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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.

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    $\begingroup$ 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
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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.

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

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