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Denote $c = \{ \xi=(\xi_n)_{n \in \mathbb{N}}: \lim_{n \rightarrow \infty}\xi_n \text{ exists} \}$ equipped with supremum norm $\| \xi \|_{\infty} = \sup_{n \in \mathbb{N}}|\xi_n|.$

In this post, Daw suggested that to view limit operator $\text{lim}$ as a continuous functional on $c$. Below is my attempt to show that $\text{lim}$ is continuous.

Define $\lim: c \rightarrow \mathbb{R}$ by $\lim(\xi) = \lim_{n \rightarrow \infty} \xi_n.$

I wish to show that $\lim$ is bounded.

Let $\xi=(\xi_n)_{n \in \mathbb{N}} \in c.$ Note that

$$|\lim(\xi)| = |\lim_{n \rightarrow \infty}\xi_n| \leq \sup_{n \in \mathbb{N}}|\xi_n| = \| \xi \|_{\infty}.$$

Therefore, $\| \lim \| = \sup\{ |\lim(\xi)|:\| \xi \|_{\infty} \leq 1 \} \leq 1,$ hence $\lim$ is bounded.

Since $\lim$ is both linear and bounded, it is continuous.

Is my proof correct?

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    $\begingroup$ looks good perhaps you could show $|\lim_{n \rightarrow \infty}\xi_n| \leq \sup_{n \in \mathbb{N}}|\xi_n| $ this bound with more details $\endgroup$ – clark May 25 '17 at 1:35
  • $\begingroup$ @clark: Actually I know the inequality holds intuitively, but do not know how to write down a formal proof. $\endgroup$ – Idonknow May 25 '17 at 1:42
  • $\begingroup$ Try, it is not hard. $\endgroup$ – reuns May 25 '17 at 1:44
  • $\begingroup$ You could say $ |\xi_n| \leq || \xi_n ||_\infty ~ \forall n$ therefore $|\lim_{n} \xi_n| = \lim_{n} |\xi_n| \leq || \xi_n ||_\infty $. which is a standard lemma. $\endgroup$ – clark May 25 '17 at 1:46
  • $\begingroup$ @mirko: yes, I mean $c $ $\endgroup$ – Idonknow May 25 '17 at 1:58

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