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Background Information:

(Folland)Theorem 5.8 - Let $\mathcal{X}$ be a normed vector space.

a.) If $M$ is a closed subspace of $\mathcal{X}$ and $x\in \mathcal{X}\setminus M$, there exists $f\in\mathcal{X}^*$ such that $f(x)\neq 0$ and $f(M) = \{0\}$. In fact, if $\delta = \inf_{y\in M}\|x - y\|$, $f$ can be taken to satisfy $\|f\| = 1$ and $f(x) = \delta$.

Question:

Let $\mathcal{X}$ be a normed vector space.

a.) If $M$ is a closed subspace and $x\in\mathcal{X}\setminus M$ then $M + \mathbb{C}x$ is closed. (Use Theorem 5.8a.)).

b.) Every finite-dimensional subspace of $\mathcal{X}$ is closed.

Proof a.) - Let $M$ be a proper closed subspace of $X$ and let $x\in X\setminus M$. There existts $f\in X^*$ such that $f(x)\neq 0$ and $f(M) = \{0\}$. Let $\{u_n + a_nx \}_{1}^{\infty}$ be a sequence in $M + Kx$ that converges to $y\in X$. Then $$f(y) = \lim\limits_{n\rightarrow \infty}f(u_n + a_n x) = \lim\limits_{n\rightarrow \infty} a_nf(x)$$ since $f$ is continuous, so $\{a_n\}_{1}^{\infty}$ converges to $a:= f(y)/f(x)$, which implies that $\{a_nx \}_{1}^{\infty}$ converges to $ax$. Therefore $$\{u_n\}_{1}^{\infty} = \{(u_n + a_nx) - a_nx\}_{1}^{\infty} \to (y - ax)$$ which lies in $M$ because $M$ is closed. It follows that $y\in M + Kx$, which shows that $M + Kx$ is closed.

Proof b.) - I have spent a considerable amount of time thinking about this but I am not sure how to proceed or how to show this result. Any hints or suggestions are greatly appreciated.

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Wolfy
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1 Answers1

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We already know (Background Information):

(Folland)Theorem 5.8 - Let $\mathcal{X}$ be a normed vector space.

a.) If $M$ is a closed subspace of $\mathcal{X}$ and $x\in \mathcal{X}\setminus M$, there exists $f\in\mathcal{X}^*$ such that $f(x)\neq 0$ and $f(M) = \{0\}$. In fact, if $\delta = \inf_{y\in M}\|x - y\|$, $f$ can be taken to satisfy $\|f\| = 1$ and $f(x) = \delta$.

Question:

Let $X$ be a normed vector space.

a.) If $M$ is a closed subspace and $x\in X \setminus M$ then $M + \mathbb{C}x$ is closed. (Use Theorem 5.8a.)).

b.) Every finite-dimensional subspace of $X$ is closed.

Proof a.) - Let $M$ be a proper closed subspace of $X$ and let $x\in X\setminus M$. There exists $f\in X^*$ such that $f(x)\neq 0$ and $f(M) = \{0\}$. Let $\{u_n + a_nx \}_{1}^{\infty}$ be a sequence in $M + Kx$ that converges to $y\in X$. Then $$f(y) = f(\lim_{n\rightarrow \infty} (u_n + a_n x))=\lim\limits_{n\rightarrow \infty}f(u_n + a_n x) = \lim\limits_{n\rightarrow \infty} a_nf(x)$$ since $f$ is continuous, so $\{a_n\}_{1}^{\infty}$ converges to $a:= f(y)/f(x)$, which implies that $\{a_nx \}_{1}^{\infty}$ converges to $ax$. Therefore $$\{u_n\}_{1}^{\infty} = \{(u_n + a_nx) - a_nx\}_{1}^{\infty} \to (y - ax)$$ which lies in $M$ because $M$ is closed. It follows that $y\in M + Kx$, which shows that $M + Kx$ is closed.

Proof b.) - Let $Y$ be a finite dimensional subspace of $X$. Let $\{e_1, \ldots, e_n\}$ be a basis of $Y$. Now, note that $M=\{0\}$ is a closed subspace of $X$. So, by item a.), $M + \mathbb{C}e_1$ is a closed subspace of $X$. Again, by item a.) we have that $M + \mathbb{C}e_1+\mathbb{C}e_2$ is a closed subspace of $X$. Repeating the argument enough times, we have that $M + \mathbb{C}e_1+\ldots+\mathbb{C}e_n$ is a closed subspace of $X$. But $M + \mathbb{C}e_1+\ldots+\mathbb{C}e_n=Y$. So we have proved that $Y$ is a closed subspace of $X$.

Ramiro
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    I see for a.) did we already know that $f$ is continuous? – Wolfy Aug 05 '16 at 20:02
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    @Wolfi, Yes, for a.) we know $f$ is a continuous linear functional. The existence of $f$ in a.) comes from the Hahn-Banach Theorem. – Ramiro Aug 06 '16 at 03:27
  • f is continuous because the function which is extended by Hahn-Banach Theorem to provide f have its kern given by M, which is closed, see https://math.stackexchange.com/questions/133678/showing-that-ker-t-is-closed-if-and-only-if-t-is-continuous . – R. W. Prado Mar 17 '21 at 22:47
  • @RenanWillianPrado , A simpler argument: the function which is extended by Hahn-Banach Theorem to provide $f$ is defined in $M$ and it is identically zero, so it is continuous. – Ramiro Mar 17 '21 at 23:06
  • I'm not sure how get $f$, which does not vanish in x, from a function that vanish everywhere. If possible, I would suggest you make the details for everyone. – R. W. Prado Mar 17 '21 at 23:54
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    @RenanWillianPrado , It is a well-known corollary of Hahn-Banach Theorem that if $M$ is a proper closed subspace of $X$ and $x\in X\setminus M$, then there exists $f\in X^$ such that $f(x)\neq 0$ and $f(M) = {0}$. Of course, $f\in X^$ means that $f$ continuous (and linear). – Ramiro Mar 18 '21 at 00:04
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    @RenanWillianPrado I see what you mean. In the proof of the corollary I mentioned, we use an "intermediate" linear function $g$ defined in span($M$, x). Such function has kernel $M$ and $g(x) \neq 0$. Such function $g$ is continuous and the simplest way to see it is continuous is because its kernel is closed. I guess it was to this intermediate function that you were talking about. I am used to apply the corollary directly. – Ramiro Mar 18 '21 at 00:22
  • That's it, Ramiro! =) The existence of a continuous $f$ is not quite obvious for a beginner in functional analysis. Hahn-Banach theorem is quite strong! That's why I like to apply it directly instead of its corollaries. – R. W. Prado Mar 18 '21 at 00:28
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    @RenanWillianPrado , I understand. But it is rather subjective. In several situations, it is more direct and practical to apply the corollaries, instead of applying the theorem in all its generality. In fact, that is why it is important to know the corollaries (as well as the theorem itself). – Ramiro Mar 18 '21 at 03:14
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    @RenanWillianPrado , It is worth mentioning that the question in the post is an exercise in Folland's book. This exercise that comes after Theorem 5.8. and the OP has explicitly given the Theorem 5.8 as known (background information). The question is supposed to be answered using the Theorem. – Ramiro Mar 18 '21 at 11:38
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    Your explanation and reasons are pretty good. You don't have to explain more. I hope my comment don't confuse anyone. I thought it was meant to use Hahn-Banach Theorem directly. Whoever is reading this comments try to understand what he said in the answer first before reading these long comments. =) – R. W. Prado Mar 18 '21 at 17:15