NEVER MIND, FIGURED THEM OUT(adsbygoogle = window.adsbygoogle || []).push({});

Definitions

(All vector spaces are over the complex field)

If [itex]\mathcal{M}[/itex] is a subspace of a normed vector space

[tex](\mathcal{X}, ||.||_{\mathcal{X}})[/tex]

then

[tex]||x + \mathcal{M}||_{\mathcal{X}/\mathcal{M}} =_{def} \mbox{inf} _{m \in \mathcal{M}}||x + m||[/tex]

defines thequotient normon the quotient space [itex]\mathcal{X}/\mathcal{M}[/itex]. We will omit the subscripts on the norms because in the following, it will always be clear as to whether we're looking at the norm of a vector in the original space or the quotient space.

A topological space isseparableiff it has a countable dense subset.

If [itex]\mathcal{X}[/itex] is a normed vector space and f is a linear functional on this space, we define:

[tex]||f||_* =_{def} \mbox{sup} _{||x|| = 1}|f(x)|[/tex]

If [itex]||f||_*[/itex] is finite, we say that f isboundedand we write [itex]\mathcal{X}^*[/itex] to denote the set of bounded linear functionals. We call it thedualof [itex]\mathcal{X}[/itex]. [itex]||.||_*[/itex] defines a norm on the set of bounded linear functionals, and as before, we will omit the subscript [itex]_*[/itex] when there is no chance of confusion.

Having given defined the vector space [itex]\mathcal{X}^*[/itex] and given it a norm, we can apply the above definitions to [itex]\mathcal{X}^*[/itex] itself to get a normed vector space [itex](\mathcal{X}^*)^*[/itex]. This "double dual" is always a complete space. If [itex]x \in \mathcal{X}[/itex], define [itex]\hat{x} : \mathcal{X}^* \to \mathbb{C}[/itex] by [itex]\hat{x}(f) = f(x)[/itex]. The map [itex]x \mapsto \hat{x}[/itex] is a linear, norm-preserving map from [itex]\mathcal{X}[/itex] into [itex](\mathcal{X}^*)^*[/itex]

Useful fact

If [itex]\mathcal{X}[/itex] is Banach, the range of [itex]x \mapsto \hat{x}[/itex] is closed in [itex](\mathcal{X}^*)^*[/itex].

Problems

1. If [itex]\mathcal{M}[/itex] is a proper closed subspace, prove that for any [itex]\epsilon > 0[/itex] there exists [itex]x \in \mathcal{X}[/itex] such that [itex]||x|| = 1[/itex] and

[tex]||x+\mathcal{M}|| \geq 1 - \epsilon[/tex]

2. If [itex]\mathcal{M}[/itex] is a finite-dimensional subspace prove there is a (topologically) closed subspace [itex]\mathcal{N}[/itex] such that

i) [itex]\mathcal{M} \cap \mathcal{N} = \{ 0\} [/itex] and

ii) [itex]\{ m + n : m \in \mathcal{M},\ n \in \mathcal{N} \} = \mathcal{X}[/itex]

3. [itex]\mathcal{X}[/itex] is a Banach and [itex]\mathcal{X}^*[/itex] is separable. Prove that [itex]\mathcal{X}[/itex] is separable too.

Attempts

1. This is actually part b) of a question. Part a) simply asked to prove that the quotient norm defined above is indeed a norm. I don't know where to go from here though.

2. I've defined [itex]\mathcal{N} = \{ x \in \mathcal{X} : ||x|| = ||x + \mathcal{M}|| \}[/itex]. I can prove that this is a (topologically) closed set satisfying i) and ii) which contains 0 and is closed under scalar multiplication. I am having trouble showing that it's closed under addition. Alternatively, I've considered [itex]\mbox{Span}(\mathcal{N})[/itex]. This is clearly a subspace and satisfies ii) by virtue of the fact that [itex]\mathcal{N}[/itex] does. I am having trouble showing that this satisfies i), and I haven't tried showing that it is closed but I suspect that the span of a closed set is closed.

3. There's a hint given

Let [itex]\{ f_n\} _1 ^{\infty}[/itex] be a countable dense subset of [itex]\mathcal{X}^*[/itex]. For each n, choose [itex]x_n \in \mathcal{X}[/itex] with [itex]||x_n|| = 1[/itex] and [itex]|f_n(x_n)| \geq \frac{1}{2}||f_n||[/itex]. Prove that the linear combinations of [itex]\{ x_n\} _1 ^{\infty}[/itex] are dense in [itex]\mathcal{X}[/itex].

This hint needs a minor justification. The set of [itex]\mathbb{C}[/itex]-linear combinations of the x_{n}is not a countable set, and hence does not witness the separability of [itex]\mathcal{X}[/itex]. However, the set of [itex](\mathbb{Q} + i\mathbb{Q})[/itex]-linear combinations is countable. It seems to me that this countable set of linear combinations is dense iff the set of [itex]\mathbb{C}[/itex]-linear combinations is dense, and this is why it's sufficient to prove that the set of [itex]\mathbb{C}[/itex]-linear combinations is dense.

I have no problem proving the existence of [itex]x_n \in \mathcal{X}[/itex] with [itex]||x_n|| = 1[/itex] and [itex]|f_n(x_n)| \geq \frac{1}{2}||f_n||[/itex]. I have no idea why their span would be dense though, here is where I need help.

Alternatively, I know that the spaces involved are separable iff they are second countable (i.e. they have a countable base). Could this lead to a simpler proof? As yet another alternative, maybe there's a way to make use of the "useful fact" given between the definitions and problem statements. Any suggestions?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Normed Vector Spaces

**Physics Forums | Science Articles, Homework Help, Discussion**