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..