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..



When can closedness of the range of an operator be checked on a positive cone?
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 $\ell^2$.)


