A [vector space](/page/Vector%20Space) knows how to add and scale, but it does not yet know what it means for two vectors to be close. That missing notion is fatal for analysis: without length there is no convergence, no continuity, no approximation, and no way to say that a sequence of functions is approaching a solution of an equation. A normed space is the first place where linear algebra becomes geometric analysis. Finite-dimensional Euclidean space is the model example, but many of the most important spaces are infinite-dimensional: spaces of continuous functions, integrable functions, and bounded sequences. The same formal definition covers all of them, and that common language is powerful because many arguments depend only on three length axioms. At the same time, infinite-dimensional normed spaces behave in ways that finite-dimensional spaces hide, so this chapter is also the point where familiar geometry begins to acquire genuinely analytical depth. ## Definition A usable notion of length on a vector space has to solve three problems at once: only the zero vector should have length zero, rescaling a vector should rescale its length by the absolute value of the scalar, and the direct route should not be longer than a route through an intermediate vector. These requirements are exactly the axioms that make linear estimates compatible with metric and topological arguments. [definition: Normed Space] Let $V$ be a vector space over $\mathbb{R}$ or over $\mathbb{C}$. A norm on $V$ is a function $\|\cdot\|:V\to [0,\infty)$ such that, for all $x,y\in V$ and all scalars $\lambda$, the following conditions hold. The positivity condition is: if $\|x\|=0$, then $x=0$. The homogeneity identity is \begin{align*} \|\lambda x\|=|\lambda|\,\|x\|. \end{align*} The triangle inequality is \begin{align*} \|x+y\|&\leq \|x\|+\|y\|. \end{align*} A normed space is a pair $(V,\|\cdot\|)$ consisting of a vector space $V$ and a norm $\|\cdot\|$ on $V$. [/definition] The real line gives the smallest nonzero example and fixes the connection with ordinary distance. It also helps separate the abstract notation from the familiar absolute value. [example: The Absolute Value Norm] For $x\in\mathbb{R}$, define $\|x\|=|x|$. If $\|x\|=0$, then $|x|=0$, so $x=0$. For $\lambda,x\in\mathbb{R}$, the multiplicative property of absolute value gives \begin{align*} \|\lambda x\|=|\lambda x|=|\lambda|\,|x|=|\lambda|\,\|x\|. \end{align*} For $x,y\in\mathbb{R}$, the ordinary triangle inequality for absolute value gives \begin{align*} \|x+y\|=|x+y|\leq |x|+|y|=\|x\|+\|y\|. \end{align*} Thus $|\cdot|$ satisfies the norm axioms on $\mathbb{R}$. The metric induced by this norm is obtained by applying the norm to the difference: \begin{align*} d(x,y)=\|x-y\|=|x-y|. \end{align*} So the norm topology on $\mathbb{R}$ is the familiar topology coming from ordinary distance on the real line. [/example] The first axiom says that the norm detects the zero vector. The second says that scaling a vector scales its length by the absolute value of the scalar. The third is the triangle inequality: it expresses the geometric idea that travelling from $0$ to $x+y$ through $x$ cannot be shorter than the direct route. The notation $\|x\|$ depends on the chosen norm. The same vector space may carry many different norms, so when the norm matters it is written explicitly as $(V,\|\cdot\|)$. ## Metric Structure ### The Induced Distance The norm immediately produces a distance function. This step matters because it lets every normed space inherit the language of metric spaces while retaining its linear operations. [definition: Metric Induced by a Norm] Let $(V,\|\cdot\|)$ be a normed space. The metric induced by the norm is the function $d:V\times V\to [0,\infty)$ defined by \begin{align*} d:V\times V\to [0,\infty),\qquad (x,y)\mapsto \|x-y\|. \end{align*} [/definition] The definition has now produced a candidate distance, but it still has to be checked before any metric language is legitimate. The next theorem is the bridge from linear length to topology: it says that the expression $\|x-y\|$ really does support open balls, convergence, and continuity. Positivity comes from positive definiteness, symmetry comes from homogeneity with scalar $-1$, and the triangle inequality for $d$ is the triangle inequality for the norm applied to $(x-z)+(z-y)$. [quotetheorem:9982] This result is the point where all metric-space language becomes available: one may now speak about open balls, convergent sequences, Cauchy sequences, continuity, and completeness in a normed space without adding any extra structure. The construction is also limited in an important way: it uses the norm, so changing the norm may change the metric unless the norms are known to be equivalent. The induced metric still remembers the linear structure, and the next elementary result records exactly how the distance behaves under the two linear operations. This compatibility is what allows metric estimates to be translated and scaled without changing their form. [quotetheorem:9983] Thus a normed space is a [metric space](/page/Metric%20Space) with extra linear structure: translating the whole picture preserves distance, and scaling the picture multiplies every distance by the absolute value of the scalar. Convergence in a normed space means convergence in the induced metric. For a sequence $(x_n)$ in $V$, the statement $x_n\to x$ means \begin{align*} \|x_n-x\|\to 0. \end{align*} The same convention applies to limits of nets or filters when those are used. ### Elementary Consequences Several useful inequalities follow from the norm axioms. For norms, the relevant estimate is the [reverse triangle inequality](/theorems/2300) \begin{align*} \bigl|\|x\|-\|y\|\bigr|\leq \|x-y\|. \end{align*} It follows directly from the triangle inequality for the norm: $\|x\|\leq \|x-y\|+\|y\|$ and symmetrically $\|y\|\leq \|x-y\|+\|x\|$. This is the form needed here, because the scalar absolute-value theorem below is not a statement about arbitrary normed spaces. This estimate says that the length function cannot oscillate faster than the vector itself moves. In particular, convergence $x_n\to x$ forces $\|x_n\|\to \|x\|$. The next observation is used constantly when checking continuity of linear maps. It records that the topology from the norm is compatible with the algebraic operations, so limits can be added and scaled inside the space. [quotetheorem:9984] The point is not merely sequential: these are continuous maps for the actual product topologies. Since the spaces involved are metric spaces, sequential tests are often enough in computations, but the theorem itself is a topological compatibility statement. ## Standard Examples The Euclidean examples provide the basic intuition. [example: Euclidean Norm] On $\mathbb{R}^n$, define \begin{align*} \|x\|_2=\left(\sum_{j=1}^n |x_j|^2\right)^{1/2}. \end{align*} Since each term $|x_j|^2$ is nonnegative, $\|x\|_2=0$ implies $\sum_{j=1}^n |x_j|^2=0$, hence $|x_j|^2=0$ for every $j$, so $x_j=0$ for every $j$ and therefore $x=0$. For $\lambda\in\mathbb{R}$ and $x\in\mathbb{R}^n$, the absolute value identities $|\lambda x_j|=|\lambda|\,|x_j|$ give \begin{align*} \|\lambda x\|_2=\left(\sum_{j=1}^n |\lambda x_j|^2\right)^{1/2}=\left(\sum_{j=1}^n |\lambda|^2|x_j|^2\right)^{1/2}=|\lambda|\left(\sum_{j=1}^n |x_j|^2\right)^{1/2}=|\lambda|\,\|x\|_2. \end{align*} For $x,y\in\mathbb{R}^n$, the scalar triangle inequality gives $|x_j+y_j|\leq |x_j|+|y_j|$ for each $j$, so \begin{align*} \|x+y\|_2^2=\sum_{j=1}^n |x_j+y_j|^2\leq \sum_{j=1}^n (|x_j|+|y_j|)^2. \end{align*} Expanding the right-hand side, \begin{align*} \sum_{j=1}^n (|x_j|+|y_j|)^2=\sum_{j=1}^n |x_j|^2+2\sum_{j=1}^n |x_j|\,|y_j|+\sum_{j=1}^n |y_j|^2. \end{align*} By the *Cauchy--Schwarz inequality*, \begin{align*} \sum_{j=1}^n |x_j|\,|y_j|\leq \left(\sum_{j=1}^n |x_j|^2\right)^{1/2}\left(\sum_{j=1}^n |y_j|^2\right)^{1/2}=\|x\|_2\|y\|_2. \end{align*} Therefore \begin{align*} \|x+y\|_2^2\leq \|x\|_2^2+2\|x\|_2\|y\|_2+\|y\|_2^2=(\|x\|_2+\|y\|_2)^2. \end{align*} Both sides are nonnegative, so taking square roots gives \begin{align*} \|x+y\|_2\leq \|x\|_2+\|y\|_2. \end{align*} Thus $\|\cdot\|_2$ is a norm on $\mathbb{R}^n$. Its induced metric is \begin{align*} d(x,y)=\|x-y\|_2=\left(\sum_{j=1}^n |x_j-y_j|^2\right)^{1/2}, \end{align*} which is the usual Euclidean distance. [/example] The same vector space has other common norms. [example: The $p$-Norms on Finite-Dimensional Space] For $1\leq p<\infty$, define \begin{align*} \|x\|_p=\left(\sum_{j=1}^n |x_j|^p\right)^{1/p} \end{align*} on $\mathbb{F}^n$, where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$. Since every term $|x_j|^p$ is nonnegative, $\|x\|_p=0$ implies $\sum_{j=1}^n |x_j|^p=0$, so $|x_j|^p=0$ for each $j$, hence $x_j=0$ for each $j$ and $x=0$. For a scalar $\lambda$, \begin{align*} \|\lambda x\|_p=\left(\sum_{j=1}^n |\lambda x_j|^p\right)^{1/p}=\left(\sum_{j=1}^n |\lambda|^p|x_j|^p\right)^{1/p}=\left(|\lambda|^p\sum_{j=1}^n |x_j|^p\right)^{1/p}=|\lambda|\|x\|_p. \end{align*} When $p=1$, the scalar triangle inequality gives $|x_j+y_j|\leq |x_j|+|y_j|$ for each $j$, and therefore \begin{align*} \|x+y\|_1=\sum_{j=1}^n |x_j+y_j|\leq \sum_{j=1}^n (|x_j|+|y_j|)=\sum_{j=1}^n |x_j|+\sum_{j=1}^n |y_j|=\|x\|_1+\|y\|_1. \end{align*} When $1<p<\infty$, *Minkowski's inequality* for finite sums gives \begin{align*} \|x+y\|_p=\left(\sum_{j=1}^n |x_j+y_j|^p\right)^{1/p}\leq \left(\sum_{j=1}^n |x_j|^p\right)^{1/p}+\left(\sum_{j=1}^n |y_j|^p\right)^{1/p}=\|x\|_p+\|y\|_p. \end{align*} Thus $\|\cdot\|_p$ is a norm for every $1\leq p<\infty$. For $p=\infty$, define \begin{align*} \|x\|_\infty=\max_{1\leq j\leq n}|x_j|. \end{align*} If $\|x\|_\infty=0$, then $0\leq |x_j|\leq \max_{1\leq k\leq n}|x_k|=0$ for each $j$, so $x_j=0$ for each $j$ and $x=0$. Homogeneity follows from \begin{align*} \|\lambda x\|_\infty=\max_{1\leq j\leq n}|\lambda x_j|=\max_{1\leq j\leq n}|\lambda|\,|x_j|=|\lambda|\max_{1\leq j\leq n}|x_j|=|\lambda|\|x\|_\infty. \end{align*} For the triangle inequality, set $M_x=\|x\|_\infty$ and $M_y=\|y\|_\infty$. For each $j$, \begin{align*} |x_j+y_j|\leq |x_j|+|y_j|\leq M_x+M_y. \end{align*} Taking the maximum over $j$ gives \begin{align*} \|x+y\|_\infty=\max_{1\leq j\leq n}|x_j+y_j|\leq M_x+M_y=\|x\|_\infty+\|y\|_\infty. \end{align*} Hence the finite-dimensional $p$-norms, including the maximum norm at $p=\infty$, all satisfy the norm axioms on $\mathbb{F}^n$. [/example] Function spaces supply the main infinite-dimensional examples. [example: Supremum Norm] Let $X$ be a set and let $B(X)$ be the vector space of bounded scalar-valued functions on $X$. For $f\in B(X)$, define \begin{align*} \|f\|_\infty=\sup_{x\in X}|f(x)|. \end{align*} This number is finite because $f$ is bounded, and it is nonnegative because $|f(x)|\geq 0$ for every $x\in X$. If $\|f\|_\infty=0$, then for every $x\in X$, \begin{align*} 0\leq |f(x)|\leq \sup_{u\in X}|f(u)|=0. \end{align*} Hence $|f(x)|=0$ for every $x\in X$, so $f(x)=0$ for every $x\in X$, and therefore $f=0$. For homogeneity, let $\lambda$ be a scalar. If $\lambda=0$, then $\lambda f=0$, so \begin{align*} \|\lambda f\|_\infty=\|0\|_\infty=0=|\lambda|\,\|f\|_\infty. \end{align*} If $\lambda\neq 0$, then for every $x\in X$, \begin{align*} |\lambda f(x)|=|\lambda|\,|f(x)|. \end{align*} Taking suprema and using that multiplication by the positive number $|\lambda|$ preserves least upper bounds gives \begin{align*} \|\lambda f\|_\infty=\sup_{x\in X}|\lambda f(x)|=|\lambda|\sup_{x\in X}|f(x)|=|\lambda|\,\|f\|_\infty. \end{align*} For the triangle inequality, let $f,g\in B(X)$. For every $x\in X$, the scalar triangle inequality gives \begin{align*} |f(x)+g(x)|\leq |f(x)|+|g(x)|. \end{align*} Since $|f(x)|\leq \|f\|_\infty$ and $|g(x)|\leq \|g\|_\infty$, we get \begin{align*} |f(x)+g(x)|\leq \|f\|_\infty+\|g\|_\infty. \end{align*} Thus $\|f\|_\infty+\|g\|_\infty$ is an upper bound for the set $\{|f(x)+g(x)|:x\in X\}$, so \begin{align*} \|f+g\|_\infty=\sup_{x\in X}|f(x)+g(x)|\leq \|f\|_\infty+\|g\|_\infty. \end{align*} Therefore $\|\cdot\|_\infty$ is a norm on $B(X)$. If $K$ is a compact [topological space](/page/Topological%20Space), then every continuous scalar-valued function on $K$ is bounded because its image is compact in $\mathbb{R}$ or $\mathbb{C}$, hence bounded. Thus $C(K)\subset B(K)$, and the same formula restricts to the supremum norm on $C(K)$. [/example] Sequence spaces form another family. [example: Sequence Norms] For $1\leq p<\infty$, the space $\ell^p$ consists of scalar sequences $x=(x_j)_{j\geq 1}$ such that \begin{align*} \sum_{j=1}^{\infty}|x_j|^p<\infty. \end{align*} Define \begin{align*} \|x\|_p=\left(\sum_{j=1}^{\infty}|x_j|^p\right)^{1/p}. \end{align*} Since each term $|x_j|^p$ is nonnegative, $\|x\|_p=0$ implies \begin{align*} \sum_{j=1}^{\infty}|x_j|^p=0. \end{align*} Hence $|x_j|^p=0$ for every $j$, so $x_j=0$ for every $j$, and therefore $x=0$. For a scalar $\lambda$, \begin{align*} \|\lambda x\|_p=\left(\sum_{j=1}^{\infty}|\lambda x_j|^p\right)^{1/p}=\left(\sum_{j=1}^{\infty}|\lambda|^p|x_j|^p\right)^{1/p}=|\lambda|\left(\sum_{j=1}^{\infty}|x_j|^p\right)^{1/p}=|\lambda|\|x\|_p. \end{align*} If $x,y\in \ell^p$, then *Minkowski's inequality for series* gives \begin{align*} \left(\sum_{j=1}^{\infty}|x_j+y_j|^p\right)^{1/p}\leq \left(\sum_{j=1}^{\infty}|x_j|^p\right)^{1/p}+\left(\sum_{j=1}^{\infty}|y_j|^p\right)^{1/p}. \end{align*} The right-hand side is finite because $x,y\in \ell^p$, so $x+y\in \ell^p$, and the same inequality says \begin{align*} \|x+y\|_p\leq \|x\|_p+\|y\|_p. \end{align*} Thus $\|\cdot\|_p$ is a norm on $\ell^p$. The space $\ell^\infty$ consists of bounded scalar sequences, and its norm is \begin{align*} \|x\|_\infty=\sup_{j\geq 1}|x_j|. \end{align*} If $\|x\|_\infty=0$, then for every $j$, \begin{align*} 0\leq |x_j|\leq \sup_{k\geq 1}|x_k|=0. \end{align*} Hence $x_j=0$ for every $j$, so $x=0$. For a scalar $\lambda$, \begin{align*} \|\lambda x\|_\infty=\sup_{j\geq 1}|\lambda x_j|=\sup_{j\geq 1}|\lambda|\,|x_j|=|\lambda|\sup_{j\geq 1}|x_j|=|\lambda|\|x\|_\infty. \end{align*} For $x,y\in \ell^\infty$, the scalar triangle inequality gives, for every $j$, \begin{align*} |x_j+y_j|\leq |x_j|+|y_j|\leq \|x\|_\infty+\|y\|_\infty. \end{align*} Thus $x+y$ is bounded, and taking the supremum over $j$ gives \begin{align*} \|x+y\|_\infty=\sup_{j\geq 1}|x_j+y_j|\leq \|x\|_\infty+\|y\|_\infty. \end{align*} Therefore $\|\cdot\|_\infty$ is a norm on $\ell^\infty$. These sequence norms turn convergence of sequences of scalars into convergence of points in a [normed vector space](/page/Normed%20Vector%20Space). [/example] These examples show why normed spaces are more than finite-dimensional geometry. They let us treat sequences and functions as points in a geometric space. Approximation of functions becomes approximation by vectors. Convergence of [Fourier series](/page/Fourier%20Series), polynomial approximation, and solutions of differential equations can then be studied using the same metric language. ## The Norm Topology The induced metric gives the usual topological notions. The main local object is the open ball, since open sets, continuity, and neighbourhood arguments are all expressed in terms of these balls. [definition: Open Ball] Let $(V,\|\cdot\|)$ be a normed space. For $x\in V$ and $r>0$, the open ball of centre $x$ and radius $r$ is the set \begin{align*} B(x,r)=\{y\in V:\|y-x\|<r\}. \end{align*} [/definition] Once balls are available, the next question is which subsets can be tested locally by these balls. Openness formalizes the idea that a point is safely inside a set: small enough errors, perturbations, or approximations do not leave the set. This is the local language in which continuity and approximation are usually phrased. [definition: Open Subset of a Normed Space] Let $(V,\|\cdot\|)$ be a normed space. A subset $U\subset V$ is open if, for every $x\in U$, there exists $r>0$ such that $B(x,r)\subset U$. [/definition] Openness describes stability under small moves from points already inside a set. For limiting arguments, the complementary question is whether a set keeps the limits of sequences it already contains. Closedness is the form of stability needed when approximations converge to an object that should remain admissible. [definition: Closed Subset of a Normed Space] Let $(V,\|\cdot\|)$ be a normed space. A subset $F\subset V$ is closed if $V\setminus F$ is open in the norm topology. [/definition] This topological definition is the one that survives in arbitrary topological vector spaces. In normed spaces, however, the metric structure gives an equivalent sequential test, and that is the form used most often in analysis. [quotetheorem:9985] The sequential criterion is often the most convenient way to prove closedness in practice. Instead of checking that every point outside the set has an open ball avoiding the set, one can take an arbitrary convergent sequence inside the set and verify that its limit remains inside. This is especially useful in function and sequence spaces, where admissibility conditions are commonly preserved by passing to limits. Open balls are too small for estimates that allow equality at the endpoint. When an argument controls an error by $\|y-x\|\leq r$ and then takes limits, the boundary case must remain part of the set being controlled. This motivates a companion to the open ball: instead of recording points strictly closer than a radius, we also include the boundary where the distance is exactly the radius. The resulting set is the basic geometric object for closed estimates in a normed space. [definition: Closed Ball] Let $(V,\|\cdot\|)$ be a normed space. For $x\in V$ and $r\geq 0$, the closed ball of centre $x$ and radius $r$ is the set \begin{align*} \overline{B}(x,r)=\{y\in V:\|y-x\|\leq r\}. \end{align*} [/definition] The open sets defined by a norm should be recognizable from local radius estimates, otherwise the topology would be hard to use in calculations. The key local question is whether membership in an [open set](/page/Open%20Set) can always be tested by fitting a small ball around each of its points. [quotetheorem:7784] This topology is called the norm topology. Once open balls are known to generate the topology, the next natural check is whether their non-strict analogues behave as closed sets. That fact lets norm estimates pass to limits without losing the boundary case $\|y-x\|=r$. [quotetheorem:9986] Closed balls are therefore stable under norm limits: if a sequence stays within a fixed error bound and converges, its limit still satisfies the same bound. This is the geometric reason closed balls appear in approximation arguments, compactness questions, and operator-norm estimates. The statement depends on using the non-strict inequality; open balls do not generally contain their boundary limits. Many estimates can be normalized by dividing by the relevant size, reducing the problem to vectors with norm at most one or less than one. Without special notation for these origin-centered radius-one sets, later statements about boundedness, operator norms, compactness, and norm comparison would repeatedly obscure the same normalization step. [definition: Unit Ball] Let $(V,\|\cdot\|)$ be a normed space. The closed unit ball of $V$ is \begin{align*} \overline{B}_V=\{x\in V:\|x\|\leq 1\}. \end{align*} The open unit ball of $V$ is \begin{align*} B_V=\{x\in V:\|x\|<1\}. \end{align*} [/definition] For many extremal arguments, vectors of size strictly less than one are not where the decisive behavior occurs; the sharp information lies on the boundary where the norm is exactly one. Naming that boundary makes it possible to discuss directions independently of scale. The next object isolates that boundary of the closed unit ball. It is the standard place to state sharp norm estimates, because every nonzero vector can be rescaled to lie on it. [definition: Unit Sphere] Let $(V,\|\cdot\|)$ be a normed space. The unit sphere of $V$ is \begin{align*} S_V=\{x\in V:\|x\|=1\}. \end{align*} [/definition] ## Boundedness and Operators ### Bounded Sets and Linear Maps Norms also define boundedness. This concept connects the size of subsets with estimates for linear maps, which are the basic morphisms between normed spaces. [definition: Bounded Subset] A subset $A$ of a normed space $V$ is bounded if there is a real number $M\geq 0$ such that \begin{align*} \|a\|\leq M \end{align*} for every $a\in A$. [/definition] For an arbitrary function, continuity is a local condition near each point. Linearity creates a stronger possibility: control near the origin might force one uniform inequality everywhere. The central question is whether topological continuity of a [linear map](/page/Linear%20Map) is exactly the same as a global bound on how much it can enlarge norms. [quotetheorem:3994] The estimate in the third condition is the usable form of continuity for linear maps: it turns topology into a single global inequality. When these conditions hold, $T$ is called a [bounded linear operator](/page/Bounded%20Linear%20Operator). The boundedness condition on a linear map is a genuine hypothesis in infinite-dimensional analysis, but finite-dimensional spaces are more rigid. There, a linear map has only finitely many coordinate directions to control, so one expects every linear map to satisfy a global norm estimate. [quotetheorem:9987] A bounded operator comes with at least one constant $C$ such that $\|Tx\|_W\leq C\|x\|_V$, but different constants may work. To measure the operator itself, we need the sharp amount by which it can stretch vectors, equivalently the largest output size produced on the unit ball. [definition: Operator Norm] Let $V$ and $W$ be normed spaces over the same scalar field, and let $\mathcal{L}(V,W)$ denote the vector space of bounded linear operators from $V$ to $W$. The [operator norm](/page/Operator%20Norm) is the function $\|\cdot\|_{\mathcal{L}(V,W)}:\mathcal{L}(V,W)\to [0,\infty)$ defined by \begin{align*} \|\cdot\|_{\mathcal{L}(V,W)}:\mathcal{L}(V,W)\to [0,\infty),\qquad T\mapsto \sup_{\|x\|_V\leq 1}\|Tx\|_W. \end{align*} [/definition] This definition packages the best constant in the inequality $\|Tx\|_W\leq C\|x\|_V$. When only one pair of spaces is in view, the shorter notation $\|T\|$ is often used for $\|T\|_{\mathcal{L}(V,W)}$. The next theorem makes precise the structural claim behind the notation: bounded operators can themselves be treated as vectors, with size measured by the operator norm. [quotetheorem:9988] This result lets estimates for individual operators become geometry inside an operator space. The simplest case is a one-dimensional scaling map, where the abstract supremum reduces to the familiar absolute value. [example: Operator Norm of a Scaling Map] Let $T:\mathbb{R}\to\mathbb{R}$ be the linear map $T(x)=ax$ for a fixed $a\in\mathbb{R}$, with the absolute value norm on both domain and codomain. By the definition of the operator norm, \begin{align*} \|T\|_{\mathcal{L}(\mathbb{R},\mathbb{R})}=\sup_{|x|\leq 1}\|T(x)\|=\sup_{|x|\leq 1}|ax|. \end{align*} For every $x$ with $|x|\leq 1$, multiplicativity of absolute value gives \begin{align*} |ax|=|a|\,|x|\leq |a|. \end{align*} Hence $|a|$ is an upper bound for the set $\{|ax|:|x|\leq 1\}$, so \begin{align*} \sup_{|x|\leq 1}|ax|\leq |a|. \end{align*} On the other hand, $x=1$ satisfies $|1|\leq 1$, and therefore \begin{align*} \sup_{|x|\leq 1}|ax|\geq |a\cdot 1|=|a|. \end{align*} Combining the two inequalities gives \begin{align*} \|T\|_{\mathcal{L}(\mathbb{R},\mathbb{R})}=|a|. \end{align*} Thus the operator norm of the scaling map is exactly the absolute value of the scaling factor, which is also its exact Lipschitz constant. [/example] Bounded subsets are the sets whose elements have uniformly controlled size. If a linear operator is bounded in the operator-norm sense, the natural test is whether it sends every such uniformly controlled family to another uniformly controlled family. The point is not just terminology. A set can contain infinitely many vectors, so controlling each vector separately would not by itself give one common bound for all of their images. What is needed is a single constant in the estimate $\|Tx\|_W\leq C\|x\|_V$; then one ball in the domain is carried into a scaled ball in the target. This raises the basic stability question for the operator-norm condition: does one global estimate automatically turn every bounded subset of the domain into a bounded subset of the target? The formal result records exactly this consequence. It is the direction most often used when applying an already-known bounded operator to bounded sequences, bounded balls, or uniformly controlled approximation families. [quotetheorem:9989] The hypothesis that the map is bounded is doing real work. A merely linear map on an infinite-dimensional normed space need not satisfy any global estimate, and then a [bounded set](/page/Bounded%20Set) can have images with no common norm bound. In applications, this theorem lets one push bounded sequences, bounded balls, and uniformly controlled approximation families through a bounded operator without rechecking the estimate element by element. ### Isometries and Isomorphisms Sometimes the right comparison between normed spaces is stronger than continuity: it preserves length exactly. This is the notion that says two normed spaces have the same metric geometry, possibly written in different coordinates. [definition: Linear Isometry] Let $V$ and $W$ be normed spaces over the same scalar field. A linear isometry from $V$ to $W$ is a linear map $T:V\to W$ such that \begin{align*} \|Tx\|_W=\|x\|_V \end{align*} for every $x\in V$. [/definition] A linear isometry may place one normed space inside another without reaching every point of the target. To say that two spaces have exactly the same metric geometry, the length-preserving map must also account for the whole target space. This motivates a stronger [equivalence relation](/page/Equivalence%20Relation) between normed spaces. The definition below adds surjectivity so that no points of the target space remain outside the length-preserving correspondence. [definition: Isometric Isomorphism of Normed Spaces] Let $V$ and $W$ be normed spaces over the same scalar field. An isometric isomorphism from $V$ to $W$ is a surjective linear isometry $T:V\to W$. [/definition] Exact preservation of lengths is often too rigid: two norms may define the same convergent sequences and continuous linear structure while assigning different numerical lengths. The right weaker equivalence requires a bijective linear correspondence whose estimates work in both directions. [definition: Isomorphism of Normed Spaces] Let $V$ and $W$ be normed spaces over the same scalar field. An isomorphism of normed spaces is a bijective linear map $T:V\to W$ such that $T$ and $T^{-1}:W\to V$ are bounded. [/definition] This definition is the operator-theoretic version of equivalent norms: it says that both spaces control each other's sizes through global estimates. Isometric isomorphisms preserve the exact geometry, while normed-space isomorphisms preserve the topology and linear structure. ## Completeness Normed spaces need not contain all limits of Cauchy sequences. This is the first major place where metric structure becomes more than decoration: an approximation scheme may produce a perfectly good [Cauchy sequence](/page/Cauchy%20Sequence) without producing an element of the space. The normed spaces where this failure cannot happen have a special name. [definition: Banach Space] A normed space is a [Banach space](/page/Banach%20Space) if it is complete with respect to the metric induced by its norm. [/definition] When a normed space is not complete, simply adjoining new points is not enough: the original distances must remain unchanged, and the enlarged space should contain exactly enough limit points to make Cauchy approximation converge. This is the universal repair supplied by a completion. [definition: Completion of a Normed Space] Let $(V,\|\cdot\|_V)$ be a normed space. A completion of $V$ is a Banach space $(\widehat{V},\|\cdot\|_{\widehat{V}})$ together with a linear isometry $i:V\to \widehat{V}$ such that $i(V)$ is dense in $\widehat{V}$. [/definition] The completion is the space obtained by adding exactly the missing limits of Cauchy sequences, up to the usual identification of completions by isometric isomorphism. Completeness is not automatic even over the allowed scalar fields $\mathbb{R}$ and $\mathbb{C}$. Let $c_{00}$ be the vector space of scalar sequences with only finitely many nonzero terms, equipped with the supremum norm. The sequence \begin{align*} x_n=\left(1,\frac{1}{2},\ldots,\frac{1}{n},0,0,\ldots\right) \end{align*} is Cauchy in the supremum norm, but its pointwise limit $(1,1/2,1/3,\ldots)$ is not in $c_{00}$. The space of polynomials on $[0,1]$ with the supremum norm is another non-complete example; its completion is $C([0,1])$ with the supremum norm. Finite dimension is the main setting where completeness stops being an extra hypothesis. The reason this deserves a theorem is that an arbitrary norm need not look Euclidean from its formula: it may be defined indirectly, by a maximum, or by a constraint. Once a basis is fixed, however, convergence of vectors is governed by finitely many scalar coordinates, and scalar completeness forces vector completeness. [quotetheorem:9990] This automatic completeness is a major reason finite-dimensional normed spaces behave more rigidly than infinite-dimensional ones. The finite-dimensional hypothesis is essential: in infinite dimension, a normed space can have Cauchy sequences whose natural limits live only in a completion. For example, the finitely supported sequence space $c_{00}$ with the supremum norm is not complete, because the Cauchy sequence with initial segments $(1,1/2,\ldots,1/n,0,\ldots)$ converges pointwise to a sequence that is not finitely supported. Thus completeness is not a formal consequence of the norm axioms; finite dimension supplies it by reducing Cauchy behavior to finitely many scalar coordinates. In later arguments this lets one take limits of finite-dimensional approximations without leaving the space, and the next norm-comparison results strengthen the same theme by showing that, in finite dimension, even the particular norm chosen does not change the convergence theory. ## Comparing Norms Finite-dimensional spaces admit many norms, but they determine the same convergence theory. The definition below captures the exact comparison needed for two norms to produce the same topology. [definition: Equivalent Norms] Let $V$ be a vector space. Two norms \begin{align*} \|\cdot\|_a,\|\cdot\|_b:V\to [0,\infty),\qquad x\mapsto \|x\|_a,\|x\|_b \end{align*} are equivalent if there are constants $m,M>0$ such that \begin{align*} m\|x\|_a\leq \|x\|_b\leq M\|x\|_a \end{align*} for all $x\in V$. [/definition] The point of the two-sided estimate is not just numerical comparison; it says that the identity map is Lipschitz in both directions between the two metric structures. Thus equivalent norms are also often called Lipschitz equivalent norms: each norm controls the other by a global constant. The next theorem makes that payoff explicit before the finite-dimensional result explains when such comparisons are automatic. [quotetheorem:9991] In infinite-dimensional spaces, proving such inequalities is a real analytical task and often fails. Finite-dimensional spaces are different: the theorem below says that no matter how a norm is chosen, it is trapped between constant multiples of any other norm. [quotetheorem:877] This statement fails in infinite dimension. For example, on $C([0,1])$, the $L^1$-type norm \begin{align*} \|f\|_1=\int_0^1 |f(t)|\,d\mathcal{L}^1(t) \end{align*} is not equivalent to the supremum norm. A concrete sequence of continuous spike functions shows the failure. Let $f_n\in C([0,1])$ satisfy $0\leq f_n\leq 1$, $f_n(0)=1$, and $f_n(t)=0$ for $t\geq 1/n$, with linear interpolation on $[0,1/n]$. Then $\|f_n\|_\infty=1$ while $\|f_n\|_1\leq 1/n$, so no lower comparison $m\|f\|_\infty\leq \|f\|_1$ can hold for all $f\in C([0,1])$. This distinction is one source of the richness of functional analysis. ## Constructions from Normed Spaces ### Subspaces and Closedness Many normed spaces are built from old ones. A linear subspace $Y\subset V$ inherits the restricted norm. The useful completeness statement requires the ambient space to be complete. Without that hypothesis, closedness inside $V$ does not by itself say whether $Y$ contains limits that would live in a larger completion. [quotetheorem:9992] A common obstruction appears when a subspace is dense but not closed. In that situation, sequences from the subspace can converge in the ambient Banach space to points outside the subspace, so the inherited norm cannot make all Cauchy approximation converge inside the subspace itself. [quotetheorem:9993] Hence non-closed subspaces of Banach spaces are not Banach in the inherited norm. ### Products Finite products can be normed in several equivalent ways. For two normed spaces $V$ and $W$, common product norms on $V\times W$ include \begin{align*} \|(v,w)\|_1=\|v\|+\|w\|. \end{align*} Another common product norm is \begin{align*} \|(v,w)\|_\infty=\max\{\|v\|,\|w\|\}. \end{align*} The reason these formulas are interchangeable for finite products is that they control exactly the same neighbourhoods. The next theorem records this as a topology statement rather than leaving it as a convention about notation. [quotetheorem:9994] The product-norm result justifies treating finite products of normed spaces as the expected product topological space. A sequence $(v_n,w_n)$ converges exactly when both coordinate sequences converge, and continuity into or out of a finite product can be checked coordinatewise. The result is specifically finite-product behavior; infinite products require additional choices and do not follow from the same simple norm comparisons. ### Quotients Quotients require a closed subspace to behave well as normed spaces. If the subspace is not closed, a nonzero coset can have distance $0$ from the origin, so the formula below may only define a seminorm. [definition: Quotient Norm] Let $(V,\|\cdot\|)$ be a normed space over $\mathbb{F}$, where $\mathbb{F}\in\{\mathbb{R},\mathbb{C}\}$, and let $Y\subset V$ be a closed linear subspace over $\mathbb{F}$. The quotient norm on $V/Y$ is the function $\|\cdot\|_{V/Y}:V/Y\to [0,\infty)$ defined by \begin{align*} \|\cdot\|_{V/Y}:V/Y\to [0,\infty),\qquad x+Y\mapsto \inf_{y\in Y}\|x-y\|. \end{align*} [/definition] The formula measures the distance from a representative $x$ to the subspace being collapsed. The theorem is needed because two hazards must be ruled out: the value must not depend on the chosen representative of the coset, and it must vanish only on the zero coset. The closedness of $Y$ is exactly what prevents a nonzero coset from having zero length. The quotient construction will be used as an actual normed-space construction, so one must also know how its basic geometry relates to the geometry of the original space. In particular, the open unit ball in the quotient should be describable in terms of the original open unit ball and the quotient map. This raises a new question about the quotient map itself: when a coset has small quotient norm, can it always be represented by an actual vector of small norm upstairs? The point is not only to verify that the quotient norm is well-defined, but to understand how neighborhoods are transported from $X$ to $X/Y$. The upcoming unit-ball theorem is needed because this transport of neighborhoods is the first place where the quotient norm is used geometrically rather than only as a formula. It identifies exactly which quotient-space vectors are close to zero by comparing them with representatives in the original space, so it turns the abstract infimum definition into a concrete statement about the quotient map. The following result answers this by identifying the quotient unit ball as the image of the original unit ball under the quotient map. It gives a practical test for which cosets have quotient norm less than one and records that passing to the quotient preserves the local scale of the space in this concrete sense. In the theorem, $D_X$ denotes the open unit ball of a normed space $X$, and $D_{X/Y}$ denotes the open unit ball of the quotient normed space $X/Y$. This is the same object previously denoted by $B_V$ when the ambient space was called $V$; the letter changes only to match the theorem's notation. The forward issue is whether the quotient map is geometrically compatible with these balls, not merely whether the quotient norm exists. If this compatibility failed, estimates in the quotient would not translate cleanly back to estimates in $X$, so the next formal result records the precise bridge between small representatives upstairs and small cosets downstairs. [quotetheorem:2631] The quotient theorem says that the quotient map is not only algebraically natural but also geometrically well aligned with the norm. Small cosets are precisely those that have small representatives upstairs, up to the open-unit-ball formulation. This is why quotient estimates can often be proved by choosing representatives in the original space and then passing down to $X/Y$. The closedness hypothesis remains essential. If the collapsed subspace is not closed, the same infimum formula can assign length zero to nonzero cosets, so the quotient construction produces only a seminormed object. That failure leads naturally to the broader language of seminorms, where zero length may occur away from the zero vector. ### Seminorms and Norms It is useful to compare norms with a nearby notion. Seminorms satisfy the linear estimates that make norms useful, but they may fail to distinguish all nonzero vectors. [definition: Seminorm] Let $V$ be a vector space over $\mathbb{R}$ or $\mathbb{C}$. A seminorm on $V$ is a function $p:V\to [0,\infty)$ such that, for all $x,y\in V$ and all scalars $\lambda$, the following conditions hold: \begin{align*} p(\lambda x)=|\lambda|p(x). \end{align*} The triangle inequality is \begin{align*} p(x+y)\leq p(x)+p(y). \end{align*} [/definition] Every norm is a seminorm. The missing condition is positive definiteness: a seminorm may have $p(x)=0$ for nonzero $x$. The failure of positive definiteness is not a small defect; it tells us that $p$ cannot separate certain vectors from the origin. To turn a seminorm into a norm, we need to locate precisely the vectors that have zero measured size, because those are the vectors that must be identified with $0$ in the quotient construction. [definition: Kernel of a Seminorm] Let $p:V\to [0,\infty)$ be a seminorm on a vector space $V$. The kernel of $p$ is the set \begin{align*} N_p=\{x\in V:p(x)=0\}. \end{align*} [/definition] The kernel is not just a set of exceptional vectors. Homogeneity and the triangle inequality force it to be a linear subspace, which means it is legitimate to quotient by it. The next theorem records the payoff: after identifying precisely the vectors that $p$ cannot distinguish from zero, the seminorm becomes a genuine norm. [quotetheorem:9995] Thus norms can often be obtained by identifying vectors that a seminorm cannot distinguish. ## Typical Uses Normed spaces organize many analytical questions. Approximation asks whether a subset is dense. For instance, polynomial approximation asks whether polynomials are dense in a function space under a chosen norm. Operator theory asks how linear maps behave with respect to norms. Spectral theory studies operators using both algebraic and metric information. Partial differential equations often place solutions in normed spaces where estimates can be proved. Probability and statistics use norms to measure convergence of random variables and functions. The chosen norm controls what kind of convergence is being discussed. [Uniform convergence](/page/Uniform%20Convergence), mean convergence, and square-mean convergence are different because they come from different norms. ## Common Pitfalls The norm is part of the structure, not extra notation. Changing the norm can change convergence, continuity, and completeness. In finite-dimensional real or complex spaces this change is mild because all norms are equivalent. In infinite-dimensional spaces it can be decisive. Another pitfall is to confuse bounded linear maps with all linear maps. The finite-dimensional theorem above says boundedness is automatic when the domain is finite-dimensional, but in infinite dimension discontinuous linear maps can exist when no further restrictions are imposed. In applications, boundedness is therefore a meaningful condition. It is also important to distinguish completeness from closedness. Completeness is a property of a metric space. Closedness is a property of a subset of a surrounding space. A closed subspace of a Banach space is Banach. A non-complete normed space can still be closed in itself. ## Beyond and Connected Topics Normed spaces sit between metric spaces and [inner product](/page/Inner%20Product) spaces. Every normed space is a metric space by $d(x,y)=\|x-y\|$. An [inner product space](/page/Inner%20Product%20Space) gives a norm by \begin{align*} \|x\|=\sqrt{\langle x,x\rangle}. \end{align*} Not every norm comes from an inner product. Over real or complex vector spaces, the parallelogram identity characterizes exactly the norms that come from an inner product through the polarization identity. For example, the Euclidean norm on $\mathbb{R}^n$ comes from an inner product. The $\ell^1$ and $\ell^\infty$ norms generally do not. This distinction matters in projection theory and [Hilbert space](/page/Hilbert%20Space) methods. Strict convexity is a later geometric refinement of normed spaces: it studies when the unit sphere contains no nontrivial line segments, so averages of distinct unit vectors fall inside the unit ball. That child topic depends on the unit ball and unit sphere language developed here. Normed spaces also lead to topological vector spaces. Every normed space is a [topological vector space](/page/Topological%20Vector%20Space). Some topological vector spaces are not normable. The normed setting is therefore a structured but flexible part of analysis. The finite-dimensional background belongs with Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra), where vector spaces and linear maps are developed before adding topology. The metric and completeness viewpoint connects with Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology), while the infinite-dimensional operator viewpoint continues in Androma, [Cambridge II Linear Analysis](/page/Cambridge%20II%20Linear%20Analysis). ## Summary A normed space is a vector space with a length function satisfying positivity, homogeneity, and the triangle inequality. The norm induces a translation-invariant metric and a compatible topology. It makes sense to discuss convergence, continuity, boundedness, completeness, and approximation. The central examples are finite-dimensional spaces, sequence spaces, and function spaces. Bounded linear operators are exactly the continuous linear maps between normed spaces. Complete normed spaces are Banach spaces. Finite-dimensional normed spaces are especially rigid, while infinite-dimensional normed spaces contain much of the structure studied in analysis. ## References For more context, see Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). For the linear background, see Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). For later functional-analytic developments, see Androma, [Cambridge II Linear Analysis](/page/Cambridge%20II%20Linear%20Analysis) and Androma, [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis).