> [!definition] > > Let $X$ be a [[Vector Space|vector space]] over $\real$ or $\complex$. $X$ is **normed** if there is a [[Function|function]] $f: X \to \real$ (norm) that satisfies: > - $|x| \ge 0$ > - $|x| = 0 \Leftrightarrow x = \zero$ > - $|cx| = |c| \cdot |x| \forall c \in \field$ > - $|x + y| \le |x| + |y|$ ([[Triangle Inequality]]) > [!theorem] Equivalent Norms > > Let $X$ be a [[Vector Space|vector space]] and $\norm{\cdot }_1, \norm{\cdot}_2$ be two norms. They are **equivalent** if there exists $C_1, C_2 > 0$ such that > $ > C_1\norm{x}_1 \le \norm{x}_2 \le C_2\norm{x}_1 \quad \forall x \in X > $ > Equivalent norms induce the same [[Topological Space|topology]]. > > *Proof*. > > ### Equivalence Relation > > Let $\norm{\cdot}_1 \sim \norm{\cdot}_2$ if there exists $C_1, C_2 > 0$ such that > $ > C_1\norm{x}_1 \le \norm{x}_2 \le C_2\norm{x}_1 > $ > then $\sim$ is an equivalence relation. > > *Proof*. $\norm{x} = \norm{x}$ for all norms, so $\sim$ is reflexive. > > If there exists $C_1, C_2 > 0$ such that > $ > C_1\norm{x}_1 \le \norm{x}_2 \le C_2\norm{x}_1 \quad \forall x \in X > $ > then > $ > \frac{1}{C_2}\norm{x}_2 \le \norm{x}_1 \le \frac{1}{C_1}\norm{x}_2 \quad \forall x \in X > $ > and $\sim$ is symmetric. > > Lastly, if > $ > C_1\norm{x}_1 \le \norm{x}_2 \le C_2\norm{x}_1 \quad \forall x \in X > $ > and > $ > C_3\norm{x}_2 \le \norm{x}_3 \le C_4\norm{x}_2 \quad \forall x \in X > $ > then > $ > C_1C_3 \norm{x}_1 \le C_3 \norm{x}_2\le \norm{x}_3 \le C_4\norm{x}_2 \le C_2C_4 \norm{x}_1 > $ > and $\sim$ is transitive. > > > ### Equivalent Topology > > Let $\norm{\cdot}_1$ and $\norm{\cdot}_2$ be norms, and $\topo_i$ be the norm topology generated with $\norm{\cdot}_i$. If $\norm{\cdot}_1 \sim \norm{\cdot}_2$, then $\topo_1 \subset \topo_2$. Since the relation is symmetric, we have $\topo_1 = \topo_2$. > > *Proof*. Denote $B_i(x, r) = \bracs{y \in \cx: \norm{x - y}_i < r}$, and let $B_1(x, r)$ be an open ball. Take $y \in B_1(x, r)$, then there exists $\varepsilon_y > 0$ such that $B_1(y, \varepsilon_y) \subset B_1(x, r)$, and > $ > B_2\paren{y, \frac{\varepsilon_y}{C_2}} \subset B_1(y, \varepsilon_y) \subset B_1(x, r) > $ > so > $ > B_1(x, r) = \bigcup_{y \in B_1(x, r)}B_2(y, \varepsilon_y/C_2) > $ > Therefore $\topo_1 \subset \topo_2$. > [!theorem] > > Let $\cx$ be a normed vector space. $\cx$ is a [[Banach Space|Banach space]] if and only if every absolutely convergent series in $\cx$ [[Limit|converges]]. > > *Proof*. Suppose that $\cx$ is a Banach space, and let $\seq{x_n} \subset \cx$ be an absolutely convergent series, then > $ > \forall \varepsilon > 0, \exists N \in \nat: \sum_{n \ge N}\norm{x_n} < \varepsilon > $ > and for any $m, n \ge N$ (assume WLG that $m \ge n$), > $ > \norm{\sum_{k = 1}^{m}x_k - \sum_{k = 1}^{n}x_k} = \norm{\sum_{k = n + 1}^{m}x_k} \le \sum_{k \ge N}\norm{x_n} < \varepsilon > $ > therefore the partial sums are [[Cauchy Sequence|Cauchy]], meaning that $\sum_{n = 1}^{\infty}x_n$ converges. > > Suppose that every absolutely convergent series in $\cx$ converges in $\cx$. Let $\seq{x_n}$ be Cauchy [[Sequence|sequence]] in $\cx$, then > $ > \forall \varepsilon > 0, \exists N \in \nat: \norm{x_m - x_n} < \varepsilon > $ > For any $k \in \nat$, choose $x_{n_k}$ such that $\norm{x_{m} - x_{n_k}} < 2^{-k}$ for all $m \ge n_k$ and $n_k > n_{k - 1}$. Let $y_k = x_{n_{k + 1}} - x_{n_k}$, then $\norm{y_k} < 2^{-k}$ and $\sum_{k = 1}^{\infty}y_k$ is an absolutely convergent series, which converges and is equal to a subsequence of $\seq{x_n}$. As $\seq{x_n}$ is Cauchy, and it has a convergent subsequence, $\seq{x_n}$ converges to the same limit. > [!theorem] > > Let $\cx$ be a normed space. Then addition and scalar multiplication are [[Continuity|continuous]] from $\cx \times \cx$ and $\cx$ with respect to the [[Product Norm|product norm]]. > > *Proof*. Let $(x, y) \in \cx^2$, then for any $\varepsilon > 0$, there exists $\delta = \varepsilon / 2$ such that for any $(x', y') \in \cx^2$ such that > $ > \norm{(x, y) - (x', y')} < \delta > $ > we have > $ > \begin{align*} > \norm{(x + y) - (x' + y')} &= \norm{(x - x') + (y - y')} \\ > &\le \norm{x - x'} + \norm{y - y'} \\ > &\le 2\norm{(x, y) - (x', y')} \\ > &< \varepsilon > \end{align*} > $ > therefore addition is continuous. > > Now let $x \in \cx$ and $\lambda \in \complex$ ($\lambda \ne 0$), then for any $\varepsilon > 0$, there exists $\delta < \varepsilon/\abs{\lambda}$ such that whenever $\norm{x - x'} < \delta$, > $ > \norm{\lambda x - \lambda x'} = \abs{\lambda}\norm{x - x'} > < \varepsilon > $ > therefore scalar multiplication is also continuous. > [!theorem] > > Let $\cx$ be a normed space and $\cm$ be a subspace of $\cx$. Then its [[Topological Closure|closure]] $\ol{\cm}$ is also a subspace of $\cx$. > > *Proof*. $\ol{\cm}$ consists of adherent points of $\cm$. Since this is a metric space, for every element $x$ of $\ol \cm$ there exists a [[Sequence|sequence]] $\seq{x_n}$ such that $x_n \to x$. > > Let $\lambda \in \complex$ and $x, y \in \ol{\cm}$ with supporting sequences $\seq{x_n}$ and $\seq{y_n}$, then since $\cm$ is a subspace, $\seq{x_n + y_n}$ and $\seq{\lambda x_n}$ are also convergent sequences in $\cm$. As addition and scalar multiplication are continuous, $x_n + y_n \to x + y$ and $\lambda x_n \to \lambda x$, both adherent points of $\cm$. Therefore $x + y, \lambda x \in \ol{\cm}$, and $\ol \cm$ is a subspace. > [!theorem] > > Let $\cx$ be a finite-dimensional normed space, and $\bracs{e_i}_1^n$ be a [[Basis|basis]] for $\cx$, and define > $ > \norm{x} = \norm{\sum_{i = 1}^{n}a_ie_i} = \sum_{i = 1}^{n}\abs{a_i} > $ > then $\norm{\cdot}$ is a norm on $\cx$. The map from coordinates to the basis $\bracs{a_i}_1^n \mapsto \sum_{i = 1}^{n}|a_i|$ is continuous, and all norms on $\cx$ are equivalent. > > Moreover, $\bracs{x \in \cx: \norm{x} = 1}$ is [[Compactness|compact]] with respect to the norm defined above. > > *Proof*. Let $x = \sum_{i}a_ie_i$ and $y = \sum_i b_ie_i$, then > $ > \norm{x + y} = \sum_{i}\abs{a_i + b_i} \le \sum_{i}|a_i| + |b_i| = \norm{x} + \norm{y} > $ > and for any $\lambda \in \complex$, > $ > \norm{\lambda x} = \sum_{i}\abs{\lambda a_i} = \sum_{i}\abs{\lambda} \cdot \abs{a_i} = \lambda \norm{x} > $ > therefore $\norm{\cdot}$ is a norm on $\cx$. > > Let $\bracs{a_i}_1^n \in \complex^n$, then > $ > \sum_{i}|a_i| \le n\max_i |a_i| \le n\sqrt{\sum_{i}\abs{a_i}^2} > $ > and $\bracs{a_i}_1^n \mapsto \sum_{i}a_ie_i$ is bounded, and therefore continuous. > > Let $\norm{\cdot}_2$ be any other norm on $\cx$, then $\norm{e_i} > 0$ for all $i$. Let $m = \min_i \norm{e_i}$ and $M = \max_i\norm{e_i}$, then > $ > \begin{align*} > m\norm{x} &= m\sum_{i}|a_i| \\ > &\le \norm{x}_2 \\ > &\le M\sum_{i}|a_i| \\ > &= M\norm{x} > \end{align*} > $ > and the norms are equivalent. > > Lastly, since the [[Unit Circle is Compact|unit "circle" is compact]], and that the map $\bracs{a_i}_1^n \mapsto \sum_{i}a_ie_i$ is continuous, $\bracs{x \in \cx: \norm{x} = 1}$ is the continuous image of a compact set, and therefore compact. > [!theorem] > > Let $\cx$ be an infinite-dimensional normed space, then > 1. There is a [[Sequence|sequence]] $\seq{x_j} \subset \cx$ such that $\norm{x_j} = 1$ for all $j$ and $\norm{x_j - x_k} \ge \frac{1}{2}$ for all $j \ne k$. > 2. $\cx$ is not [[Locally Compact|locally compact]]. > > *Proof*. First show that the closed unit sphere is not compact through a sequence with no Cauchy subsequence, then translate that sphere to whatever neighbourhood that might be compact. Since all closed subsets of a compact set is compact, that neighbourhood cannot be compact. > > ### "Ortho"normal Set > > While there is no inner product, it is still possible to find vectors that are "reasonably far from each other" on the surface of the unit sphere using quotient spaces. > > Let $\cx$ be an infinite-dimensional normed space, then there is a sequence $\seq{x_j} \subset \cx$ such that $\norm{x_j} = 1$ for all $j$ and $\norm{x_j - x_k} \ge \frac{1}{2}$ for all $j \ne k$. > > *Proof*. Choose $x_1 \in \cx$ such that $\norm{x_1} = 1$, and $\cm_1 = \complex x_1$, then $\cm_1$ is closed. > > Let $n \in \nat$, and choose $x_n \in \cx$ such that $\norm{x_n} = 1$ and $\norm{x_n} \ge \frac{1}{2}$ in the [[Quotient Norm|quotient space]] $\cx/\cm_{n - 1}$, then $\norm{x_n - x_m} \ge \frac{1}{2}$ for all $x_m \in \cm_{n - 1}$. Take $\cm_n = \cm_{n - 1} + \complex x_n$, then $\cm_n$ is another closed subspace. > > Repeating this process for all $n \in \nat$ yields the desired sequence. > > > ### Unit Sphere > > Let $\cx$ be an infinite-dimensional normed space, then $S = \bracs{x \in \cx: \norm{x} \le 1}$ is not compact. > > *Proof*. Let $\seq{x_n}$ such that $\norm{x_n} = 1$ for all $n$ and $\norm{x_m - x_n} \ge \frac{1}{2}$ for all $m, n \in \nat$. Then $\seq{x_n} \subset S$ has no Cauchy and no convergent subsequence, and $S$ is not compact. > > > ### Translation > > Let $\cx$ be an infinite-dimensional normed space, then for any $\lambda > 0$ and $x \in \cx$, $\lambda S + x$ is not compact. > > *Proof*. Let $\seqi{U}$ be an [[Open Cover|open cover]] of $\lambda S + x$, then $\seqi{\lambda^{-1}(U_i - x)}$ is an open cover for $S$. Since the latter has no finite subcover, the former has no finite subcover neither. Therefore $\lambda S + x$ is not compact. > > > ### Local Compactness > > Let $\cx$ be an infinite-dimensional normed space, then $\cx$ is not compact. > > *Proof*. Suppose that $\cx$ is locally compact. Let $x \in \cx$ and $K$ be a compact neighbourhood of $x$. Then there exists $\lambda > 0$ such that $\lambda S + x \subset K^o$. Since a closed subset of a compact set is compact, $\lambda S + x$ would be compact, which contradicts the fact that it isn't. Therefore the assumption that $\cx$ is locally compact is false.