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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''?

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  • $\begingroup$ The text says ''A linear map $u\colon X\to Y$ between Banach spaces is bounded below if there is a positive number $\delta$ such that $\| u(x) \| \ge \delta \| x \| \ (x \in X)$''. $\endgroup$ – Ichiko Mar 11 '18 at 8:04
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If $X$ and $Y$ are Banach spaces, bounded from below implies bounded from above.

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 $\|Te_n\| = n$.

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  • $\begingroup$ Does the range of the map $T$ really included in $\ell^2$? $\endgroup$ – Ichiko Apr 8 '18 at 15:20
  • $\begingroup$ @Ichiko You are right, $T$ is not well defined. If we consider the same map $T : c_{00} \to c_{00}$ then it's well defined, but $c_{00}$ is no longer a Banach space. $\endgroup$ – mechanodroid Apr 8 '18 at 15:56
  • $\begingroup$ @Ichiko I fixed this answer in case you are still interested. It turns out that if $X$ and $Y$ are Banach spaces, bounded from below indeed implies bounded from above. Otherwise it does not hold. $\endgroup$ – mechanodroid May 15 '18 at 18:21

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