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 4*: 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 5*: 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
> 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 6*: 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
> 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 1*: 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 7*: 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 2*: 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.
