# Normed Vector Spaces

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NEVER MIND, FIGURED THEM OUT

Definitions

(All vector spaces are over the complex field)

If $\mathcal{M}$ is a subspace of a normed vector space

$$(\mathcal{X}, ||.||_{\mathcal{X}})$$

then

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

defines the quotient norm on the quotient space $\mathcal{X}/\mathcal{M}$. 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 $\mathcal{X}$ is a normed vector space and f is a linear functional on this space, we define:

$$||f||_* =_{def} \mbox{sup} _{||x|| = 1}|f(x)|$$

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

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

Useful fact

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

Problems

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

$$||x+\mathcal{M}|| \geq 1 - \epsilon$$

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

i) $\mathcal{M} \cap \mathcal{N} = \{ 0\}$ and
ii) $\{ m + n : m \in \mathcal{M},\ n \in \mathcal{N} \} = \mathcal{X}$

3. $\mathcal{X}$ is a Banach and $\mathcal{X}^*$ is separable. Prove that $\mathcal{X}$ 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 $\mathcal{N} = \{ x \in \mathcal{X} : ||x|| = ||x + \mathcal{M}|| \}$. 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 $\mbox{Span}(\mathcal{N})$. This is clearly a subspace and satisfies ii) by virtue of the fact that $\mathcal{N}$ 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 $\{ f_n\} _1 ^{\infty}$ be a countable dense subset of $\mathcal{X}^*$. For each n, choose $x_n \in \mathcal{X}$ with $||x_n|| = 1$ and $|f_n(x_n)| \geq \frac{1}{2}||f_n||$. Prove that the linear combinations of $\{ x_n\} _1 ^{\infty}$ are dense in $\mathcal{X}$.

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

I have no problem proving the existence of $x_n \in \mathcal{X}$ with $||x_n|| = 1$ and $|f_n(x_n)| \geq \frac{1}{2}||f_n||$. 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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StatusX
Homework Helper
It's usually helpful to have some nice simple example in mind, even if it has features that aren't present in the general case. If you can translate all the steps you do in the simple example into the formal, general language, then you're done.

So for the first one, think of a nice inner product space. Then ||x+M|| is just the length of the projection of x onto $M^\perp$, and what you want to find is a unit vector that's (almost) in $M^\perp$. Where can you get this? In an inner product space, we can always decompose an arbitrary vector into components in a subspace M and its complement $M^\perp$. So, pick an x which isn't in M, which will have ||x+M||=a>0 since M is closed. If there was a v in M with ||x-v||=a, x-v would be exactly the projection of x onto $M^\perp$. We don't know such a v exists, but for any e>0, there is a v in M with ||x-v||<a+e, and so this roughly means x-v is nearly in $M^\perp$. So you should be able to show y=(x-v)/||x-v|| has ||y+M|| nearly 1.

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Homework Helper
Thanks StatusX, that worked. And we figured out the rest.