In Murphy's book, ''bounded below'' is defined on a linear map $u\colon X\to Y$ between Banach spaces, not on a bounded linear operator.But continuity of $u$ is used when proving closedness of $u(X)$.
Do I misunderstand the definition of ''bounded below''? Or does continuity of $u$ follow from ''bounded below''?
Best Answer
Then answer is no, as stated here.
Below is a tentative proof which is wrong since the image of a bounded-from-below linear map doesn't have to be closed.
Let $A : X \to Y$ be bounded from below so there exists $m >0$ such that $\|Ax\| \ge m\|x\|, \forall x \in X$.
In particular, $A$ is injective, so its corestriction $A : X \to \operatorname{Im} A$ is a bijection. Hence $A^{-1} : \operatorname{Im} A \to X$ exists and is again a linear bijection. For $y = Ax \in \operatorname{Im A}$ we have
$$\|A^{-1}y\| = \|A^{-1}Ax\| = \|x\| \le \frac1m \|Ax\| = \frac1m \|y\|$$
so $A^{-1}$ is bounded.
Furthermore, $A$ bounded from below implies that $\operatorname{Im} A$ is a closed subspace of $Y$. Therefore, $\operatorname{Im} A$ is also a Banach space.
The Bounded Inverse Theorem now implies that $A = (A^{-1})^{-1} : X \to \operatorname{Im} A$ is a bounded linear map. Therefore $A : X \to Y$ is bounded.
The statement doesn't hold if the spaces are not Banach.
Consider the linear map $T : c_{00} \to c_{00}$ defined as $T(x_n)_{n=1}^\infty = (nx_n)_{n=1}^\infty$ for every sequence $(x_n)_{n=1}^\infty \in \ell^2$.
We have
$$\left\|T(x_n)_{n=1}^\infty\right\|_2^2 = \sum_{n=1}^\infty n^2 |x_n|^2 \ge \sum_{n=1}^\infty |x_n|^2 = \left\|(x_n)_{n=1}^\infty\right\|_2^2$$
so $T$ is bounded from below. However, $T$ is clearly not bounded from above since $\|Te_n\| = n$.