notes/docs/mathematics/functional-analysis/normed-spaces/normed-spaces.md

Normed spaces

Definition 1: a vector space X is a normed space if a norm \| \cdot \|: X \to F is defined on X, satisfying

1. \forall x \in X: \|x\| \geq 0,
2. \|x\| = 0 \iff x = 0,
3. \forall x \in X, \alpha \in F: \|\alpha x\| = |\alpha| \|x\|,
4. \forall x, y \in X: \|x + y\| \leq \|x\| + \|y\|.

Also called a normed vector space or normed linear space.

Proposition 1: a norm on a vector space X defines a metric d on X given by

d(x,y) = |x - y|,

for all x, y \in X and is called a metric induced by the norm.

??? note "Proof:"

Will be added later.

Furthermore, there is a category of normed spaces with interesting properties which is given in the following definition.

Definition 2: a Banach space is a complete normed space with its metric induced by the norm.

If we define the norm \| \cdot \| of the Euclidean vector space \mathbb{R}^n by

|x| = \sqrt{\sum_{j=1}^n |x(j)|^2},

for all x \in \mathbb{R}^n, then it yields the metric

d(x,y) = |x - y| = \sqrt{\sum_{j=1}^n |x(j) - y(j)|^2},

for all x, y \in \mathbb{R}^n which imposes completeness. Therefore (\mathbb{R}^n, \|\cdot\|) is a Banach space.

This adaptation also works for C, l^p and l^\infty, obviously. Obtaining that \mathbb{R}^n, C, l^p and l^\infty are all Banach spaces.

Lemma 1: a metric d induced by a norm on a normed space (X, \|\cdot\|) satisfies

1. \forall x, y \in X, \alpha \in F: d(x + \alpha, y + \alpha) = d(x,y),
2. \forall x, y \in X, \alpha \in F: d(\alpha x, \alpha y) = |\alpha| d(x,y).

??? note "Proof:"

We have

$$
    d(x + \alpha, y + \alpha) = \|x + \alpha - (y + \alpha)\| = \|x - y\| = d(x,y),
$$

and

$$
    d(\alpha x, \alpha y) = \|\alpha x - \alpha y\| = |\alpha| \|x - y\| = |\alpha| d(x,y).
$$

By definition, a subspace M of a normed space X is a subspace of X with its norm induced by the norm on X.

Definition 3: let M be a subspace of a normed space X, if M is closed then M is a closed subspace of X.

By definition, a subspace M of a Banach space X is a subspace of X as a normed space. Hence, we do not require M to be complete.

Theorem 1: a subspace M of a Banach space X is complete if and only if M is a closed subspace of X.

??? note "Proof:"

Will be added later.

Convergence in normed spaces follows from the definition of convergence in metric spaces and the fact that the metric is induced by the norm.

Convergent series

Definition 4: let (x_k)_{k \in \mathbb{N}} be a sequence in a normed space (X, \|\cdot\|). We define the sequence of partial sums (s_n)_{n \in \mathbb{N}} by

s_n = \sum_{k=1}^n x_k,

if s_n converges to s \in X, then

\lim_{n \to \infty} \sum_{k=1}^n x_k,

is convergent, and s is the sum of the series, writing

s = \lim_{n \to \infty} \sum_{k=1}^n x_k = \sum_{k=1}^\infty x_k = \lim_{n \to \infty } s_n.

If the series

\sum_{k=1}^\infty |x_k|,

is convergent in F, then the series is absolutely convergent.

From the notion of absolute convergence the following theorem may be posed.

Theorem 2: absolute convergence of a series implies convergence if and only if (X, \|\cdot\|) is complete.

??? note "Proof:"

Will be added later.

Schauder basis

Definition 5: let (X, \|\cdot\|) be a normed space and let (e_k)_{k \in \mathbb{N}} be a sequence of vectors in X, such that for every x \in X there exists a unique sequence of scalars (\alpha_k)_{k \in \mathbb{N}} such that

\lim_{n \to \infty} |x - \sum_{k=1}^n \alpha_k e_k| = 0,

then (e_k)_{k \in \mathbb{N}} is a *Schauder basis of (X, \|\cdot\|).

The expansion of a x \in X with respect to a Schauder basis (e_k)_{k \in \mathbb{N}} is given by

x = \sum_{k=1}^\infty \alpha_k e_k.

Lemma 2: if a normed space has a Schauder basis then it is seperable.

??? note "Proof:"

Will be added later.

Completion

Theorem 3: for every normed space (X, \|\cdot\|_X) there exists a Banach space (Y, \|\cdot\|_Y) that contains a subspace W that satisfies the following conditions

1. W is a normed space isometric with X.
2. W is dense in Y.

??? note "Proof:"

Will be added later.

The Banach space (Y, \|\cdot\|_Y) is unique up to isometry.

Finite dimension

Lemma 3: let \{x_k\}_{k=1}^n with n \in \mathbb{N} be a linearly independent set of vectors in a normed space (X, \|\cdot\|), then there exists a c > 0 such that

\Big| \sum_{k=1}^n \alpha_k x_k \Big| \geq c \sum_{k=1}^n |\alpha_k|,

for all \{\alpha_k\}_{k=1}^n \in F.

??? note "Proof:"

Will be added later.

As a first application of this lemma, let us prove the following.

Theorem 4: every finite-dimensional subspace M of a normed space (X, \|\cdot\|) is complete.

??? note "Proof:"

Will be added later.

In particular, every finite dimensional normed space is complete.

Proposition 2: every finite-dimensional subspace M of a normed space (X, \|\cdot\|) is a closed subspace of X.

??? note "Proof:"

Will be added later.

Another interesting property of finite-dimensional vector space X is that all norms on X lead to the same topology for X. That is, the open subsets of X are the same, regardless of the particular choice of a norm on X. The details are as follows.

Definition 6: a norm \|\cdot\|_1 on a vector space X is equivalent to a norm \|\cdot\|_2 on X if there exists a,b>0 such that

\forall x \in X: a |x|_1 \leq |x|_2 \leq b |x|_1.

This concept is motivated by the following proposition.

Proposition 3: equivalent norms on X define the same topology for X.

??? note "Proof:"

Will be added later.

Using lemma 3 we may now prove the following theorem.

Theorem 5: on a finite dimensional vector space X any norm \|\cdot\|_1 is equivalent to any other norm \|\cdot\|_2.

??? note "Proof:"

Will be added later.

This theorem is of considerable importance. For instance, it implies that convergence or divergence of a sequence in a finite dimensional vector space does not depend on the particular choice of a norm on that space.