I know that $||T(x)|| \geq C ||x||$ for some $C$ is equivalent. I am looking for less analytic conditions, maybe something to do with the topological structure of $X$. Does anyone know some?
By $T$ closed I mean that the image of $T$ is closed in $Y$.
Addenum:
My reason for doing this is:
I am trying to argue that the Schwartz class S is not normable by displaying it as a dense subspace of L1, and then trying to argue that (given normability) the inclusion can be upgraded to surjectivity (which is nonsense), by showing that the range is closed. Since I don't know what the norm looks like, I wanted to make use of some less analytical property of S in order to show that the range is closed.
No, $\|Tx\| \ge C \|x\|$ is not equivalent. The range of a bounded linear operator $T: X \to Y$ is closed iff there is $C>0$ such that $\|Tx\| \ge C \|x + \text{ker}(T)\|$, i.e. that the operator $\widetilde{T}: X / \text{ker}(T) \to Y$ given by $\widetilde{T} \circ \pi = T$ is an isomorphism onto its range (where $\pi$ is the quotient map of $X$ to $X/\text{ker}(T)$.
It is also a theorem that $T$ has closed range iff its dual $T^*: Y^* \to X^*$ has closed range.
I'm not sure what you're getting at with "topological structure of $X$". You could say, for example, that if $T$ is injective and $Y$ has no subspaces isomorphic to $X$, then $T$ can't have closed range.