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Let $D \subseteq C([0,1])$ be a dense countable subset of the continuous functions from $[0,1]$ to $\mathbb{R}$. Let $L\colon D \rightarrow \mathbb R$ be a linear and bounded functional with norm $\leq 1$. How can I extend it to a continuous functional over all $C([0,1])$ with norm ${}\leq 1$?

For every point $x \not \in D$ I can define $L(x)$ as the limit of a sequence in $D$ converging to $x$. But is this limit well defined?

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    $\begingroup$ In case you are interested, this works in a much more general setting. If you look at what is actually being used here, you are only using that $\mathbb{R}$ is complete (along with what the norm structure on the two spaces). $\endgroup$ – Francis Adams Oct 19 '13 at 16:30
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Yes: we can check that this definition does not depend on the choice of the approximating sequence (if $x_n\to x$ and $y_n\to x$, consider $(L(x_1),L(y_1),L(x_2),L(y_2),\dots)$: this is a Cauchy sequence hence the subsequences $(L(x_{k}))$ and $(L(y_k))$ have the same limit.

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  • $\begingroup$ $x_n \rightarrow x$ means that $\| x_n\|_{\infty} \rightarrow \| x \|_{\infty}$? $\endgroup$ – User11111 Oct 19 '13 at 16:26
  • $\begingroup$ I implies that ($x_n\to x$ means $\lVert x_n-x\rVert\to 0$ which is stronger). $\endgroup$ – Davide Giraudo Oct 19 '13 at 16:27

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