreas' response I realized I should probably ask not for general conditions but any sufficient condition that would give the above property, in particular in special cases like the motivating example above..

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I'm reluctant to say no when the question is "Are there any natural, general conditions ..." but I'll take my chances. The following seems to be a counterexample with every nice property I can think of. Let X and Y both be the Hilbert space $\ell^2$, and label an orthonormal basis as $\{e_n,f_n:n\in\mathbf N\}$. Let $T$ be the (diagonal) operator that sends each $e_n$ to itself and sends each $f_n$ to $f_n/n$. Note that this isn't closed; its range doesn't contain the vector $\sum_0^\infty (f_n/n)$ (because the pre-image would be $\sum_0^\infty f_n$ which isn't in $\ell^2$) but it does contain all the finite partial sums $\sum_0^M (f_n/n)$ that converge to it. Now let $C$ be the cone consisting of those vectors $\sum_0^\infty (a_ne_n+b_nf_n)$ where $|b_n|\leq a_n$ for all $n$. On vectors of this form, the inequality $k\Vert x\Vert \leq \Vert Tx\Vert$ holds with $k=1/\sqrt 2$. (To prove it, consider each of the 2-dimensional spaces spanned by one $e_n,f_n$ pair.) Finally, each vector in $\ell^2$ is the difference of two vectors from $C$. (Again, to prove it, look in those 2-dimensional subspaces and check that there any vector $v$ is the difference of two vectors from $C$, each having at most twice the norm of $v$. The norm bound is used to assemble the 2-dimensional facts into the corresponding fact for $\ell^2$.)

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Andreas, thanks, your example certainly works. However, I am interested particularly in positive results. I realized already when was writing the question that it might be difficult to convey that. –  Piotr Nowak Jul 6 '10 at 4:02

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