Continuous linear operator

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.