Dual space is separable

Question

Let $X$ be a Banach space and $Y$ be a closed subspace of $X$. Suppose that $X^*$ is separable. Prove that $Y^*$ is separable.

Attempt: Since $X^*$ is separable then we can conclude that $X$ is separable. If I could prove that $Y$ is reflexive (which I don't think is true) I could easily deduce that $Y^*$ is separable.

Any hint would be appreciated.

Well why would $Y$ be reflexive? this would be like claiming that every subspace of a space with separable dual is reflexive. But what about $X=Y=c_0$? we have that $X^=\ell^1$ which is separable but $X^{*}=\ell^\infty$ which is not canonically isomorphic to $c_0$. — Just dropped in, Jun 02 '21 at 22:30

score 5 · Accepted Answer · answered Jun 02 '21 at 22:36

5

Let $\{\varphi_n\}\subset X^*$ be a countable set that is dense in $X^*$. We claim that the restrictions $\{\varphi_n\vert_Y\}\subset Y^*$ form a dense subset of $Y^*$. Indeed, let $\psi\in Y^*$ be a bounded functional and let $\varepsilon>0$. Our goal is to find $m\in\mathbb{N}$ so that $\|\psi-\varphi_m\vert_Y\|<\varepsilon$, so this is what we are trying to do now.

By the Hahn-Banach extension therorem we can find a functional $\Psi\in X^*$ so that $\Psi\vert_Y=\psi$. Now since $\{\varphi_n\}$ is dense in $X^*$ we can find $m\in\mathbb{N}$ so that $\|\Psi-\varphi_m\|<\varepsilon$. In other words, $$\sup_{x\in X,\|x\|\leq1}|\Psi(x)-\varphi_m(x)|<\varepsilon$$

Therefore, $$\sup_{y\in Y,\|y\|\leq1}|\psi(y)-\varphi_m(y)|<\varepsilon$$ so $\|\psi-\varphi_m\vert_Y\|<\varepsilon$, as we wanted.

answered Jun 02 '21 at 22:36

Just dropped in

16,664

Very nice answer. Where was the hypothesis that $Y$ is closed used? – George Giapitzakis Jun 02 '21 at 22:44
@GiorgosGiapitzakis nowhere! – Just dropped in Jun 02 '21 at 22:45
Interesting. The question was given as is in a past exam. I think it is subtly used to guarantee that $Y^*$ is a Banach space though. – George Giapitzakis Jun 02 '21 at 22:46
1

@GiorgosGiapitzakis sometimes an extra assumption is added so the setting feels more natural somehow. You are doing the right thing by looking for the part where the assumption is being used. – Just dropped in Jun 02 '21 at 22:48
Thanks again for the answer! – George Giapitzakis Jun 02 '21 at 22:48

Dean Miller · Answer 2 · 2025-04-21T07:40:01.907

Another way to solve this problem is to define $\varphi \colon X^{*}\to Y^{*}$ by $\varphi (f) := f\vert_{Y}$. Then $\varphi$ is a bounded linear operator and is surjective by a Hahn-Banach theorem extension theorem. Since continuous functions map separable subsets to separable subsets (see here) and $X^{*}$ is separable, it follows that $Y^{*}$ is separable. Note that the assumptions that $X$ is complete and $Y$ is closed are not needed.

Dual space is separable

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