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NEVER MIND, FIGURED THEM OUT

(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 the

A topological space is

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 is

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]

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

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.

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

This hint needs a minor justification. The set of [itex]\mathbb{C}[/itex]-linear combinations of the x

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?

**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 the

**quotient norm**on 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 is

**separable**iff 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 is

**bounded**and we write [itex]\mathcal{X}^*[/itex] to denote the set of bounded linear functionals. We call it the**dual**of [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?

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