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I was looking at a question about the dual space of the completion of a normed linear space. It's clear to me that the dual space of the completion agrees with the dual space of its completion, which can be seen by extending and restriction linear functionals defined on the original space and completion, respectively--these will be inverses. However, I cannot see how to show that they are each an isometry (in this case) without resorting to the Hahn Banach Theorems. Does anyone know of anyway to see this just via the definition of the norm of the dual space?

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If $X$ is a normed space, $\tilde{X}$ its completion, and $f\in \tilde{X}^*$, then $$\{|f(x)|:x\in X,\|x\|\leq1\}$$ is dense in $$\{|f(x)|:x\in\tilde{X},\|x\|\leq1\},$$ and thus $$\sup\{|f(x)|:x\in X,\|x\|\leq1\}=\sup\{|f(x)|:x\in\tilde{X},\|x\|\leq1\}.$$

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  • $\begingroup$ Thank you! This follows from the density of $X$ in $\tilde{X}$, correct? That's the only way I see to verify the density claim. $\endgroup$ – mathgenesis22813 Sep 19 '17 at 1:48
  • $\begingroup$ Yes, the density of $X$ in $\tilde X$ implies the density of $\{|f(x)|:x\in X,\|x\|\leq1\}$ in $\{|f(x)|:x\in \tilde{X},\|x\|\leq1\}$. $\endgroup$ – Aweygan Sep 19 '17 at 2:00

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