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 $X$.

  • $\begingroup$ 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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.