diff --git a/docs/en/mathematics/functional-analysis/normed-spaces/compactness.md b/docs/en/mathematics/functional-analysis/normed-spaces/compactness.md index e69de29..1cf5f59 100644 --- a/docs/en/mathematics/functional-analysis/normed-spaces/compactness.md +++ b/docs/en/mathematics/functional-analysis/normed-spaces/compactness.md @@ -0,0 +1,60 @@ +# Compactness + +> *Definition 1*: a metric space $X$ is **compact** if every sequence in $X$ has a convergent subsequence. A subset $M$ of $X$ is compact if every sequence in $M$ has a convergent subsequence whose limit is an element of $M$. + +A general property of compact sets is expressed in the following proposition. + +> *Proposition 1*: a compact subset $M$ of a metric space $(X,d)$ is closed and bounded. + +??? note "*Proof*:" + + Will be added later. + +The converse of this proposition is generally false. + +??? note "*Proof*:" + + Will be added later. + +However, for a finite dimensional normed space we have the following proposition. + +> *Proposition 2*: in a finite dimensional normed space $(X, \|\cdot\|)$ a subset $M \subset X$ is compact if and only if $M$ is closed and bounded. + +??? note "*Proof*:" + + Will be added later. + +A source of interesting results is the following lemma. + +> *Lemma 1*: let $Y$ and $Z$ be subspaces of a normed space $(X, \|\cdot\|)$, suppose that $Y$ is closed and that $Y$ is a strict subset of $Z$. Then for every $\alpha \in (0,1)$ there exists a $z \in Z$, such that +> +> 1. $\|z\| = 1$, +> 2. $\forall y \in Y: \|z - y\| \geq \alpha$. + +??? note "*Proof*:" + + Will be added later. + +Lemma 1 gives the following remarkable proposition. + +> *Proposition 3*: if a normed space $(X, \|\cdot\|)$ has the property that the closed unit ball $M = \{x \in X | \|x\| \leq 1\}$ is compact, then $X$ is finite dimensional. + +??? note "*Proof*:" + + Will be added later. + +Compact sets have several basic properties similar to those of finite sets and not shared by non-compact sets. Such as the following. + +> *Proposition 4*: let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and let $T: X \to Y$ be a continuous mapping. Let $M$ be a compact subset of $(X,d_X)$, then $T(M)$ is a compact subset of $(Y,d_Y)$. + +??? note "*Proof*:" + + Will be added later. + +From this proposition we conclude that the following property carries over to metric spaces. + +> *Corollary 1*: let $M \subset X$ be a compact subset of a metric space $(X,d)$ over a field $F$, a continuous mapping $T: M \to F$ attains a maximum and minimum value. + +??? note "*Proof*:" + + Will be added later. \ No newline at end of file diff --git a/docs/en/mathematics/functional-analysis/normed-spaces/linear-operators.md b/docs/en/mathematics/functional-analysis/normed-spaces/linear-operators.md index e69de29..38c5e5a 100644 --- a/docs/en/mathematics/functional-analysis/normed-spaces/linear-operators.md +++ b/docs/en/mathematics/functional-analysis/normed-spaces/linear-operators.md @@ -0,0 +1,2 @@ +# Linear operators + diff --git a/docs/en/mathematics/functional-analysis/normed-spaces/normed-spaces.md b/docs/en/mathematics/functional-analysis/normed-spaces/normed-spaces.md index e69de29..a011836 100644 --- a/docs/en/mathematics/functional-analysis/normed-spaces/normed-spaces.md +++ b/docs/en/mathematics/functional-analysis/normed-spaces/normed-spaces.md @@ -0,0 +1,198 @@ +# 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*. + +> *Definition 2*: 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**. + +Furthermore, there is a category of normed spaces with interesting properties which is given in the following definition. + +> *Definition 3*: 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 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 +> +> $$ +> s = \lim_{n \to \infty} \sum_{k=1}^n x_k = \sum_{k=1}^\infty x_k = \lim_{n \to \infty } s_n. +> $$ +> +> 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 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 +> +> $$ +> \lim_{n \to \infty} \|x - \sum_{k=1}^n \alpha_k e_k\| = 0, +> $$ +> +> 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 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. \ No newline at end of file