Constructions:: *[[Constructions|Used in the construction of...]]* Specializations:: *[[Specializations]]* Justifications and Intuition:: *[[Justifications and Intuition]]* Examples:: [[norm#^a059e1|norm > examples]] Nonexamples:: [[on the circle, continuous functions are dense in Riemann integrable functions with respect to the 1-seminorm|1-seminorm]] Properties:: *[[Properties]]* > [!definition] Definition. (norm) > Let $X$ be a [[vector space]] over the [[complex numbers]] or [[real numbers]]. A **norm** on $X$ is a map $\vert \vert \cdot \vert \vert : X \to [0, \infty)$ satisfying > 1. $\vert \vert x+y \vert \vert \leq \vert \vert x \vert \vert + \vert \vert y \vert \vert$ for every $x,y \in X$. > 2. $\vert \vert cx \vert \vert = \vert c \vert \vert \vert x \vert \vert$ for every $c \in \mathbb{R}$ and $x \in X$. > 3. $\vert \vert x \vert \vert=0 \iff x=0$. > > We say two norms $\|\cdot\|_{\alpha}$, $\|\cdot\|_{\beta}$ on $X$ are **equivalent** if $\|\cdot\|_{\alpha} \asymp \|\cdot\|_{\beta}$, i.e., if there exist $c, C>0$ such that $c \|\cdot\|_{\alpha} \leq \|\cdot\|_{\beta} \leq C\|\cdot\|_{\alpha}.$We say two norms $\|\cdot\|_{\alpha}$, $\|\cdot\|_{\beta}$ on $X$ are **topologically equivalent** if they induce[^1] the same topology. In general,[^2] $\|\cdot\|_{\alpha} , \|\cdot\|_{\beta , }\text{ equivalent} \stackrel{\Rightarrow}{\nLeftarrow} \|\cdot\|_{\alpha}, \|\cdot\|_{\beta} \text{ topologically equivalent}.$ > The pair $(X, \|\cdot\|)$ is called a **normed (vector) space**. [[seminorm|As with]] $\mathsf{Seminorm}$, there are many (nonequivalent) ways to define a [[category]] $\mathsf{Norm}$ whose objects are normed spaces over $\mathbb{F}$. Here are two: > 1. As a [[subcategory|full subcategory]] $\mathsf{Norm_{T}}$ [[topological vector space|of]] $\mathsf{TVS}$[^4]; > 2. As a [[subcategory]] $\mathsf{Norm_{M}}$ [[metric space|of]] $\mathsf{Met}$ and $\mathsf{Vect}$, where morphisms are [[Lipschitz continuous|nonexpansive]] [[linear map|linear mappings]] (whence an [[isomorphism]] is a [[surjection|surjective]] ($\iff$ [[bijection|bijective]]) [[bijection|]] [[linear map|linear]] [[Lipschitz continuous|isometry]]). > > Note that $\mathsf{Norm_{M}}$ is a [[subcategory]] of $\mathsf{Norm_{T}}$. [^4]: That is, morphisms are [[continuous]] ([[characterizing continuity of linear maps|equivalently,]] [[operator norm|bounded]]) [[linear map|linear maps]] (whence an [[isomorphism]] is a [[linear map|linear]] [[homeomorphism]]). > [!equivalence] > $\|\cdot\|_{\alpha} \asymp \|\cdot\|_{\beta}$ iff any [[sequence]] [[Cauchy sequence|Cauchy]] wrt one norm is [[Cauchy sequence|Cauchy]] wrt the other. > > > [!proof]- Proof. > > First suppose $\|\cdot\|_{\alpha} \asymp \|\cdot\|_{\beta}$; obtain $C>0$ for which $\|\cdot\|_{\beta} \leq C \|\cdot\|_{\alpha}$. Suppose $(x_{k})$ is $\alpha$-Cauchy. Fix $\varepsilon>0$. Obtain $N$ such that for all $k,j \geq N$ we have $\|x_{k}-x_{j}\|_{\alpha}<\varepsilon /C$. Then for all $k,j \geq N$ we have $ \|x_{k}-x_{j}\|_{\beta} \leq { {C} \| x_{k}-x_{j}\|_{\alpha} }_{ }<\varepsilon,$ > > and so $(x_{k})$ is $\beta$-Cauchy. > > > > - [ ] The converse is optional and I don't have time for it right now. > > [!note] Remark. > Equivalent to (3) is the following: 3'. $\vert \vert x-y \vert \vert \geq \ \ \vert \ \vert \vert x \vert \vert - \vert \vert y \vert \vert \ \vert$. ^note [^2]: Indeed, if $\|\cdot\|_{\alpha} \asymp \|\cdot\|_{\beta}$, then given any $\beta$-basic open set $B_{r}^{(\beta)}(x_{0})$ in $X$ we have $ B_{cr}^{(\alpha)}(x_{0}) \subset B_{r}^{(\beta)}(x_{0}) \subset B_{Cr}^{(\alpha)}(x_{0})$ and then apply [[basis-nestling characterization of comparing topologies]]. On the other hand, the norms $|\cdot|$, $\frac{|\cdot|}{|\cdot|+1}$ on $\mathbb{R}$ induce the same topology but there is no $C \geq 0$ such that $|\cdot|\leq C \frac{|\cdot|}{|\cdot|+1}$ e.g. because the inequality $(C+1 )\leq C \frac{C+1}{C+2}$ does not hold. [^1]: The norm $\|\cdot\|$ on $X$ induces a [[metric]] on $X$ [[metric topology|and in turn]] a [[topological space|topology]], making $X$ into [[topological vector space]]. (Check continuity of vector space operations.) > [!basicexample] > >- [[canonical norm induced by inner product]] > - [[Euclidean-Sup Norm Inequality]] > - Not all norms are derived from [[inner product|inner product]]s! For example, on $\mathbb{R}^n$ we have the oft-convenient [[sup norm]]. > [!basicproperties] > - [[norms are convex functions]] > - The [[closure]] of a [[linear subspace]] of a normed vector space is again a [[linear subspace]] > > > > [!proof]- Proof. > > Let $W$ be the [[linear subspace]] in question. Since $W \subset \overline{W}$, $0 \in \overline{W}$. Let $v,w \in \overline{W}$. Fix $\varepsilon>0$; per the [[neighborhood-basis characterization of set closure]], we want to show that $B_{\varepsilon}(v+w) \cap W \neq \emptyset$. [[neighborhood-basis characterization of set closure|Each of]], each of $v$ and $w$ can be made arbitrarily close to some respective elements $v'$ and $w'$ of $W$, say, $\|w-w'\| < \frac{\varepsilon}{2}$ and $\|v-v'\|< \frac{\varepsilon}{2}$. Then > > $\begin{align} > > \|(w+v)-(w'+v')\| &= \|(w-w')+(v-v')\| \\ > > & \|w-w'\| + \|v-v'\| \\ > > & < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon , > > \end{align}$ > > meaning $w'+v' \in B_{\varepsilon}(v+w)$. Since $W$ is a [[linear subspace]], $w'+v' \in W$ also; it follows that $w+v \in \overline{W}$. > > > > Next let $c \in \mathbb{F}$ and $v \in \overline{W}$. Again fix $\varepsilon>0$; we want to show that $B_{\varepsilon}(cv) \cap W \neq \emptyset$. Since $v \in \overline{W}$, we can write $\|v-v'\|< \frac{\varepsilon}{c}$ for some $v' \in W$. Since $W$ is a [[linear subspace]], $cv' \in W$. And $\|cv - cv'\|=\|c\| \|v - v'\|< c \left( \frac{\varepsilon}{c} \right)=\varepsilon,$ > > as required. > > > > - For all nonzero $f \in V$, $\|f\|= \max \{ |\varphi(f)|: \varphi \in V^{\vee} \text{ and } \|\varphi\|=1 \},$ > where $V^{\vee}$ denotes the [[dual vector space|dual space]] to $V$ and $\|\varphi\|$ is the [[operator norm]] of $\varphi$. > > > [!proof]- Proof. > > It is always true that $|\varphi(f)| \leq \|\varphi\| \|f\|=\|f\|$ when $\|\varphi\|=1$. Below we apply the [[Hahn-Banach Extension Theorem]] to obtain $\varphi \in V^{\vee}$ to attain equality. Let $U$ be the one-dimensional [[linear subspace]] of $V$ given by $U:= \{ \alpha f: \alpha \in \mathbb{F} \}=\span(f)$ > > Define $\psi: U \to \mathbb{F}$ by $\psi(\alpha f)=\alpha \|f\|$ > > for $\alpha \in \mathbb{F}$. Then $\psi$ is a [[dual vector space|linear functional]] on $U$ with $\|\psi \|=1$ and $\psi(f)=\|f\|$. The [[Hahn-Banach Extension Theorem]] implies that there exists an extension of $\psi$ to a [[dual vector space|linear functional]] $\varphi$ on $V$ with $\|\varphi\|=1$, completing the proof. Thus, $\|f\|=\varphi(f)$ as required. > > - If $X$ is a [[dimension|finite-dimensional]] [[vector space]], then all norms on $X$ are equivalent (and in turn topologically equivalent). todo: https://math.mit.edu/~stevenj/18.335/norm-equivalence.pdf) > > > *(The two-epsilon trick)* If $(X, \|\cdot\|)$ is a normed space and $f_{k} \to f$ in [[norm]] such that there exists $g_{k, m} \to f_{k}$ in norm as $m \to \infty$, then there is a single sequence $(h_{n})$ chosen from the family $\{ g_{k,m} \}_{k,m=1}^{\infty}$ such that $h_{n} \to f$ in [[norm]]. > > > [!proof]- Proof. > > Fix $\varepsilon>0$. Obtain $K$ large enough that $\|f_{k}-f\|<\varepsilon/2$ for all $k \geq K$. Then obtain $M$ for which $\|g_{K,M}- f_{M}\|< \varepsilon / 2$. > > > > By the triangle inequality: $\| g_{K,M}-f \|\leq \|g_{K,M}-f_{K}\|+ \|f_{K}-f\| < \varepsilon.$ > > This is usually enough when one just wants an approximation argument. But we can harvest an explicit sequence $(h_{n}) \to f$ as follows. Choose $k_{n}$ so that $\|f-f_{k_{n}}\|<2^{-(n+1)}$. Then choose $m_{n}$ so that $\|g_{k_{n}, m_{n}}- f_{k_{n}}\|<2^{-(n+1)}$. Then Set $h_{n}=g_{k_{n}, m_{n}}$. Then $\|h_{n}-f\|=\|g_{k_{n}, m_{n}} -f_{k_{n}}\|+\|f_{k_{n}}-f\|< 2^{-n}$. Taking the limit $n \to \infty$ on both sides gives the result.