Let $\phi_n:X \rightarrow \mathbb{R}$ be a sequence $(n=0,1,\dots)$ of linear functionals on X, where X is Banach and for which the map $\Phi:X\ni x \longrightarrow(\phi_n(x))_{n=0}^{\infty}\in \ell^1$ is well defined. Prove that: $$\Phi \text{ is continous } \iff\phi_n \text{ is continuous for any } n\ge0$$

My attempt

Proof from left to right is obvious. From right to left I know I should use Banach Steinhaus theorem but I can't see how it helps.

  • 2
    $\begingroup$ Find operators $\Phi_n$ that are continuous, and satisfy $\Phi_n(x) \to \Phi(x)$ for all $x\in X$. $\endgroup$ – Daniel Fischer Jun 10 '14 at 12:04
  • $\begingroup$ Duplicate (almost) of this. $\endgroup$ – David Mitra Jun 10 '14 at 12:30

Set $\Phi_n:X\to\ell_1$, $x\mapsto(\phi_1(x),\dots,\phi_n(x),0,0,\dots)$. Then $\Phi$ is the strong limit of $\Phi_n$ as $n\to\infty$ and if all $\Phi_n$ are bounded, then they are uniformly bounded by Banach-Steinhaus (because they converge on each vector $x\in X$) and hence the limit, $\Phi$, is a bounded operator as well.

  • $\begingroup$ all is clear now, thanks $\endgroup$ – luka5z Jun 10 '14 at 12:11

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.