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''?

  • $\begingroup$ 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

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 $\|Te_n\| = n$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.