Prove that $X^\ast$ separable implies $X$ separable

Can someone tell me if I got the following right:

Assume $X$ to be a normed vector space over $\mathbb{R}$. Prove that if the dual space $X^\ast$ is separable then $X$ is separable as well.

I'm supposed to use the following hint: First show that for each $x_n^\ast$ we may choose a unit vector $x_n \in X$ such that $x_n^\ast (x_n) \geq \frac{\| x_n^\ast \|}{2}$. Then show that $\overline{Y} = \overline{span_\mathbb{Q} \{ x_n\}} = X$.

Assume for all $x_n \in X$ with $\| x_n \|_X = 1$ that

$$x_n^\ast (x_n) < \frac{\| x_n^\ast \|}{2} \iff 2 x_n^\ast (x_n) < \sup_{x_n; \| x_n \|_X = 1 } \{ | x_n^\ast(x_n) | \}$$

Claim: Then $\exists r \in \mathbb{R}: x_n^\ast(x_n) < r < \sup_{\dots}\{\dots \}$:

(i) $x_n^\ast (x_n) \neq 0$: because $\| x_n \| = 1$ and $x_n^\ast$ linear

(ii) if $x_n^\ast (x_n) < 0$ then $2 x_n^\ast (x_n) < x_n^\ast (x_n) < x_n^\ast(- x_n) < 2 x_n^\ast(-x_n) < \sup_{\dots} \{ \dots \}$

(iii) $x_n^\ast(x_n) > 0$ then $\forall x_n: x_n^\ast (x_n) < 2 x_n^\ast (x_n) < \sup$

$\implies$ the $\sup$ is not the l.u.b., contradiction, so the claim $x_n^\ast (x_n) \geq \frac{\| x_n^\ast \|}{2}$ is true.

(ii) Using:

$x \in \overline{Y} \iff \lnot \exists$ bounded linear functional $f^\ast$ such that $f^\ast (y ) = 0 \forall y \in Y$ and $f^\ast (x) \neq 0$

Let $f^\ast \in X^\ast$. Then $\{ x_n^\ast \}$ dense in $X^\ast \implies$

$$\forall \varepsilon > 0 \exists x_n^\ast : \| x_n^\ast - f^\ast \| = \sup_{x \in X: \| x \| \leq 1} \{ |x_n^\ast (x) - f^\ast(x)| \} < \varepsilon$$

But for $x_n \in Y \subset X (\| x_n \| = 1)$ we know $$| x_n^\ast (x_n)| \geq x_n^\ast (x_n) \geq \frac{\| x_n \|}{2} > 0$$

so if $f^\ast (y) = 0 \forall y \in Y$ then

$$0 < \frac{\| x_n \|}{2} \leq |x_n^\ast (x_n)| = |x_n^\ast (x_n) - f^\ast(x_n)|$$

$$\implies \exists \varepsilon: \sup |x_n^\ast (x_n) - f^\ast(x_n)| > \varepsilon$$

$$\implies \lnot \exists f^\ast \in X^\ast : \forall y \in Y: f^\ast (y) = 0$$

$$x \in \overline{Y}$$

Thanks for your help.

• By dual, do you mean algebraic or topological? – Asaf Karagila Nov 15 '11 at 15:35
• @AsafKaragila: If $X$ is a vector space over some field $K$ then $X^\ast$ is the (vector) space of all linear functionals $X \rightarrow K$. – Rudy the Reindeer Nov 15 '11 at 15:37
• "$x^*_n(x_n)$" isn't exactly good notation. Consider using $\varphi$ (\varphi) or other letters for functionals. – kahen Nov 15 '11 at 15:38
• Trust me :) Seriously: you're exploiting the operator norm here. $\sup_{\|x\|=1} |x^\ast (x)|$. This is finite if and only if $x^\ast$ is continuous. That's what's called the dual space. You won't consider the algebraic dual of a Banach space or a topological vector space anytime soon. – t.b. Nov 15 '11 at 20:35
• ...so assume $x_{n}^\ast \neq 0$. Then: Why do you want to go by contradiction? Your strategy boils down to arguing that the norm of $x_{n}^\ast$ can't be what it is therefore there must be an $x_n$ as you want. As I wrote: The claim follows directly from the very definition of the supremum. You have $0\lt\|x_{n}^\ast\| =\sup_{\|x\| = 1}|x_{n}^\ast(x)|$. In particular for every $\varepsilon > 0$ there must be a $y_{\varepsilon}$ such that $|x_{n}^\ast(y_\varepsilon)| \geq \|x_{n}^\ast\| - \varepsilon$. Take $\varepsilon = \|x_{n}^\ast\|/2$ and choose $x_n$ to be such a $y_\varepsilon$. – t.b. Nov 16 '11 at 13:33

To cut a long story short: I don't really follow what you're doing in “(i) (by contradiction)” because the notation is quite confusing, but the argument for “(ii) Using:...” is basically okay, even if only barely readable.

A few remarks:

1. You don't say what $\{x_{n}^\ast\}_{n=1}^\infty$ is. Clearly, there is the implicit assumption that $D = \{x_{n}^\ast\}_{n=1}^\infty$ is a countable norm-dense subset of the Banach space dual $X^\ast$ of $X$.

Always state what you assume! For instance, your choice of $Y$ later on indicates that you work with a real Banach space (who should know that this is assumed if you don't state it?)

2. Fix $n$. By definition of the norm on $X^\ast$ we have $$\|x_{n}^\ast\|_{X^\ast} = \sup_{\Vert x\Vert=1} |x_{n}^\ast (x)|.$$ The definition of the supremum provides us with an $x$ such that $\|x\| = 1$ and $\|x_{n}^\ast\|_{X^\ast} \geq |x_{n}^\ast(x)| \geq \|x_{n}^\ast\|_{X^\ast} / 2$ (if $x_{n}^\ast = 0$ take any $x$ with $\|x\| = 1$).

Multiplying $x$ with an appropriate scalar $\lambda = e^{i\alpha}$ (or $\pm 1$ if indeed we work over the reals), we can put $x_n = \lambda x$ in order to get $x_{n}^\ast(x_n) \geq \|x_{n}^\ast\|_{X^\ast}/2$. Do this for every $n$ to get a set $S = \{x_n\}_{n=1}^{\infty} \subset X$ such that for each $n$ we have $\|x_{n}\|_{X} = 1$ and $x_{n}^\ast(x_n) \geq \|x_{n}^\ast\|_{X^\ast}/2$.

This takes care of your point (i).

3. Let $S =\{x_n\}_{n=1}^\infty \subset X$ be the set chosen in 2 and let $Y=\operatorname{span}_{\mathbb{Q}}\,{S} \subset X$ be the set of rational linear combinations of elements of $S$ (if we should happen to work over the conplex numbers take $\mathbb{Q}+i\mathbb{Q}$ instead of $\mathbb{Q}$). Then $Y$ is countable (say that!). We want to show that $Y$ is dense in $X$, so that $X$ is indeed separable.

Note that the closure $\overline{Y}$ of $Y$ is a linear subspace of $X$ (why is that?).

Assume towards a contradiction that $\overline{Y}$ is not all of $X$. By Hahn-Banach we can find a continuous linear functional $0 \neq x^\ast \in X^\ast$ such that $x^\ast|_\overline{Y} = 0$. By norm-density of $D$ in $X^\ast$, we can find an $x_{n}^\ast$ in $D$ such that $\Vert x^\ast - x_{n}^\ast \Vert_{X^\ast} \lt 1/4$ . Because $x_n$ is in $\overline{Y}$ we have $x^\ast(x_n) = 0$ and thus $$\frac{1}{2} \leq |x_{n}^\ast(x_n)| = |x_{n}^\ast(x_n) - x^\ast(x_n)| \leq \Vert x_{n}^\ast- x^\ast\Vert_{X^\ast} \, \|x_{n}\| \lt \frac14,$$ which is a contradiction. Done.

Note. I'm not sure if that was a typo, but actually we don't need the extra argument in 2. where we show that we can find $x_n$ such that $x_{n}^\ast(x_n) \geq \|x_{n}^\ast\|/2$ (without modulus signs).

4. You start the proof of (ii) by saying “Then $\{x_n\}$ dense in $X$” — wait, what?! are you assuming what you want to prove? — Ah, no, okay, you mean $\{x_{n}^\ast\}$ dense in $X^\ast$. pheew. You then conclude the argument more or less the same way as I did but your exaggerated formalism makes this extremely hard to follow.

Finally, a few things I have told you over and over again:

1. Get rid of all the symbols: $\forall, \exists, \implies, \iff, \lnot$ when you're writing stuff.

They are unpleasant to look at and hard to decipher.

Write prose. Use formulas only when you absolutely need them.

2. Please make complete, grammatically correct sentences. There is not a single complete sentence in your whole solution!

The idea of point 1. is that it forces you to observe point 2. and if you observe point 2. you are forced to eliminate all the quantifiers, implication sings and other logical symbols from your writing.

Please do have a look at

• True. I wasn't actually doing it on purpose to annoy you. I'll try to not use symbols anymore from now on. I do remember that you have told me this before, although I only remember one time, not several. Off to read the 2 links in your answer. – Rudy the Reindeer Nov 15 '11 at 19:51
• Where does your inequality $1/2 \leq |x_n^*(x_n)|$ come from in the last string? This is not what we assume... Also, where did you even use the definition of $Y$? I think there must be something missing. – Johnny Apple Oct 21 '15 at 20:11

I’m going to comment only on what you wrote for (i):

Assume for all $x_n \in X$ with $\| x_n \|_X = 1$ that

$$x_n^\ast (x_n) \lt \frac{\| x_n^\ast \|}{2} \iff 2 x_n^\ast (x_n) \lt \sup_{x_n; \| x_n \|_X = 1 } \{| x_n^\ast(x_n)|\}$$

Claim: Then $\exists r \in \mathbb{R}: x_n^\ast(x_n) \lt r \lt \sup_{\dots}\{\dots \}$:

(i) $x_n^\ast (x_n) \neq 0$: because $\| x_n \| = 1$ and $x_n^\ast$ linear

(ii) if $x_n^\ast (x_n) \lt 0$ then $2 x_n^\ast (x_n) \lt x_n^\ast (x_n) \lt x_n^\ast(- x_n) \lt 2 x_n^\ast(-x_n) \lt \sup_{\dots} \{ \dots \}$

(iii) $x_n^\ast(x_n) \gt 0$ then $\forall x_n: x_n^\ast (x_n) \lt 2 x_n^\ast (x_n) \lt \sup$

$\implies$ the $\sup$ is not the l.u.b., contradiction, so the claim $x_n^\ast (x_n) \geq \frac{\| x_n^\ast \|}{2}$ is true.

The first problem, and it’s a major one, is that right off the bat you’re using $n$ for two completely different things. On the one hand it’s apparently supposed to be the index of a particular, fixed member of some countable norm-dense subset of $X^*$, though you never actually said that. On the other hand it’s a dummy index picking out members of $X$ of norm $1$; this would be a bad idea even if you weren’t already using $n$ for something else, since there’s no reason to suppose that there are only countably many elements of $X$ of norm $1$. You should have begun something like this:

Let $D=\{x_n^*:n\in\mathbb{N}\}$ be a countable norm-dense subset of $X^*$. Fix $x_n^*\in D$, and assume for each $x\in X$ with $\|x\|=1$ that $$x_n^*(x)<\frac{\|x_n^*\|}2.$$

Note that I stopped before your $\iff$ symbol: that’s because what follows it is not part of your assumption, but rather an inference from your assumption, so it does not belong in the same clause with assume that. Once you’ve clearly stated your assumption, then you can go on and draw conclusions:

This implies that $$2x_n^*(x)<\sup_{\|y\;\|=1}\|x_n^*(y)\|$$ for each $x\in X$ with $\|x\|=1$.

Note that I had to use a different letter for the dummy variable ($y$) in the supremum from the one used for the specific $x$ of norm $1$ in the surrounding statement: they refer to different objects.

The line that begins Claim makes no sense even after the ellipsis at the end is properly filled in. (Note, by the way, that this is something that you should have done yourself, so that the reader needn’t guess; with cut-and-paste it’s completely trivial.) You still have $n$ meaning two different things, and it’s not clear whether you’re talking talking about a specific $x$ of norm $1$ or all of them together. I suspect that you meant this:

Claim: There is an $r\in\mathbb{R}$ such that $$x_n^*(x)<r<\sup_{\|y\;\|=1}|x_n^*(y)|$$ for each $x\in X$ of norm $1$.

Presumably what follows is supposed to be a proof of the claim. Say so.

Proof of Claim: Fix $x\in X$ of norm $1$. Since $\|x_n^*\|$ is linear, $x_n^*(x)\ne 0$.

This doesn’t appear to follow. In fact, there seems to be nothing in your argument to here to preclude the possibility that $x_n^*$ is the zero functional.

If $x_n^*(x)>0$, then $$x_n^*(x)<2x_n^*(x)<\sup_{\|y\;\|=1}|x_n^*(y)|\;,$$ and if $x_n^*(x)<0$, then $$2x_n^*(x)<x_n^*(x)<x_n^*(-x)<2x_n^*(-x)<\sup_{\|y\;\|=1}|x_n^*(y)|\;.$$

This does not in fact prove the claim; you’ve merely shown that for each $x\in X$ of norm $1$, $$x_n^*(x)<\|x_n^*\|=\sup_{\|y\;\|=1}|x_n^*(y)|\;,$$ i.e., that $x_n^*$ does not attain its norm on $\{y\in X:\|y\|=1\}$.

Had you made the effort to write it intelligibly, thinking about what you were actually saying, you might have noticed that much of it makes no sense and that it does not in fact do what you wanted it to do. At the very least you would have made it possible for others to spot the problems and help with them much more easily.

Remember: A proof is just a particular kind of expository prose. It should consist of paragraphs of sentences. Yes, it will often contain special symbols, but DON’T use symbols just for the sake of using them. The object is to convince the reader that something is true, and you can’t do that if you can’t make yourself understood.

• Hi Brian. Thanks for your helpful answer! I edited the case where I try to show that $x_n^\ast (x_n)$ does not equal $0$. Is case (i) now correct? – Rudy the Reindeer Nov 16 '11 at 8:28
• And is it okay to say "dense" instead of "norm dense"? – Rudy the Reindeer Nov 16 '11 at 8:30
• @Matt: Yes, as long as the context is clear, dense is fine, but since you’ve a tendency to be a bit imprecise, I think that you should probably err in the direction of saying too much rather than too little. No, (i) still doesn’t work. Take $X$ to be $\mathbb{R}^2$ with the usual dot product, and let $x^*$ be the linear functional that maps $\langle a,b\rangle$ to $a$. Let $y=\langle 0,1\rangle\in X$. Then $\|x^*\|_{X^*}=1=\|y\|_X$ and yet $x^*(y)=0$. – Brian M. Scott Nov 16 '11 at 12:30
• Thank you. Now I wonder how I could make such a flawed argument : ( (the answer is because my head is messed up) Regarding the "norm dense": I'd never heard it before. It probably means w.r.t. the norm but then is there ever a normed space with a metric different from the metric induced by the norm? I guess the answer is yes, otherwise the word wouldn't exist. – Rudy the Reindeer Nov 17 '11 at 8:03