finished metric spaces
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chapters/multivar_calc.tex
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chapters/multivar_calc.tex
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\documentclass[../script.tex]{subfiles}
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%! TEX root = ../../script.tex
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\begin{document}
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\chapter{Multivariable Calculus}
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\subfile{sections/partial_total_diff.tex}
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\end{document}
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323
chapters/sections/conv_of_funcseq.tex
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chapters/sections/conv_of_funcseq.tex
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\documentclass[../../script.tex]{subfiles}
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% !TEX root = ../../script.tex
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\begin{document}
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\section{Convergence of Function sequences}
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\begin{defi}[Pointwise convergence]
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Let $M$ be a set, $f_n: M \rightarrow \field ~~\forall n \in \natn$ and $f: M \rightarrow \field$.
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The sequence $\seq{f}$ is said to be pointwise convergent to $f$ if
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\[
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\limn f_n(x) = f(x) ~~\forall x \in M
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\]
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\end{defi}
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\begin{eg}
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Consider
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\begin{align*}
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f_n: [0, 1] &\longrightarrow \realn \\
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x &\longmapsto \begin{cases}
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1 - nx, & x \in [0, \frac{1}{n}] \\
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0, & \text{else}
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\end{cases}
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\end{align*}
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\begin{center}
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\begin{tikzpicture}[domain=0:1]
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\begin{axis}[xmin=0, ymin=0, xmax=1, ymax=1, samples=50]
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\addplot[black, thick] {1-x} node[above right, pos=0.6] {$f_1$};
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\addplot[black, thick] {1-2*x} node[above right, pos=0.4] {$f_2$};
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\addplot[black, thick] {1-3*x} node[right, pos=0.3] {$f_3$};
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\addplot[black, thick] {1-4*x} node[left, pos=0.23] {$f_4$};
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\end{axis}
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\end{tikzpicture}
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\end{center}
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The $f_n$ are continuous for all $n \in \natn$ and converge pointwise to
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\begin{align*}
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f: [0, 1] &\longrightarrow \realn \\
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x &\longmapsto \begin{cases}
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1, & x = 0 \\
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0, & x \ne 0
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\end{cases}
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\end{align*}
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$f$ is not continuous.
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\end{eg}
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\begin{rem}
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Let $M$ be a set. Then
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\[
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B(M) = \set[\exists K \in \realn: ~~|f(x)| < K ~~\forall x \in M]{f_n: M \longrightarrow \field}
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\]
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is a linear subspace of the space of all functions $M \rightarrow \field$. We can define the supremum norm
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\begin{align*}
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\supnorm{\cdot}: B(M) &\longrightarrow \realn \\
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f &\longmapsto \sup_{x \in M}\set{\abs{f(x)}}
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\end{align*}
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\end{rem}
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\begin{proof}
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We will now proof that $\supnorm{\cdot}$ is a norm.
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It is defined, because
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\begin{equation}
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\supnorm{f} = 0 \implies \abs{f(x)} = 0 ~~\forall x \in M
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\end{equation}
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This implies
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\begin{equation}
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f(x) = 0 ~~\forall x \in M \implies f = 0
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\end{equation}
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The triangle inequality is proven by first considering
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\begin{equation}
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\abs{f(x)} \le \supnorm{f} ~~\forall f \in B(M) ~\forall x \in M
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\end{equation}
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Let $f, g \in B(M)$, then
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\begin{equation}
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\abs{f(x) + g(x)} \le \abs{f(x)} + \abs{g(x)} \le \supnorm{f} + \supnorm{g} ~~\forall x \in M
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\end{equation}
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Which implies
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\begin{equation}
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\supnorm{f + g} = \sup_{x \in M} \abs{f(x) + g(x)} \le \supnorm{f} + \supnorm{g}
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\end{equation}
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\end{proof}
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\begin{defi}[Uniform convergence]
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A sequence of bounded functions $\seq{f}$,
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\[
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f_n: M \longrightarrow \field
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\]
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is said to be uniformly convergent to $f: M \rightarrow \field$ if its norm converges.
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\[
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\supnorm{f_n - f} \conv{n \rightarrow \infty} 0
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\]
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\end{defi}
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\begin{rem}
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Formally, pointwise convergence means
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\[
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\forall \epsilon > 0 ~\forall x \in M ~\exists N \in \natn ~\forall n \ge N: ~~\abs{f_n(x) - f(x)} < \epsilon
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\]
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and uniform convergence means
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\[
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\forall \epsilon > 0 ~\exists N \in \natn ~\forall x \in M ~\forall n \ge N: ~~\abs{f_n(x) - f(x)} < \epsilon
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\]
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\end{rem}
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\begin{thm}
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The function space $B(M)$ is complete.
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\end{thm}
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\begin{proof}
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Let $\anyseqdef[f]{B(M)}$ be a Cauchy sequence in terms of $\supnorm{\cdot}$. Firstly, we have for some fixed $x \in M$
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\begin{equation}
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\abs{f_n(x) - f_m(x)} \le \supnorm{f_n - f_m}
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\end{equation}
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Since $\seq{f}$ is a Cauchy sequence, $(f_n(x))$ is also a Cauchy sequence in $\field_0$. Because $\field$ is complete,
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$(f_n(x))$ converges, and we define
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\begin{equation}
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f(x) = \limn f_n(x)
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\end{equation}
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thus $\seq{f}$ converges pointwise to $f$. Let $\epsilon > 0$. Then
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\begin{equation}
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\exists N \in \natn: ~~\supnorm{f_n \cdot f_m} < \epsilon ~~\forall n, m \ge N
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\end{equation}
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Then $\forall x \in M, ~\forall n, m \ge N$ we have
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\begin{equation}
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\abs{f_n(x) - f_m(x)} \le \supnorm{f_n - f_m} < \epsilon
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\end{equation}
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We can find the limit for $m \rightarrow \infty$
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\begin{equation}
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\abs{f(x) - f_n(x)} \le \epsilon
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\end{equation}
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and
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\begin{equation}
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\supnorm{f} = \sup_{x \in M}\abs{f} \le \sup_{x \in M} \abs{f(x) - f_n(x)} + \sup_{x \in M} \abs{f_n(x)} = \epsilon + \supnorm{f_n}
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\end{equation}
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Thus, $f$ is bounded. Furthermore
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\begin{equation}
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\supnorm{f - f_n} = \sup_{x \in M} \abs{f(x) - f_n(x)} \le \epsilon
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\end{equation}
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which in turn implies
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\begin{equation}
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\supnorm{f - f_n} \conv{n \rightarrow \infty} 0
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\end{equation}
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\end{proof}
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\begin{defi}
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Let $\metric$ be a metric space, then $C_b(X)$ is said to be the space of all continuous bounded functions.
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\end{defi}
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\begin{rem}
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If $X$ is compact (e.g. a bounded, closed subset of $\realn^n$) then all continuous functions are bounded. We then write $C(X)$ for $C_b(X)$.
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\end{rem}
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\begin{thm}
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Let $\metric$ be a metric space. $C_b(X)$ is closed in $B(X)$. In other words, every uniformly convergent sequence of continuous functions converges to a continuous function.
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\end{thm}
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\begin{proof}
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Let $\anyseqdef[f]{C_b(X)}$ be a sequence that uniformly converges to $f \in B(X)$.
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Let $x \in X$ and $\epsilon > 0$, then
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\begin{equation}
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\exists N \in \natn: ~~\supnorm{f - f_n} y \frac{\epsilon}{3} ~~\forall n \ge N
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\end{equation}
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Choose a fixed $n \ge N$. Since $f_n$ is continuous, this means that
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\begin{equation}
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\exists \delta > 0: ~~\abs{f_n(x) - f_n(y)} < \frac{\epsilon}{3} ~~\forall y \in \oball[\delta](x)
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\end{equation}
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Then we have for all such $y$
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\begin{equation}
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\begin{split}
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\abs{f(x) - f(y)} &\le \abs{f(x) - f_n(x)} + \abs{f_n(x) - f_n(y)} + \abs{f_n(y) - f(y)} \\
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&\le 2 \cdot \supnorm{f - f_n} + f_n(x) - f_n(y) < \epsilon
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\end{split}
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\end{equation}
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This proves the continuity of $f$ in $x$. Since $x \in X$ was chosen arbitrarily, $f$ is continuous everywhere.
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\end{proof}
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\begin{defi}
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Let $x_0 \in \field$ and $\anyseqdef[a]{\field}$. Then
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\[
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\series{n} a_n (x - x_0)^n
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\]
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is called a power series around $x_0$. The number
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\[
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\rho := \sup\set[\series{n} a_n (x - x_0)^n \text{ converges}]{\abs{x - x_0}}
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\]
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is the convergence radius.
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\begin{center}
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\begin{tikzpicture}
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\draw[->, thick] (0, 0) -- (0, 4) node [above] {};
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\draw[->, thick] (0, 0) -- (4, 0) node [right] {};
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\draw[fill] (2, 2) circle [radius=1pt] node [below right] {$x_0$};
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\draw[dashed] (2, 2) circle [radius=1.2cm];
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\draw[<->, rotate around={45:(2, 2)}] (2, 2) -- node[above] {$\rho$} (3.2, 2);
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\draw[fill, rotate around={100:(2, 2)}] (3.2, 2) circle [radius=1pt] node [above] {$x$};
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\end{tikzpicture}
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\end{center}
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\end{defi}
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\begin{rem}
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All results so far (including proofs) can be extended to $\realn^n$-valued functions, or functions with values in a Banach space in general.
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\end{rem}
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\begin{thm}
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Let $\series{n} a_n (x - x_0)^n$ be a power series with convergence radius $\rho \in [0, \infty) \cup \set{\infty}$.
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If $\abs{x - x_0} < \rho$ then the series converges absolutely, for $\abs{x - x_0} > \rho$ it diverges.
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\[
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\frac{1}{\rho} = \limsupn \sqrt[n]{\abs{a_n}}
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\]
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\end{thm}
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\begin{proof}
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W.l.o.g. choose $x_0 = 0$:
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For $\abs{x} > \rho$ the series diverges by definition. If $\abs{x} < \rho$ then there exists $y \in \field$ such that
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$\abs{x} < \abs{y} \le \rho$ and $\series{n} a_n y^n$ convergent. Especially, $(a_n y^n)$ is a null sequence.
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This means $\exists C > 0$ such that $\abs{a_n y^n} \le C ~~\forall n \in \natn$
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\begin{equation}
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\series{n} \abs{a_n x^n} = \series{n} \abs{a_n y^n} \abs{\frac{x}{y}}^n \le C \cdot \series{n} \abs{\frac{x}{y}}^n < \infty
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\end{equation}
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This statement only holds for $\rho > 0$.
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\end{proof}
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\begin{rem}
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\begin{enumerate}[(i)]
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\item We have
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\[
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\rho = \sup \set[\series{n}\abs{a_n}a^n \text{ converges}]{a \in [0, \infty)}
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\]
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\item If the following limit exists, then
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\[
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\rho = \limn \frac{\abs{a_n}}{\abs{a_{n+1}}}
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\]
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\end{enumerate}
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\end{rem}
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\begin{eg}
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The series
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\[
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\series{n} x^n
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\]
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is convergent on $(-1, 1)$, so $\rho = 1$. The limit function is
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\[
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x \longmapsto \frac{1}{1 - x}
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\]
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\end{eg}
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\begin{thm}
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Let $\series{n} a_n (x - x_0)^n$ be a power series with convergence radius $\rho > 0$.
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Let $0 < a < \rho$. Then this power series converges uniformly on $\cball[a](x_0)$. Especially
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\begin{align*}
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f: \oball[\rho](x_0) &\longrightarrow \realn \\
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x &\longmapsto \series{n} a_n (x_n - x_0)^n
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\end{align*}
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\end{thm}
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\begin{proof}
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W.l.o.g. choose $x_0 = 0$. Let $0 < a < \rho$. We know that $\series{n} a_nx^n$ converges on $\cball[a](0)$.
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\begin{center}
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\begin{tikzpicture}
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\draw[->, thick] (0, -3) -- (0, 3);
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\draw[->, thick] (-3, 0) -- (3, 0);
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\draw[fill] (0, 0) circle [radius=1pt];
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\draw[solid] (0, 0) circle [radius=1.8cm];
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\draw[dashed] (0, 0) circle [radius= 2.4cm] node[above right=2.4cm] {$\rho$};
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\node at (2, 0.2) {$a$};
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\end{tikzpicture}
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\end{center}
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Define
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\begin{equation}
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\begin{split}
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f_n: \cball[a](0) &\longrightarrow \field \\
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x &\longmapsto x^n ~~\forall n \in \natn
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\end{split}
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\end{equation}
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We can see that
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\begin{equation}
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\supnorm{f} = \sup_{x \in \cball[a](0)} \abs{f_n} = \sup_{x \in \cball[a](0)} = a^n
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\end{equation}
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and thus
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\begin{equation}
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\series{n} a_n f_n \implies \series{n} \supnorm{a_n f_n} = \series{n} \abs{a_n}^n < \infty
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\end{equation}
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because $a < \rho$. The series $\series{n} a_n f_n$ is absolutely convergent in $C(\cball[a](0))$.
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Since $C(\cball[a](0))$ is complete, $\series{n} a_n f_n$ is convergent because the partial sums $\series[N]{n} a_n f_n$ are continuous $\forall N \in \natn$.
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Therefore $f$ is also continuous on $\cball[a](0)$. Let $x \in \oball[\rho](0)$. Then there exists some $a > 0$ such that $\abs{x} < a < \rho$.
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Thus, $f$ is continuous on $\cball[a](0)$. Since $\cball[a](0)$ contains a neighbourhood of $x$, and continuity is a local property,
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$f$ is also continuous in $x$. Because $x \in \oball[\rho](0)$ was chosen arbitrarily, $f$ is continuous.
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\end{proof}
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\begin{rem}
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$\exp$, $\sin$, $\cos$ are continuous.
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\end{rem}
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\begin{eg}
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The statements above can be extended to Banach space-valued power series (e.g. matrix-valued functions).
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The norm on $\realn^{n \times n}$ is
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\[
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\norm{A} = \sup\set[{\forall x \in \oball[1](0)}]{\norm{Ax}}
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\]
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Define
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\[
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\exp(A) := \series{0} \frac{A^n}{n!}
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\]
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This converges $\forall A \in \realn^{n \times n}$, because
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\begin{align*}
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\series{n} \norm{\frac{A^n}{n!}} &= \series{n} \frac{1}{n!} \norm{A^n} \le \series{n} \frac{1}{n!} \norm{A}^n \\
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&= \exp(\norm{A}) < \infty
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\end{align*}
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Thus, $\series{n} \frac{A^n}{n!}$ converges absolutely. Now consider the function
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\begin{align*}
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\realn &\longrightarrow \realn^{n \times n} \\
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t &\longmapsto \exp(At)
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\end{align*}
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This is a matrix-valued power series
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\[
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\exp(At) = \series{n} \frac{(At)^n}{n!} = \series{n} \frac{A^n}{n!} t^n
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\]
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with a convergence radius of $\rho = \infty$. In this case $\exp(A + B)$ doesn't necessarily have to equal $\exp(A) \cdot \exp(B)$.
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\end{eg}
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\end{document}
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10
chapters/sections/partial_total_diff.tex
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10
chapters/sections/partial_total_diff.tex
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\documentclass[../../script.tex]{subfiles}
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%! TEX root = ../../script.tex
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\begin{document}
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\section{Partial and Total Differentiability}
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\begin{defi}
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\end{defi}
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\end{document}
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@ -304,7 +304,7 @@ Let $V$ be a vector space. A non-empty set $W \subset V$ is called a vector subs
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\begin{eg}
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Consider
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\[
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W = \set[chi \in \realn]{(\chi, \chi) \in \realn^2}
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W = \set[\chi \in \realn]{(\chi, \chi) \in \realn^2}
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\]
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This is a subspace, because
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\[
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\subfile{sections/seq_ser_limits.tex}
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\subfile{sections/open_closed_sets.tex}
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\subfile{sections/continuity.tex}
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\subfile{sections/conv_of_funcseq.tex}
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\end{document}
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BIN
script.pdf
BIN
script.pdf
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@ -156,5 +156,6 @@
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\subfile{chapters/linear_algebra.tex}
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\subfile{chapters/real_analysis_2.tex}
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\subfile{chapters/topo_of_metr_spaces.tex}
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\subfile{chapters/multivar_calc.tex}
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\end{document}
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