I've encontered a problem that appears simple, almost like it's a definition of a bounded operator, but with a reversed inequality sign... and I can't seem to find my way to a solution.
Any suggestions appreciated!
Note: $X$ and $Y$ are Banach (or even Hilbert) spaces.
Problem:
$ A:X \to Y $ is a bounded linear operator. Let $ A(X) = Y$, and $ y_n \to y_0 $ in $ Y $.
Prove that there exists $ c > 0 $, and a sequence $ x_n $ in $ X$, so that $ x_n \to x_0 $ , $ Ax=y $ , and $ \lVert x_n\rVert \le c\lVert y_n\rVert$
One approach is to consider $\dot{A} : X/\mathcal{N}(A)\rightarrow Y$, which is continuous and a linear bijection between $X/\mathcal{N}(A)$ and $Y$. So the inverse of $\dot{A}$ is continuous, and the statement of that continuity gives you what you want.