$A$ is a bounded linear operator from $X$ to $Y$ (both Banach spaces). Show that if there exists $k > 0$ such that $\|Ax\| \geq k\|x\|$, for all $x$ then $\operatorname{range}(A)\,$ is closed.
My approach: I've tried to show that the complement of $\operatorname{range}(A)$ is open but I haven't had much luck and at this point I feel like I'm banging my head against a wall. Can anyone offer a hint or suggest an approach? It's appreciated.
Edit: Okay, suppose $Ax$$_{n}$ $\rightarrow$ $y$ in $Y$ then from $$\|Ax_{n} - Ax_{m}\|= \|A(x_{n} - x_{m})\| \geq k\|x_{n} - x_{m}\|,$$ we see that the sequence {$x_{n}$} is Cauchy in a complete space (so it converges). If $x_{n}$ $\rightarrow$ $x$ then $Ax_{n}$ $\rightarrow$ $Ax$ and so $y = Ax$. Now, I feel like this would mean that range$(A)$ must be closed. I'm having some trouble articulating why this is the case. If somebody could spell it out for me I'd appreciate it. Thanks for the help thus far.
I am just repeating your argument, which is tottaly correct:
If $Ax_n\to y\in Y$, then $Ax_n$ is a Cauchy sequence, and as $$ \|Ax_m-Ax_n\|\ge k\|x_m-x_n\|, $$ then $\{x_n\}$ is also a Cauchy sequence, and hence converges. Let $x_n\to x$, then $Ax_n\to Ax$, as $A$ is bounded, and therefore $y=Ax$. Thus range$(A)$ is closed.