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Let $Z\leq X$ be a dense subspace. Prove that their duals are equals.

I need to complete my idea. I know that $Z^*\subset X^*$. By the way, if $f\in Z^*$, I think that I can use the density to extend $f$ to $X$.

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  • $\begingroup$ Yes, that's the right idea. The key point is that if $f$ is a continuous linear functional vanishing on a dense subspace of something, then it is identically $0$. $\endgroup$ – user61527 Jun 25 '14 at 23:16
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They are not quite equal, but they are isometrically isomorphic (so equal for all practical purposes). The key lemma to use is:

If $f$ is a continuous linear functional and $W \subseteq X$ is a dense subspace such that $f|_W = 0$, then $f = 0$.

The proof of this is based on density: Choose a sequence $x_n \in W$ converging to $x \in X$ and compute $f(x)$ via continuity.

So it follows that the functionals on $W$ can be identified with the functionals on $X$, and go from there. Note that the lemma gives uniqueness of the extension, and continuity lets you extend in the first place.

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