From ec3ad8cb31f292452fb95463c7f9f8911c59b553 Mon Sep 17 00:00:00 2001 From: Luc Date: Sun, 4 Aug 2024 22:01:11 +0200 Subject: [PATCH] Added the sections convergence, completeness and completion to functional analysis. --- config/en/mkdocs.yaml | 3 +- .../metric-spaces/completeness.md | 173 ++++++++++++++++++ .../metric-spaces/completion.md | 20 ++ .../metric-spaces/convergence.md | 59 ++++++ 4 files changed, 254 insertions(+), 1 deletion(-) create mode 100644 docs/en/mathematics/functional-analysis/metric-spaces/completion.md diff --git a/config/en/mkdocs.yaml b/config/en/mkdocs.yaml index 8bf4268..d56a48f 100755 --- a/config/en/mkdocs.yaml +++ b/config/en/mkdocs.yaml @@ -102,7 +102,8 @@ nav: - 'Definition': mathematics/functional-analysis/metric-spaces/definition.md - 'Topological notions': mathematics/functional-analysis/metric-spaces/topological-notions.md - 'Convergence': mathematics/functional-analysis/metric-spaces/convergence.md - - 'Completeness'mathematics/functional-analysis/metric-spaces/completeness.md + - 'Completeness': mathematics/functional-analysis/metric-spaces/completeness.md + - 'Completion': mathematics/functional-analysis/metric-spaces/completion.md - 'Normed spaces': - 'Vector spaces': mathematics/functional-analysis/normed-spaces/vector-spaces.md - 'Definition': mathematics/functional-analysis/normed-spaces/definition.md diff --git a/docs/en/mathematics/functional-analysis/metric-spaces/completeness.md b/docs/en/mathematics/functional-analysis/metric-spaces/completeness.md index e69de29..cc60456 100644 --- a/docs/en/mathematics/functional-analysis/metric-spaces/completeness.md +++ b/docs/en/mathematics/functional-analysis/metric-spaces/completeness.md @@ -0,0 +1,173 @@ +# Completeness + +> *Definition 1*: a sequence $(x_n)_{n \in \mathbb{N}}$ in a metric space $(X,d)$ is a **Cauchy sequence** if +> +> $$ +> \forall \varepsilon > 0 \exists N \in \mathbb{N} \forall n,m > N: \quad d(x_n, x_m) < \varepsilon. +> $$ + +A convergent sequence $(x_n)_{n \in \mathbb{N}}$ in a metric space $(X,d)$ is always a Cauchy sequence since + +$$ + \forall \varepsilon > 0 \exists N \in \mathbb{N}: \quad d(x_n, x) < \frac{\varepsilon}{2}, +$$ + +for all $n > N$. By axiom 4 of the definition of a metric space we have for $m, n > N$ + +$$ + d(x_m, x_n) \leq d(x_m, x) + d(x, x_n) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon, +$$ + +showing that $(x_n)$ is Cauchy. + +> *Definition 2*: a metric space $(X,d)$ is **complete** if every Cauchy sequence in $X$ is convergent. + +Therefore, in a complete metric space every Cauchy sequence is a convergent sequence. + +> *Proposition 1*: let $M \subset X$ be a nonempty subset of a metric space $(X,d)$ and let $\overline M$ be the closure of $M$, then +> +> 1. $x \in \overline M \iff \exists (x_n)_{n \in \mathbb{N}} \text{ in } M: x_n \to x$, +> 2. $M \text{ is closed } \iff M = \overline M$. + +??? note "*Proof*:" + + To prove statement 1, let $x \in \overline M$. If $x \notin M$ then $x$ is an accumulation point of $M$. Hence, for each $n \in \mathbb{N}$ the ball $B(x,\frac{1}{n})$ contains an $x_n \in M$ and $x_n \to x$ since $\frac{1}{n} \to 0$ as $n \to \infty$. Conversely, if $(x_n)_{n \in \mathbb{N}}$ is in $M$ and $x_n \to x$, then $x \in M$ or every neighbourhood of $x$ contains points $x_n \neq x$, so that $x$ is an accumulation point of $M$. Hence $x \in \overline M$. + + Statement 2 follows from statement 1. + +We have that the following statement is equivalent to statement 2: $x_n \in M: x_n \to x \implies x \in M$. + +> *Proposition 2*: let $M \subset X$ be a subset of a complete metric space $(X,d)$, then +> +> $$ +> M \text{ is complete} \iff M \text{ is a closed subset of } X +> $$ + +??? note "*Proof*:" + + Let $M$ be complete, by proposition 1 statement 1 we have that + + $$ + \forall x \in \overline M \exists (x_n)_{n \in \mathbb{N}} \text{ in } M: x_n \to x. + $$ + + Since $(x_n)$ is Cauchy and $M$ is complete, $x_n$ converges in $M$ with the limit being unique by statement 1 in [lemma 1](). Hence, $x \in M$ which proves that $M$ is closed because $x \in \overline M$ has been chosen arbitrary. + + Conversely, let $M$ be closed and $(x_n)$ Cauchy in $M$. Then $x_n \to x \in X$ which implies that $x \in \overline M$ by statement 1 in proposition 1, and $x \in M$ since $M = \overline M$ by assumption. Hence, the arbitrary Cauchy sequence $(x_n)$ converges in $M$. + +> *Proposition 3*: let $T: X \to Y$ be a map from a metric space $(X,d)$ to a metric space $(Y,\tilde d)$, then +> +> $$ +> T \text{ is continuous in } x_0 \in X \iff x_n \to x_0 \implies T(x_n) \to T(x_0), +> $$ +> +> for any sequence $(x_n)_{n \in \mathbb{N}}$ in $X$ as $n \to \infty$. + +??? note "*Proof*:" + + Suppose $T$ is continuous at $x_0$, then for a given $\varepsilon > 0$ there is a $\delta > 0$ such that + + $$ + \forall \varepsilon > 0 \exists \delta > 0: \quad d(x, x_0) < \delta \implies \tilde d(Tx, Tx_0) < \varepsilon. + $$ + + Let $x_n \to x_0$ then + + $$ + \exists N \in \mathbb{N} \forall n > N: \quad d(x_n, x_0) < \delta. + $$ + + Hence, + + $$ + \forall n > N: \tilde d(Tx_n, Tx_0) < \varepsilon. + $$ + + Which means that $T(x_n) \to T(x_0)$. + + Conversely, suppose that $x_n \to x_0 \implies T(x_n) \to T(x_0)$ and $T$ is not continuous. Then + + $$ + \exists \varepsilon > 0: \forall \delta > 0 \exists x \neq x_0: \quad d(x, x_0) < \delta \quad \text{ however } \quad \tilde d(Tx, Tx_0) \geq \varepsilon, + $$ + + in particular, for $\delta = \frac{1}{n}$ there is a $x_n$ satisfying + + $$ + d(x_n, x_0) < \frac{1}{n} \quad \text{ however } \quad \tilde d(Tx_n, Tx_0) \geq \varepsilon, + $$ + + Clearly $x_n \to x_0$ but $(Tx_n)$ does not converge to $Tx_0$ which contradicts $Tx_n \to Tx_0$. + +## Completeness proofs + +To show that a metric space $(X,d)$ is complete, one has to show that every Cauchy sequence in $(X,d)$ has a limit in $X$. This depends explicitly on the metric on $X$. + +The steps in a completeness proof are as follows + +1. take an arbitrary Cauchy sequence $(x_n)_{n \in \mathbb{N}}$ in $(X,d)$, +2. construct for this sequence a candidate limit $x$, +3. prove that $x \in X$, +4. prove that $x_n \to x$ with respect to metric $d$. + +> *Proposition 4*: the Euclidean space $\mathbb{R}^n$ with $n \in \mathbb{N}$ and the metric $d$ defined by +> +> $$ +> d(x,y) = \sqrt{\sum_{j=1}^n \big(x(j) - y(j) \big)^2}, +> $$ +> +> for all $x,y \in \mathbb{R}^n$ is complete. + +??? note "*Proof*:" + + Will be added later. + +A similar proof exists for the completeness of the Unitary space $\mathbb{C}^n$. + +> *Proposition 5*: the space $C([a,b])$ of all **real-valued continuous functions** on a closed interval $[a,b]$ with $a +> $$ +> d(x,y) = \max_{t \in [a,b]} |x(t) - y(t)|, +> $$ +> +> for all $x, y \in C$ is complete. + +??? note "*Proof*:" + + Will be added later. + +While $C$ with a metric $d$ defined by + +$$ + d(x,y) = \int_a^b |x(t) - y(t)| dt, +$$ + +for all $x,y \in C$ is incomplete. + +??? note "*Proof*:" + + Will be added later. + +> *Proposition 6*: the space $l^p$ with $p \geq 1$ and the metric $d$ defined by +> +> $$ +> d(x,y) = (\sum_{j \in \mathbb{N}} | x(j) - y(j) |^p)^\frac{1}{p}, +> $$ +> +> for all $x,y \in l^p$ is complete. + +??? note "*Proof*:" + + Will be added later. + +> *Proposition 7*: the space $l^\infty$ with the metric $d$ defined by +> +> $$ +> d(x,y) = \sup_{j \in \mathbb{N}} | x(j) - y(j) |, +> $$ +> +> for all $x,y \in l^\infty$ is complete. + +??? note "*Proof*:" + + Will be added later. diff --git a/docs/en/mathematics/functional-analysis/metric-spaces/completion.md b/docs/en/mathematics/functional-analysis/metric-spaces/completion.md new file mode 100644 index 0000000..28f8ee4 --- /dev/null +++ b/docs/en/mathematics/functional-analysis/metric-spaces/completion.md @@ -0,0 +1,20 @@ +# Completion + +> *Definition 1*: let $(X,d)$ and $(\tilde X, \tilde d)$ be metric spaces, then +> +> 1. a mapping $T: X \to \tilde X$ is an **isometry** if $\forall x, y \in X: \tilde d(Tx, Ty) = d(x,y)$. +> 2. $(X,d)$ and $(\tilde X, \tilde d)$ are **isometric** if there exists a bijective isometry $T: X \to \tilde X$. + +Hence, isometric spaces may differ at most by the nature of their points but are indistinguishable from the viewpoint of the metric. + +Or in other words, the metric space $(\tilde X, \tilde d)$ is unique up to isometry. + +> *Theorem 1*: for every metric space $(X,d)$ there exists a complete metric space $(\tilde X, \tilde d)$ that contains a subset $W$ that satisfies the following conditions +> +> 1. $W$ is a metric space isometric with $(X,d)$. +> 2. $W$ is dense in $X$. + +??? note "*Proof*:" + + Will be added later. + diff --git a/docs/en/mathematics/functional-analysis/metric-spaces/convergence.md b/docs/en/mathematics/functional-analysis/metric-spaces/convergence.md index e69de29..ed42390 100644 --- a/docs/en/mathematics/functional-analysis/metric-spaces/convergence.md +++ b/docs/en/mathematics/functional-analysis/metric-spaces/convergence.md @@ -0,0 +1,59 @@ +# Convergence + +> *Definition 1*: a sequence $(x_n)_{n \in \mathbb{N}}$ in a metric space $(X,d)$ is **convergent** if there exists an $x \in X$ such that +> +> $$ +> \lim_{n \to \infty} d(x_n, x) = 0. +> $$ +> +> $x$ is the **limit** of $(x_n)$ and is denoted by +> +> $$ +> \lim_{n \to \infty} x_n = x, +> $$ +> +> or simply by $x_n \to x$, $(n \to \infty)$. + +We say that $(x_n)$ *converges to* $x$ or *has the limit* $x$. If $(x_n)$ is not convergent then it is **divergent**. + +We have that the limit of a convergent sequence must be a point of $X$. + +> *Definition 2*: a non-empty subset $M \subset X$ of a metric space $(X,d)$ is **bounded** if there exists an $x_0 \in X$ and an $r > 0$ such that $M \subset B(x_0,r)$. + +Furthermore, we call a sequence $(x_n)$ in $X$ a **bounded sequence** if the corresponding point set is a bounded subset of $X$. + +> *Lemma 1*: let $(X,d)$ be a metric space then +> +> 1. a convergent sequence in $X$ is bounded and its limit is unique, +> 2. if $x_n \to x$ and $y_n \to y$ then $d(x_n, y_n) \to d(x,y)$, $(n \to \infty)$. + +??? note "*Proof*:" + + For statement 1, suppose that $x_n \to x$. Then, taking $\varepsilon = 1$, we can find an $N$ such that $d(x_n, x) < 1$ for all $n > N$. Which shows that $(x_n)$ is bounded. Suppose that $x_n \to x$ and $x_n \to z$ then by axiom 4 of the definition of a metric space we have + + $$ + d(x_n, x) \leq d(x_n, z) + d(x, z) \to 0, + $$ + + as $n \to \infty$ and by axiom 2 of the definition of a metric space it follows that $x = z$. + + For statement 2, we have that + + $$ + d(x_n,y_n) \leq d(x_n, x) + d(x, y) + d(y, y_n), + $$ + + by axiom 4 of the definition of a metric space. Hence we obtain + + $$ + d(x_n, y_n) - d(x, y) \leq d(x_n, x) + d(y_n, y), + $$ + + such that + + $$ + |d(x_n, y_n) - d(x, y)| \leq d(x_n, x) + d(y_n, y) \to 0 + $$ + + as $n \to \infty$. +