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Let $X$ be a normed space over field $\mathbb{K}$ and let $X_0$ be a linear subspace of $X$. I have to prove that:

$X=\overline X_0 \iff$ For every linear continuous functional $\phi : X\rightarrow \mathbb{K}$ we have $\phi |_{X_0} \equiv0 \implies \phi \equiv 0$

My attempt

"$\implies$"

Assume, that there exists $x\in X\setminus X_0$ such that $\phi(x) \ne 0$

$X_0$ is dense in $X$ so for any $\epsilon>0$ I can find $y\in X_0$ such that $\|x-y\|<\epsilon$. So I can find $y\in X_0$ arbitrarly close to $x$ and $\phi(y)=0$. But $\phi$ is continuous. Contradiction.

I have no idea about the second part...

luka5z
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2 Answers2

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Suppose that there is some non-empty open set $U$ such that $U \cap \overline{X_0} = \emptyset$. We can take $U$ to be of the form $U=B(\hat{x},\epsilon)$ for some $\epsilon >0$. Since $X_0 $ is a subspace, we have $\hat{x} \ne 0$.

The Hahn Banach theorem gives the existence of some continuous functional $\phi$ and some $\alpha$ such that $\operatorname{re}\phi(u) < \alpha \le \operatorname{re} \phi(y)$ for all $u \in U, y \in \overline{X_0}$.

Use the fact that $X_0$ is a subspace and the form of $U$ to conclude that $\phi(y) = 0$ for all $y \in X_0$, but $\phi(\hat{x}) < 0$.

Addendum: Proof using the version of Hahn Banach in the comments below.

Let $\hat{x}$ be as above and let $W = \{ \lambda \hat{x} + y | \lambda \in \mathbb{K}, y \in \overline{X_0} \}$. Note that any $w \in W$ has a unique representation $w = \lambda \hat{x} + y$.

Define $f: W \to \mathbb{K}$ by $f(\lambda \hat{x} + y ) = \lambda$. Note that $\ker f = \overline{X_0}$ is closed (in $W$), hence $f$ is continuous (as a map $W \to \mathbb{K}$), and $f(\hat{x}) = 1$.

The theorem referred to in the comments lets us extend $f$ to $\tilde{f}: X \to \mathbb{K}$ without increasing the norm (in particular, $\tilde{f} \in X^*$, which is what we care about here).

Hence we have a functional $\tilde{f}$ which is zero on $\overline{X_0}$, but non-zero on $X$.

copper.hat
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  • can you elaborate on the part with Hahn Banach? Why we have this inequality? – luka5z May 28 '14 at 19:06
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    Is this what you are asking about: http://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem#Hahn.E2.80.93Banach_Separation_Theorem ? – copper.hat May 28 '14 at 19:52
  • we didn't have this theorem on our classes. is it possible to prove this without that theorem? – luka5z May 28 '14 at 20:10
  • Do you have some version of the Hahn Banach theorem? – copper.hat May 28 '14 at 20:16
  • Yes, this one: http://www.maths.manchester.ac.uk/~nikita/31002/hahn-banach.pdf – luka5z May 28 '14 at 20:19
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    I added some notes relating the above to your version. Note that the only issue in applying the theorem is to show that the $f$ constructed above is continuous. – copper.hat May 28 '14 at 22:01
  • @copper.hat, Can you explain the fact that $X_0$ is a subspace and the form of $U$ to conculde that $\phi(y)=0$ for all $y \in X_0$? – User124356 May 04 '21 at 22:16
  • $X_0$ is a subspace by assumption. If $\phi(y) \neq 0$ then by taking a scalar multiple of $y$ (since $X_0$ is a subspace) we have $\phi(t y) = t \phi(y)$ and so the range of $\phi$ over $X_0$ is the entire scalar field. This would contradict the form of $\phi$ given by the Hahn Banach theorem, hence we must have $\phi(y) = 0$ for $y \in X_0$. – copper.hat May 05 '21 at 00:34
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The reverse part follows from a corollary of Hahn Banach Theorem: Assuming the second part, we need to prove $\bar{Y}$=X.

Suppose on contrary: $\bar{Y}$ $\subset$ X.

Then there exist a $x_0$ $\in$ X $\backslash \bar{Y}$. Observe that $\bar{Y}$ is a closed subspace of X. Then by HBT, there exists f $\in X^*$ such that f($x_0$)=d($x_0$,$\bar{Y}$) and f(y)=0, $\forall$ y $\in$ $\bar{Y}$. Now that we have f $\equiv$ 0 on X $\implies$ d($x_0$,$\bar{Y}$)=0, which clearly implies $x_0$ $\in$ $\bar{Y}$. A contradiction!

Krishna Kumar M.V.
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