Topological vector space

A topological vector space is a vector space $E$ over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, equipped with a topology such that vector addition and scalar multiplication are jointly continuous:The domains carry the product topology, and the scalar field carries its usual topology. Joint continuity permits both inputs to vary at once. Here the definition does not impose the Hausdorff condition; we state it separately when needed. ^[ref].

Every normed vector space is a topological vector space: the triangle inequality gives $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$, whilecontrols scalar multiplication near any $({\lambda }_0,x_0)$. More generally, a seminorm topology supplies many measurements of smallness, rather than one norm. Translation by any vector is a homeomorphism, so a neighbourhood basis at zero determines all neighbourhoods by translation.

The compatibility matters: on a nonzero real vector space, the discrete topology makes vector addition continuous, but not scalar multiplication. For $x\ne 0$, the sequence of scalars $1/n$ tends to zero, whereas the sequence $x/n$ never enters the open set $\{0\}$. Thus choosing a topology arbitrarily need not respect the linear algebra.

This framework lets functional analysis choose the appropriate notion of limit of a sequence. Norm topology measures one global error, weak topology tests linear functionals, and the Fréchet space of smooth functions controls every order of derivative. In each case the operations forming linear combinations are continuous, making approximation compatible with the underlying equations.

A dense subspace of a topological vector space $X$ is a linear subspace $D$ whose closure is all of $X$. It permits continuous operators to be determined or extended from values on $D$. For a normed space, every vector in $X$ is a limit of vectors from $D$.

A linear map from a metrizable space carrying a vector-space topology is continuous if it takes every null sequence to a null sequence. If continuity fails, choose a point in each shrinking basic neighborhood whose image stays outside a fixed neighborhood of zero. These points converge to zero but their images do not. For a Fréchet space described by increasing seminorms $p_j$, a discontinuous functional admits $p_j(f_j)\le 1/j$ and $|u(f_j)|\ge 1$.

A subset $B$ of a topological vector space is bounded when every neighborhood $U$ of zero absorbs it: $B\subseteq tU$ for all sufficiently large positive $t$. In a locally convex space defined by seminorms $p_{\lambda }$, this is equivalent to ${\sup }_{x\in B}{p}_{\lambda }(x)<\infty$ for every $\lambda$. In particular, a bounded subset of the Schwartz space has a uniform bound for every Schwartz seminorm; it need not have uniformly bounded supports.

A strict inductive limit topology on an increasing union $E={\bigcup }_n{E}_n$ is the finest locally convex topology compatible with the inclusions of the stages $E_n$. A linear map from $E$ is continuous exactly when each restriction to $E_n$ is continuous.

A topological vector space is metrizable when its topology is induced by a metric. At the origin it then has a countable neighborhood basis, so continuity of a linear map can be tested on sequences.

Every finite-dimensional real vector space has a unique Hausdorff topology making it a topological vector space. Choosing a basis identifies it with ordinary Euclidean space; every linear map between such spaces is continuous.

On a finite-dimensional vector space over a complete valued field, any two norms induce the same topology. Choosing a basis reduces the claim to comparison with the maximum of the absolute values of the coordinates.

For $x\in {\mathbb{R}}^n$, Cauchy-Schwarz inequality applied to $(|x_i|)$ and $(1)$ gives the upper bound. Equal absolute coordinate values attain its factor $\sqrt{n}$. Expanding $({\sum }_i|x_i|{)}^2$ gives the lower bound, and a vector with only one nonzero coordinate attains its factor one. Thus both constants are optimal. These Lp norms define the same finite-dimensional topology despite having different unit balls.

A locally convex space is a topological vector space with a neighbourhood basis at zero consisting of convex sets. Equivalently, its topology is a seminorm topology generated by some family $\mathcal{P}$ of seminorms. We allow non-Hausdorff spaces here; a separating family of seminorms gives precisely the Hausdorff case. Every normed vector space is locally convex, but smooth convergence generally requires a whole family of measurements.

The geometric and measurement descriptions fit together: seminorm sublevel sets are convex, while the Minkowski functional of an open, convex zero neighbourhood invariant under all scalar multiplications with $|\lambda |\le 1$ is a continuous seminorm. Such neighbourhoods can be chosen as a neighbourhood basis. ^[ref].

Local convexity makes Hahn-Banach theorem and separation of a point and an open convex set available. It turns geometric distinctions into continuous linear functionals, connecting convex sets to the continuous dual space. For a seminorm topology generated by $\mathcal{P}$, a linear functional $\ell$ is continuous exactly when there are $p_1,\dots ,p_m\in \mathcal{P}$ and $C>0$ withThus one continuous linear functional needs only finitely many of the available measurements. For example, point evaluation of a derivative on the space of smooth functions on a compact interval is controlled by a suitable derivative seminorm.

Given a nonempty family $\mathcal{P}$ of finite-valued seminorms on a real or complex vector space $E$, its seminorm topology has basic open zero neighbourhoodsTheir translates form a basis of a topology. Only finite intersections are required: an infinite intersection might impose a stronger topology. Equivalently, put $q(x)={\max }_i{p}_i(x)/{\epsilon }_i$; then $q$ is a seminorm and $U=\{q<1\}$. See Günther Hörmann, Advanced Functional Analysis, section 4, neighbourhood systems from seminorms for the general construction.

Each defining seminorm is continuous, since $|p(x)-p(y)|\le p(x-y)$. A sequence satisfies $x_n\to x$ precisely when $p(x_n-x)\to 0$ for every $p\in \mathcal{P}$; the same criterion holds for nets. Finitely many measurements can each meet a prescribed tolerance after a common index.

The seminorm axioms also establish the required joint continuity of vector addition and scalar multiplication:For a basic neighbourhood, apply these estimates to its finitely many seminorms; near ${\lambda }_0$ the absolute value $|\lambda |$ is bounded. This gives a locally convex space even when some directions are invisible to every measurement.

A family of seminorms separates points when every nonzero vector $z$ has $p(z)>0$ for some member $p$. Equivalently, its kernels of seminorms have intersection $\{0\}$, or any two distinct vectors $x,y$ have $p(x-y)>0$ for some $p$. Individual seminorms can have large kernels of seminorms; the whole family must have no common invisible direction.

The associated seminorm topology is Hausdorff if and only if the family separates points. Indeed, if $p(x-y)=a>0$, the neighbourhoods $\{u:p(u-x)<a/3\}$ and $\{v:p(v-y)<a/3\}$ are disjoint by the triangle inequality. Conversely, if every $p$ vanishes on a nonzero $z$, then zero and $z$ belong to exactly the same open sets and cannot be separated. This criterion keeps separation distinct from completeness: detecting all directions does not guarantee that all Cauchy sequences have limits.

Here the set is a nonempty compact metrizable convex subset of a Hausdorff locally convex space. The ambient space can be infinite-dimensional. Continuous affine functions separate its points, and weak-star compact dual balls with separable preduals provide examples. This general setting is needed for Choquet theorem.

Every point of a compact metrizable convex subset of a Hausdorff locally convex space is the barycenter of a Borel probability measure concentrated on its extreme points. The theorem is an existence assertion; uniqueness requires additional hypotheses such as a simplex structure.

For a real function in the $L^{\infty }$ unit ball, push uniform measure on $t\in [0,1]$ forward by the displayed sign-valued functions. In the weak-star topology this is a Borel probability concentrated on the extreme points, and Fubini's theorem shows that its barycenter is $f$. Weak-star compactness and separability of the predual $L^1$ place the example within Choquet theorem.

Maximize the integral of a continuous strictly convex function among probability measures with a fixed barycenter. The supporting measure lemma for affine upper envelopes implies that the maximizing measure has zero integral of the nonnegative envelope gap. Strict convexity makes this gap positive at every nonextreme point, so the measure is concentrated on the extreme boundary. Metrizability makes that boundary Borel and supplies the continuous strictly convex function.

For a bounded real function on a compact convex set, this envelope is the infimum of its continuous affine majorants. It is finite, concave and upper semicontinuous, and dominates the function. For continuous functions it describes the maximal integral among probability measures with a specified barycenter. Its homogeneity, subadditivity and affine-translation identities give the supporting measure lemma for affine upper envelopes.

The sublinear functional $p(g)=\mu (\overline{g})$ has a linear supporting functional taking the value $p(f)$ at a specified continuous $f$, by the Hahn-Banach theorem. It is positive and normalized because $p$ has these values on constants. The Riesz-Markov-Kakutani representation theorem gives the probability $\nu$. Testing affine functions and their negatives shows that $\mu$ and $\nu$ have the same barycenter.

The barycenter of a probability measure on a compact convex set is the point satisfying the displayed affine integral identities. It exists by approximating the measure by finitely supported measures and using compactness of their finite convex combinations. It is unique because continuous affine functions separate points. In a Banach-space setting it agrees with the appropriate vector integral when that integral is defined.

A seminorm on a real or complex vector space $E$ is a finite-valued function $p:E\to [0,\infty )$ satisfying absolute homogeneity and the triangle inequality:In particular $p(0)=0$ and $p(-x)=p(x)$. A seminorm is a norm exactly when $p(x)=0$ implies $x=0$: this is the extra definiteness of a norm axiom. The zero function is a valid seminorm. The missing definiteness is useful when a measurement deliberately ignores some directions, such as an individual coordinate or the constant term of a function whose derivative is measured.

For example, on $C^1([0,1])$ the function $p(f)={\sup }_{t\in [0,1]}|f^{\prime }(t)|$ is a seminorm that vanishes on every constant function. Adding the value measurement gives the norm $q(f)=|f(0)|+p(f)$: if $q(f)=0$, the fundamental theorem of calculus gives $f(t)=f(0)+{\int }_0^t{f}^{\prime }(s)ds=0$. A family of such partial measurements defines a seminorm topology.

The kernel of a seminorm is its zero set, a linear subspace by absolute homogeneity and the triangle inequality. The estimate $|p(x)-p(y)|\le p(x-y)$ shows that adding an element of $\ker p$ does not change $p$. Hence $\Vert x+\ker p\Vert =p(x)$ defines a norm on the quotient vector space $E/\ker p$.

On $E$ itself, $d_p(x,y)=p(x-y)$ is a pseudometric; it is a metric exactly when $p$ is a norm. In particular, zero distance in one pseudometric need not mean equality of vectors. A separating family of seminorms removes the common zero-distance ambiguity.

On ${\mathbb{R}}^2$, the coordinate seminorms $p_x$ and $p_y$ measure horizontal and vertical components separately. Their kernels of seminorms are the vertical and horizontal axes. Neither is a norm, but together they form a separating family of seminorms. A basic zero neighbourhood isEach condition alone is an unbounded strip; their intersection is an open rectangle. For $a=b=1$, this is the unit open ball of the maximum norm $\Vert (x,y){\Vert }_{\infty }=\max \{|x|,|y|\}$. For unequal tolerances it is the unit open ball of the weighted norm $\max \{|x|/a,|y|/b\}$. Thus the two coordinate measurements generate the usual topology of the plane.

For a numerical check, take $a=1$, $b=1/2$: $(0.8,0.4)$ passes both tests, $(0.8,0.7)$ passes only the horizontal test, and $(1.2,0.4)$ passes only the vertical test. A direction missed by one seminorm is detected by the other.

Given a bilinear pairing $⟨\cdot ,\cdot ⟩$ and a seminorm $p$, its dual seminorm isIt may be infinite when $y$ does not annihilate the kernel of $p$. Whenever it is finite, the defining inequality gives $|⟨x,y⟩|\le p(x)p^{\ast }(y)$.

A Fréchet space is a Hausdorff locally convex space whose topology is induced by a complete translation-invariant metric. Translation invariance means $d(x+z,y+z)=d(x,y)$; completeness means every Cauchy sequence converges in the space. Use a translation-invariant metric for this condition: an arbitrary compatible metric can change which sequences are Cauchy sequences. ^[ref].

Equivalently, the topology comes from a countable separating family of seminorms $(p_k{)}_{k\ge 0}$ and is complete in their common sense: if $p_k(x_n-x_m)\to 0$ as $n,m\to \infty$ for every fixed $k$, then some $x\in E$ satisfies $p_k(x_n-x)\to 0$ for every $k$. A useful compatible translation-invariant metric isIt is finite because the weights sum to one. The separating condition makes it a metric; without that condition the formula gives only a pseudometric. Its convergence in a metric space and Cauchy sequence conditions are exactly the conditions for all the $p_k$ separately: control finitely many terms, then bound the remaining tail by $2^{-(K+1)}$. Replacing $p_k$ by ${\max }_{0\le j\le k}{p}_j$ gives an increasing family with the same topology.

Every Banach space is a Fréchet space, using its norm as the single measurement. The space of smooth functions on a compact interval is a Fréchet space with derivative seminorms but with no single norm inducing its smooth topology. This is why Fréchet spaces are useful in differential equations: one can require accuracy in every order of derivative, while each particular estimate uses only finitely many orders.

The name honors Maurice Fréchet. His doctoral thesis was published as ^[ref]; an ^[ref] is available. This identifies a historical source without attributing the modern locally convex definition to that paper.

Here $C^{\infty }([0,1])$ means the real- or complex-valued functions continuous on $[0,1]$, smooth on $(0,1)$, with every interior derivative having a continuous extension to the endpoints; endpoint values of derivatives are these extensions. On this space of smooth functions on a compact interval, defineThese finite-valued seminorms increase with $k$ and separate points. In fact every $p_k$ is a norm, since it includes $p_0(f)=\Vert f{\Vert }_{\infty }$. The terminology seminorm includes norms; it does not assert that every measurement has a nontrivial kernel of a seminorm. Smooth convergence means $p_k(f_n-f)\to 0$ for every fixed $k$, equivalently uniform convergence of every derivative separately. It does not demand one tolerance or one convergence index that works for all orders simultaneously. The maximum convention is equivalent to the derivative supremum norm convention using a finite sum, since $p_k(f)\le {\sum }_{j=0}^k\Vert f^{(j)}{\Vert }_{\infty }\le (k+1)p_k(f)$.

For a worked example, $f(t)=t^3$ gives $p_0(f)=1$, $p_1(f)=3$, $p_2(f)=6$, and $p_k(f)=6$ for $k\ge 3$. The linear functional $\ell (f)=f^{\prime \prime }(1/2)$ obeys $|\ell (f)|\le p_2(f)$ and is continuous. Similarly, differentiation is a continuous linear operator $D:f\mapsto f^{\prime }$ for smooth convergence, because $p_k(Df)\le p_{k+1}(f)$.

The space of smooth functions on a compact interval is complete for this seminorm topology. A Cauchy sequence has uniform limits $g_j$ for its derivatives of each order. Passing to the limit in the fundamental theorem of calculus identitygives $g_j(t)-g_j(0)={\int }_0^t{g}_{j+1}(s)ds$, so $g_j^{\prime }=g_{j+1}$ and $g_0$ belongs to the same space. This is the completeness of the smooth-function metric argument; the metric there and the increasing-$p_k$ metric above induce the same smooth topology and the same Cauchy sequences.

Smooth convergence on a compact interval requires uniform convergence of the functions and of their derivatives of every order separately, as measured by the derivative seminorms. The need for several scales is visible inFor $n\ge 2$, $p_0(f_n)=1/n$ because $t=\pi /(2n)$ lies in $[0,1]$. The derivative has supremum norm one for every $n$, already at $t=0$. Thus $f_n$ converges uniformly to zero, but fails the test for smooth convergence to zero. Andreas Buttenschoen, Applied Functional Analysis, spaces of continuous and differentiable functions discusses this example and the family of smoothness measurements.

More generally, for any fixed $K\ge 0$, $h_n(t)=\sin (nt)/n^{K+1}$ has $p_K(h_n)\le 1/n$ but $p_{K+1}(h_n)=1$ for $n\ge 2$. If a norm $N$ induced the smooth topology, its continuity would give $N(f)\le C{p}_K(f)$ for some finite $K$, while continuity of $p_{K+1}$ for $N$ would give $p_{K+1}(f)\le C^{\prime }N(f)$. Substituting $h_n$ contradicts these two bounds. Smooth convergence therefore cannot be reduced to one norm. Its beauty is that every finite order can be controlled without fixing a final order in advance.

For an absorbing convex set $U$ containing zero, its Minkowski functional is $p_U(x)=\inf \{t>0:x\in tU\}$. If $U$ is open and convex, then $p_U$ is sublinear and $U=\{x:p_U(x)<1\}$.

For a bounded convex open set $U$ containing zero and invariant under $x\mapsto -x$, its Minkowski functional is a norm. Inner and outer Euclidean balls give positive finite bounds; symmetry and scaling give absolute homogeneity; convexity gives the triangle inequality. Openness makes $U=\{p_U<1\}$, while boundedness prevents nonzero vectors from having zero gauge.

If $C$ is a nonempty open convex subset of a real locally convex space and $x_0\notin C$, a continuous linear functional strictly separates them: after choosing a sign, $f(c)<f(x_0)$ for every $c\in C$. Apply the Hahn-Banach theorem to the Minkowski functional of a translate of $C$.

If $A$ and $B$ are disjoint nonempty convex subsets of a real locally convex space and $A$ is open, then some continuous linear functional $f$ satisfiesfor all $a\in A$ and $b\in B$.

If finitely many open convex subsets $K_1,\dots ,K_n$ of a locally convex space have empty intersection, some continuous linear map $T:X\to {\mathbb{R}}^{n-1}$ preserves that fact: ${\bigcap }_{i}T(K_i)=\varnothing$.

A dual pair $(E,F)$ consists of a real vector space $E$ and a vector space $F$ of linear functionals that separates the points of $E$. Its weak topology $\sigma (E,F)$ is the coarsest topology making every $f\in F$ continuous, and its continuous dual is exactly $F$.

For a separated dual pair, the continuous linear functionals for its weak topology are exactly evaluations by members of the paired space. Continuity bounds a functional on a neighbourhood defined by finitely many evaluations; scalar rescaling makes it vanish on their common kernel. It then factors through their finite-dimensional evaluation map and is a finite linear combination of them.

Let $Z\subseteq X^{\ast }$ separate points of a Banach space $X$. If the closed unit ball of $X$ is compact for $\sigma (X,Z)$ and $Z$ has the norm inherited from $X^{\ast }$, then the evaluation map $X\to Z^{\ast }$ is an isometric isomorphism. Goldstine theorem makes its unit-ball image weak-star dense, while the assumed compactness makes that image closed.

For $A\subseteq E$ in a dual pair $(E,F)$, define $A^{\circ }=\{f\in F:f(a)\le 1\text{ for all }a\in A\}$. Then

The product topology on $X\times Y$ is generated by the seminorms $(x,y)\mapsto p(x)$ and $(x,y)\mapsto q(y)$. Its continuous dual is canonically $X^{\ast }\oplus Y^{\ast }$.

A continuous linear map is a linear map that is continuous for the topologies on its domain and codomain. Between normed vector spaces, continuity is equivalent to boundedness.

A bounded linear operator between Banach spaces is completely continuous when it maps weakly convergent sequences to norm-convergent sequences. Equivalently, it maps weakly compact sets to norm-compact sets: the Eberlein-Šmulian theorem gives the forward direction, while a weakly convergent sequence together with its limit is weakly compact and gives the reverse direction. The inclusion $C(\mathbb{T})\to L^2(\mathbb{T})$ for finite Lebesgue measure is completely continuous by pointwise convergence, the Uniform boundedness principle and the dominated convergence theorem, but is not a compact operator because of the orthogonal Fourier modes.

Let $T:V\to W$ be a bounded linear operator with $V$ a Banach space. If every $y\in W$ has an approximate preimage $x$ with $\Vert x\Vert \le R\Vert y\Vert$ and $\Vert Tx-y\Vert \le k\Vert y\Vert$, where $k<1$, then every $y$ has an exact preimage with the displayed norm bound. Repeatedly correct the residual: the corrections have norms at most $R{k}^j\Vert y\Vert$ and form an absolutely convergent series; its image under $T$ equals $y$. This proves surjectivity but does not provide a bounded linear right inverse.

For $1\le p<\infty$, a bounded linear operator $T:E\to F$ between Banach spaces is absolutely p-summing if $({\sum }_j\Vert T{x}_j{\Vert }^p{)}^{1/p}\le C{\sup }_{\varphi \in B_{E^{\ast }}}({\sum }_j|\varphi (x_j){|}^p{)}^{1/p}$ for every finite family. The least $C$ is its p-summing norm ${\pi }_p(T)$. Direct substitution gives the ideal inequality ${\pi }_p(BTA)\le \Vert B\Vert {\pi }_p(T)\Vert A\Vert$ for compatible bounded linear operators.

A bounded linear operator $T:E\to F$ is 2-summing if ${\sum }_j\Vert T{x}_j{\Vert }^2\le C^2{\sup }_{\varphi \in B_{E^{\ast }}}{\sum }_j|\varphi (x_j){|}^2$ for every finite family; ${\pi }_2(T)$ is the least $C$. For operators between Hilbert spaces, this is equivalent to being a Hilbert-Schmidt operator, with ${\pi }_2(T)=\Vert T{\Vert }_{\mathrm{HS}}$. The upper bound follows by applying the Hilbert-Schmidt norm ideal inequality to the column operator of the family; testing finite orthonormal families gives the reverse bound.

For a finite-dimensional normed vector space $E$, represent a finite family by its column operator $A:{\ell }_2^m\to E$. Its operator norm is the family's weak square norm. Since $A=AQ$ for the orthogonal projection onto $(\ker A{)}^{\perp }$, summing $\Vert A{e}_j{\Vert }^2\le \Vert A{\Vert }^2\Vert Q{e}_j{\Vert }_2^2$ proves the upper bound $\sqrt{\dim E}$. In coordinates adapted to a maximal inscribed Euclidean or Hermitian ellipsoid, the John contact decomposition gives ${\sum }_r{c}_r{u}_r{u}_r^{\ast }=I$ and ${\sum }_r{c}_r=\dim E$. The vectors $\sqrt{c_r}{u}_r$ have total squared norm $\dim E$ and weak square norm one, giving equality. This holds over both real and complex scalars.

For an absolutely p-summing operator $T:E\to F$, there is a probability measure $\mu$ on the dual unit ball with its weak-star topology such that $\Vert Tx\Vert \le {\pi }_p(T)(\int |\varphi (x){|}^{p}d\mu (\varphi ){)}^{1/p}$. Thus $T$ factors through the closed span of the evaluation functions in $L^p(\mu )$, with the final operator having norm at most ${\pi }_p(T)$. Conversely this factorization implies the summing inequality by summing under the integral. Existence of $\mu$ follows from finite-dimensional Hahn-Banach separation theorem: failure for finitely many vectors would contradict the summing inequality for a nonnegative weighted family. Compactness of probability measures then combines the finite constraints. The final operator need only be defined on the indicated closed subspace, not on all of $L^p(\mu )$.

An absolutely summing operator is an absolutely p-summing operator with $p=1$. The identity on the infinite-dimensional l2 sequence space is not absolutely summing: the first $n$ coordinate vectors have sum of norms $n$, while their weak 1-norm is $\sqrt{n}$.

A bounded linear operator is strictly singular if it is bounded below on no infinite-dimensional subspace. Equivalently, every infinite-dimensional subspace contains unit vectors with arbitrarily small images. On a hereditarily indecomposable Banach space, an operator which is arbitrarily small on unit vectors in every finite-codimensional subspace is strictly singular: construct an infinite basic span on which its operator norm is small, then use the zero-angle property of hereditary indecomposability.

A bounded linear operator $T:L^1(\Omega )\to L^1(\Omega )$ preserves positivity when it takes nonnegative functions to nonnegative functions. By linearity it also preserves the pointwise order: $u\le v$ implies $Tu\le Tv$. For example, integration against a nonnegative kernel preserves positivity, as does the identity operator. This is order positivity, which differs from a positive quadratic form on a Hilbert space.

On a bounded connected Lipschitz domain, let $u\ge 0$ belong to the BV space, and let $T$ be a positivity-preserving operator with $T1\ne 0$. Writing $u=c1+(u-c1)$, $c=u_{\Omega }\ge 0$, the triangle inequality and Poincaré inequality for total variation give $c\Vert T1{\Vert }_1\le \Vert Tu{\Vert }_1+\Vert T\Vert \Vert u-c{\Vert }_1\le \Vert Tu{\Vert }_1+C_{\Omega }\Vert T\Vert \mathrm{TV}(u)$. Thus forward-image and variation bounds control the missing constant mode, and hence the full $BV$ norm.

For a matrix with nonnegative entries, row sums at most $R$ and column sums at most $C$ imply an ${\ell }^2$ operator norm at most $\sqrt{RC}$. Indeed the Cauchy-Schwarz inequality gives $|{\sum }_j{A}_{ij}{x}_j{|}^2\le ({\sum }_j{A}_{ij}){\sum }_j{A}_{ij}|x_j{|}^2$. Summing over $i$ bounds the result by $RC{\sum }_j|x_j{|}^2$. Integral kernels and positive weight functions give analogous forms. The symmetric case has equal row and column bounds.

An invertible bounded linear operator has a bounded inverse when $\Vert u\Vert \le C\Vert Ku\Vert$ for all $u$.

A linear map $U:C(K,\mathbb{R})\to C(K,\mathbb{R})$ is positive when $f\ge 0$ implies $Uf\ge 0$. This is positivity for the pointwise order, not positivity of a quadratic form. On a compact space, $|Uf|\le U|f|\le \Vert f{\Vert }_{\infty }U1$, so $\Vert U\Vert =\Vert U1{\Vert }_{\infty }$. This makes positive maps especially useful for uniform approximation.

If a linear operator on real continuous functions preserves every zero, evaluate $f-f(x)1$ at $x$. The zero-preserving hypothesis forces $Lf(x)=f(x)L1(x)$, so the operator is multiplication by the continuous function $L1$. Positivity further makes that multiplier nonnegative; linearity and zero preservation alone already give the multiplication representation.

A bounded linear operator between normed spaces is contractive if its operator norm is at most one. This norm property is distinct from a strict contraction mapping, whose Lipschitz constant must be less than one.

The operator range is the image of a linear map $K$. It need not be closed in an infinite-dimensional Hilbert space; that failure makes the Moore–Penrose inverse of an operator unbounded on its domain.

Let $D$ be a dense linear subspace of a normed space $X$, let $Y$ be a Banach space, and let $T:D\to Y$ be linear with $\Vert Tx\Vert \le C\Vert x\Vert$. Then $T$ has a unique bounded extension $\tilde{T}:X\to Y$, defined by $\tilde{T}x={\lim }_{n}T{x}_n$ for any sequence $x_n\in D$ converging to $x$, and $\Vert \tilde{T}\Vert \le C$.

Let $c_{00}$ have the ${\ell }^2$ norm, take $D=c_{00}\subset X={\ell }^2$, and let $T:D\to c_{00}$ be the identity. It is bounded but has no continuous extension ${\ell }^2\to c_{00}$, because continuity and density would force the extension to be the identity on every element of ${\ell }^2$.

If $X$ is a normed vector space and $Y$ is a Banach space, then the bounded linear maps $\mathcal{B}(X,Y)$ form a Banach space under the operator norm. Completeness of the domain is unnecessary.

A continuous linear functional is a continuous linear map from a topological vector space to its scalar field. The collection of all such functionals is the continuous dual space.

A linear functional on a normed vector space is continuous exactly when its kernel is closed. The forward direction follows because zero is closed in the scalar field. Conversely, if $l\ne 0$, choose $v$ with $l(v)=1$ and write $M=\ker l$. Closedness gives $d=\mathrm{dist}(v,M)>0$. Since $x/l(x)-v\in M$ when $l(x)\ne 0$, $|l(x)|\le \Vert x\Vert /d$. The zero functional is continuous separately. Thus the criterion does not require completeness or the closed graph theorem.

For a function space on a set $X$ and a point $x\in X$, the point evaluation functional is ${\delta }_x(f)=f(x)$. It is bounded on $C(X)$ with the supremum norm.

A linear map from a metrizable topological vector space is continuous exactly when it maps every sequence converging to zero to a sequence converging to zero. If continuity fails, a decreasing countable neighborhood basis supplies a null sequence whose images stay outside one fixed neighborhood of zero.

An isomorphism of Banach spaces is a bijective bounded linear map whose inverse is bounded. The bounded inverse theorem makes boundedness of the inverse automatic.