Continuous linear operator
A continuous linear operator, also known as a bounded linear operator, is a linear transformation T: V \to W between normed vector spaces V and W that is continuous with respect to the topologies induced by their norms.[1] This means that if a sequence \{v_n\} in V converges to v \in V, then T(v_n) converges to T(v) in W.[1] Equivalently, T is continuous if and only if it is bounded, i.e., there exists a constant C \geq 0 such that \|T(v)\|_W \leq C \|v\|_V for all v \in V.[1][2] The smallest such C is the operator norm \|T\| = \sup_{\|v\|_V \leq 1} \|T(v)\|_W, which equips the space B(V, W) of all bounded linear operators from V to W with a norm, making it a normed vector space.[1] If W is a Banach space, then B(V, W) is itself a Banach space.[1][2] Continuous linear operators are central to functional analysis, enabling the study of infinite-dimensional spaces and their duals, where the dual space X^* consists of all continuous linear functionals on X.[3] Key theorems include the open mapping theorem, which states that a surjective continuous linear operator between Banach spaces is an open map; if the operator is also injective, then it admits a bounded inverse; the closed graph theorem, asserting that a linear operator with a closed graph between Banach spaces is continuous; and the uniform boundedness principle, which guarantees that pointwise bounded families of operators are uniformly bounded.[2][3] These operators arise in applications such as integral operators on function spaces, like T f(x) = \int_0^1 K(x,y) f(y) \, dy for continuous kernel K, which are bounded with norm at most \|K\|_\infty, and in spectral theory for self-adjoint operators on Hilbert spaces.[1] They underpin solutions to partial differential equations, quantum mechanics (e.g., deriving the Heisenberg uncertainty principle), and approximation theory in Banach spaces.[3]Definitions and preliminaries
Linear operators on normed spaces
In functional analysis, a linear operator between two normed vector spaces X and Y over the same field \mathbb{K} (typically \mathbb{R} or \mathbb{C}) is a function T: X \to Y satisfying T(\alpha x + \beta y) = \alpha T(x) + \beta T(y) for all scalars \alpha, \beta \in \mathbb{K} and vectors x, y \in X.[4] This linearity preserves the vector space structure, making the set \mathcal{L}(X, Y) of all such operators a vector space under pointwise addition and scalar multiplication.[5] Normed spaces equip these operators with additional structure via their norms, enabling the study of their magnitudes and behaviors without invoking topology at this stage. The operator norm of a linear operator T \in \mathcal{L}(X, Y) quantifies its "size" and is defined as \|T\| = \sup \left\{ \|T x\|_Y : x \in X, \|x\|_X \leq 1 \right\} = \sup \left\{ \frac{\|T x\|_Y}{\|x\|_X} : x \in X, x \neq 0 \right\}, which is always finite or infinite, though the focus here is on cases where it is finite.[6] This norm induces a normed space structure on \mathcal{L}(X, Y), turning it into a normed vector space itself, and satisfies the submultiplicative property \|S \circ T\| \leq \|S\| \|T\| for composable operators S and T.[4] Representative examples illustrate these concepts. Consider the differentiation operator D on the normed space of polynomials of degree at most n, P_n(\mathbb{R}), equipped with the sup norm \|p\| = \sup_{x \in [-1,1]} |p(x)|; here D: P_n \to P_{n-1} maps p \mapsto p' and satisfies linearity, with \|D\| = n on this finite-dimensional setting.[7] Another example is the Volterra integration operator V on the space C[0,1] of continuous functions on [0,1] with the sup norm, defined by (V f)(x) = \int_0^x f(t) \, dt; this is linear and has operator norm \|V\| = 1. The foundational ideas for linear operators on normed spaces trace back to David Hilbert's 1904 investigations into integral equations, where he analyzed mappings akin to these operators in infinite-dimensional settings to solve physical problems like potential theory.[8] Continuity of such operators, as a topological property, will be addressed separately using the norms defined here.Notion of continuity
In topology, a map f: X \to Y between topological spaces X and Y is continuous at a point x_0 \in X if for every neighborhood V of f(x_0) in Y, there exists a neighborhood U of x_0 in X such that f(U) \subseteq V.[9] The map f is continuous on X if it is continuous at every point x \in X.[9] When X and Y are metric spaces with metrics d_X and d_Y, respectively, this topological notion aligns with the familiar \epsilon-\delta definition from analysis: f is continuous at x_0 if for every \epsilon > 0, there exists \delta > 0 such that d_X(x, x_0) < \delta implies d_Y(f(x), f(x_0)) < \epsilon for all x \in X.[10] In the context of normed linear spaces, where the metrics are induced by norms \|\cdot\|_X and \|\cdot\|_Y, the definition takes the form: f is continuous at x_0 if for every \epsilon > 0, there exists \delta > 0 such that \|x - x_0\|_X < \delta implies \|f(x) - f(x_0)\|_Y < \epsilon.[11] For a linear operator T: X \to Y between normed linear spaces, continuity at a single point implies continuity at every point in X, owing to the homogeneity and additivity of T.[12] In particular, T is continuous on X if and only if it is continuous at the zero vector $0 \in X.[12] This reduces to the condition that for every \epsilon > 0, there exists \delta > 0 such that \|x\|_X < \delta implies \|T(x)\|_Y < \epsilon.[1] \|x\|_X < \delta \implies \|T(x)\|_Y < \epsilonCharacterizations of continuity
Sequential continuity
In topological vector spaces, a linear operator T: X \to Y is said to be sequentially continuous at a point x_0 \in X if, for every sequence (x_n) in X converging to x_0, the sequence (T(x_n)) converges to T(x_0) in Y.[1] This notion aligns with the general ε-δ definition of continuity when specialized to sequences, providing a practical characterization in spaces where sequences suffice to probe limits. For linear operators between normed spaces, sequential continuity at any point x_0 is equivalent to sequential continuity at the origin $0. To see this, suppose T is sequentially continuous at x_0. For any sequence (y_n) converging to $0, the sequence (y_n + x_0) converges to x_0, so T(y_n + x_0) \to T(x_0). By linearity, T(y_n) + T(x_0) \to T(x_0), which implies T(y_n) \to 0. Conversely, if T is sequentially continuous at $0, and (x_n) \to x_0, then (x_n - x_0) \to 0, so T(x_n - x_0) \to 0, and thus T(x_n) = T(x_n - x_0) + T(x_0) \to T(x_0) by linearity.[2] Normed spaces are first-countable, as they carry a metric topology generated by the norm, and in such spaces, sequential continuity coincides with the general notion of continuity for maps, including linear operators.[13] Moreover, if (x_n) \to x_0 in a normed space X and T: X \to Y is sequentially continuous, then \|T(x_n)\| \to \|T(x_0)\| in the normed codomain Y, since convergence of vectors implies convergence of their norms via the reverse triangle inequality: \big| \|T(x_n)\| - \|T(x_0)\| \big| \leq \|T(x_n) - T(x_0)\| \to 0.[2]Continuity at zero
A fundamental property of linear operators between normed vector spaces is that their continuity at a single point, specifically the origin, determines continuity across the entire domain. Let X and Y be normed vector spaces over the same field, and let T: X \to Y be a linear operator. Then T is continuous if and only if it is continuous at $0 \in X.[14] To see this, first note that if T is continuous on X, then it is continuous at every point, including the origin, by the definition of continuity on the whole space. Conversely, suppose T is continuous at $0. For any x \in X and \varepsilon > 0, there exists \delta > 0 such that \|h\| < \delta implies \|T h\| < \varepsilon, since T(0) = 0. Now consider continuity at x: for the same \varepsilon > 0, if \|h\| < \delta, then \|T(x + h) - T(x)\| = \|T h\| < \varepsilon. Thus, the same \delta works for every x, establishing continuity at x. This \delta depends only on \varepsilon and not on x, reflecting the uniform nature of the continuity implied by linearity. In some formulations, to explicitly construct a modulus of continuity at x, one may scale as \delta(\varepsilon, x) = \delta\left( \frac{\varepsilon}{1 + \|x\|} \right), ensuring the neighborhood scales appropriately with the position, though the unscaled \delta suffices due to the translation invariance from linearity.[14][15] This equivalence highlights the simplifying role of linearity in topological vector spaces more broadly, where continuity at zero implies continuity everywhere via neighborhood translation: if V is a neighborhood of T(x), then U = V - T(x) is a neighborhood of $0, and continuity at zero yields a neighborhood W of $0 with T(W) \subseteq U, so T(W + x) = T(W) + T(x) \subseteq V.[15] Unlike linear operators, this property does not hold for nonlinear maps. For example, consider the function f: \mathbb{R} \to \mathbb{R} defined by f(x) = x if x is rational and f(x) = 0 if x is irrational. This f is continuous at $0 because |f(x) - f(0)| = |f(x)| \leq |x| for all x, so given \varepsilon > 0, take \delta = \varepsilon. However, f is discontinuous at every x \neq 0, as sequences of rationals and irrationals approaching x yield limits x and $0, respectively, which differ.[16]Open mapping characterizations
The open mapping theorem provides a key topological characterization of continuous linear operators between Banach spaces. Specifically, if T: X \to Y is a continuous linear surjection between Banach spaces X and Y, then T maps open sets in X to open sets in Y.[17] This result, originally established by Stefan Banach, asserts that such operators are open mappings, meaning the image of every nonempty open subset of X under T contains a nonempty open subset of Y.[18] The theorem highlights the interplay between continuity, surjectivity, and topological openness in complete normed spaces. A broader characterization holds for linear operators between general topological vector spaces: a linear operator T: X \to Y is continuous if and only if it maps open sets in X to relatively open sets in its image T(X), or equivalently, T is open onto its range.[15] This equivalence follows from the fact that continuity of T implies the preimage of any open set in Y is open in X, and restricting to the subspace topology on T(X) preserves openness of images. Conversely, if T is open onto T(X), then for any open V \subset Y, the set T^{-1}(V) = T^{-1}(V \cap T(X)) is open because V \cap T(X) is open in T(X) and its preimage under T is open by assumption. This characterization extends the pointwise notion of continuity at zero to a global topological property without requiring completeness.[17] The proof of the open mapping theorem relies on the Baire category theorem applied to the completeness of Banach spaces. To show that T is open, it suffices to verify that T maps the open unit ball B_X(0,1) in X onto a set containing an open neighborhood of the origin in Y. Consider the sets T(B_X(0, n)) for n \in \mathbb{N}; since T is surjective, their union covers Y. By the Baire category theorem, one of these sets has nonempty interior, implying that the image of the unit ball absorbs a neighborhood of zero in Y. More precisely, there exists c > 0 such that B_Y(0,1) \subseteq c \, T(B_X(0,1)), where B_Y(0,1) and B_X(0,1) denote the open unit balls in Y and X, respectively. This inclusion ensures that T maps open neighborhoods to open sets, confirming openness.[17] The constant c quantifies how the operator "covers" the codomain, providing a measure of the surjectivity in topological terms.Boundedness equivalence
Bounded implies continuous
A linear operator T: X \to Y between normed vector spaces X and Y is said to be bounded if there exists a constant M > 0 such that \|T(x)\|_Y \leq M \|x\|_X for all x \in X.[19] This condition ensures that the operator does not amplify norms excessively across the entire domain. The following theorem establishes that boundedness implies continuity for linear operators.[19] Theorem. Let T: X \to Y be a bounded linear operator between normed vector spaces. Then T is continuous at every point in X, and in particular at the origin. To prove continuity at the origin, fix \varepsilon > 0. Since T is bounded, choose \delta = \varepsilon / M > 0. For any x \in X with \|x\|_X < \delta, linearity and boundedness yield \|T(x)\|_Y \leq M \|x\|_X < M \delta = \varepsilon. Thus, T is continuous at $0. For an arbitrary point x_0 \in X, consider \|T(x) - T(x_0)\|_Y = \|T(x - x_0)\|_Y. If \|x - x_0\|_X < \delta, then \|T(x - x_0)\|_Y < \varepsilon by the preceding argument, so T is continuous at x_0.[19] Boundedness imposes a uniform control on the operator's action over the whole space via the constant M, which is the operator norm \|T\| = \sup_{\|x\|_X \leq 1} \|T(x)\|_Y, in contrast to general continuity that requires only local behavior without such global uniformity.[19]Continuous implies bounded
A fundamental result in functional analysis establishes that continuity and boundedness are equivalent for linear operators between normed vector spaces. Specifically, if T: X \to Y is a linear operator between normed spaces X and Y, then T is continuous if and only if it is bounded, meaning there exists a constant M \geq 0 such that \|T x\|_Y \leq M \|x\|_X for all x \in X. The direction "bounded implies continuous" follows from the definition of continuity via \epsilon-\delta, as the bound directly controls the image of small neighborhoods. The converse direction, that continuity implies boundedness, is proved using the property of continuity at the origin.[1] To see this, since T is continuous at $0 \in X, for \epsilon = 1 > 0, there exists \delta > 0 such that if \|x\|_X < \delta, then \|T x\|_Y < 1. Now consider an arbitrary x \in X with x \neq 0. Define z = \frac{\delta / 2}{\|x\|_X} x; then \|z\|_X = \delta / 2 < \delta, so \|T z\|_Y < 1. But T z = \frac{\delta / 2}{\|x\|_X} T x, which implies \|T x\|_Y = \frac{2 \|x\|_X}{\delta} \|T z\|_Y < \frac{2 \|x\|_X}{\delta}. For x = 0, the inequality holds trivially. Thus, \|T x\|_Y \leq \frac{2}{\delta} \|x\|_X for all x \in X, so T is bounded with operator norm \|T\| \leq 2/\delta. This bound shows that the supremum of \|T x\|_Y over the unit ball in X is finite.[3] This equivalence characterizes linear operators on normed spaces, where the norm provides a quantitative measure of boundedness. In more general topological vector spaces without a compatible norm, such as those equipped with the indiscrete topology, the implication holds in the weaker sense that continuous linear operators map bounded sets (in the topological sense) to bounded sets, but the specific norm-boundedness fails to apply due to the absence of a norm structure.[20]Bounded on neighborhoods
A linear operator T: X \to Y between normed linear spaces X and Y is bounded on a neighborhood U of the origin if there exists a finite constant M > 0 such that \|T x\|_Y \leq M for all x \in U, or equivalently, \sup \{ \|T x\|_Y : x \in U \} < \infty.[1] This local form of boundedness captures the behavior of T near zero without requiring control over the entire space. In normed spaces, boundedness on some neighborhood of the origin implies global boundedness on the unit ball, and thus continuity of T. Specifically, suppose T is bounded on the open ball B_X(0, r) = \{ x \in X : \|x\|_X < r \} for some r > 0, with \sup_{\|x\|_X < r} \|T x\|_Y = M < \infty. For arbitrary x \in X with x \neq 0, consider y = \frac{r}{2 \|x\|_X} x. Then \|y\|_X < \frac{r}{2} < r, so \|T y\|_Y \leq M. By linearity, T y = \frac{r}{2 \|x\|_X} T x, which rearranges to \|T x\|_Y = \frac{2 \|x\|_X}{r} \|T y\|_Y \leq \frac{2 M}{r} \|x\|_X. Thus, T satisfies \|T x\|_Y \leq K \|x\|_X for all x \in X with K = \frac{2 M}{r} < \infty, establishing global boundedness.[1][20] This implication relies on the homogeneity and scaling properties of norms, which allow the local bound to be extended unitarily across the space. In normed spaces, the converse also holds: every globally bounded (and hence continuous) linear operator is bounded on every neighborhood of the origin, as neighborhoods are scaled versions of the unit ball. However, in more general topological vector spaces that are not locally bounded (such as certain non-metrizable or non-locally convex spaces), continuous linear operators are always locally bounded but need not be globally bounded, highlighting a distinction absent in the normed setting.[21]Continuous linear functionals
Hahn-Banach theorem applications
The Hahn-Banach theorem provides a cornerstone for extending linear functionals while preserving certain bounds, playing a pivotal role in the theory of continuous linear operators on normed spaces. Independently established by Hans Hahn in 1927 and Stefan Banach in 1929, the theorem asserts that if X is a normed vector space over \mathbb{R} or \mathbb{C}, M \subseteq X is a subspace, f: M \to \mathbb{K} (where \mathbb{K} = \mathbb{R} or \mathbb{C}) is a linear functional, and p: X \to [0, \infty) is a sublinear functional satisfying |f(x)| \leq p(x) for all x \in M, then there exists a linear extension \tilde{f}: X \to \mathbb{K} such that |\tilde{f}(x)| \leq p(x) for all x \in X.[22][23] This extension is not unique in general but can be chosen to maintain continuity when f is continuous and p is continuous, such as the seminorm induced by the space's norm.[24] A primary application of the Hahn-Banach theorem arises in the context of continuous linear functionals, enabling the extension of any such functional defined on a subspace to the entire space without altering its continuity properties. Specifically, if \phi: M \to \mathbb{C} is a continuous linear functional on a subspace M of a normed space X, then it admits an extension \tilde{\phi}: X \to \mathbb{C} that remains continuous on X with the same operator norm \|\tilde{\phi}\| = \|\phi\|.[23] To achieve this, one sets p(x) = \|\phi\| \cdot \|x\|, which is sublinear since it satisfies p(\lambda x) = |\lambda| p(x) for scalars \lambda and p(x + y) \leq p(x) + p(y), ensuring the bound |\phi(m)| \leq p(m) holds for m \in M by the continuity of \phi.[24] This preservation of the norm is crucial, as it guarantees that the extended functional respects the original's boundedness, linking directly to the equivalence between continuity and boundedness for linear operators.[23] The operator norm of a continuous linear functional \phi: X \to \mathbb{C} on a normed space X is defined as \|\phi\| = \sup_{\|x\| \leq 1} |\phi(x)|, which equals \sup_{x \neq 0} \frac{|\phi(x)|}{\|x\|} and satisfies |\phi(x)| \leq \|\phi\| \cdot \|x\| for all x \in X.[23] In Hahn-Banach extensions, this norm is explicitly preserved: if \phi is defined on M with \|\phi\| = \sup_{\|m\| \leq 1, m \in M} |\phi(m)|, the extension \tilde{\phi} satisfies \|\tilde{\phi}\| = \|\phi\|, ensuring no inflation of the bound during extension.[24] This norm-preserving property underpins the construction of dual spaces by allowing arbitrary continuous functionals on subspaces to be lifted to the full space, facilitating the identification of the dual X^* with bounded linear functionals while maintaining algebraic and topological consistency.[23]Dual spaces
The dual space of a normed linear space X, denoted X^*, consists of all continuous linear functionals on X. Each element \phi \in X^* is a linear map \phi: X \to \mathbb{K} (where \mathbb{K} is the scalar field, typically \mathbb{R} or \mathbb{C}) that is continuous with respect to the norm topology on X. The dual space X^* is equipped with the dual norm defined by \|\phi\| = \sup \left\{ \frac{|\phi(x)|}{\|x\|} : x \in X, \, x \neq 0 \right\}, which is equivalent to \sup \{ |\phi(x)| : \|x\| \leq 1 \}. This norm makes X^* a Banach space, regardless of whether X itself is complete.[25] The Hahn-Banach theorem provides a method for constructing non-zero elements in X^* by extending linear functionals while preserving continuity and norm bounds. The double dual X^{**} is the dual of X^*, and there is a canonical isometric embedding J: X \to X^{**} given by Jx(\phi) = \phi(x) for \phi \in X^*, which identifies X with a closed subspace of X^{**}.[25] A normed space X is reflexive if and only if this embedding J is surjective, meaning X is isometrically isomorphic to X^{**}. Hilbert spaces, such as L^2 spaces, are reflexive, as their duals coincide with themselves via the Riesz representation theorem (though the focus here is on the general embedding property). In contrast, the space c_0 of sequences converging to zero is not reflexive, since its double dual properly contains it.[25] The dimension of X^* is at least that of X, with equality holding if and only if X is finite-dimensional. For infinite-dimensional spaces, the dimension of X^* is typically strictly larger; for instance, the dual of \ell^\infty, the space of bounded sequences, has uncountable dimension exceeding that of \ell^\infty itself.[26]Examples and counterexamples
Standard examples
One fundamental example of a continuous linear operator is the identity operator \mathrm{Id}: X \to X on a normed vector space X, defined by \mathrm{Id}(x) = x for all x \in X. This operator is continuous, as it is bounded with operator norm \|\mathrm{Id}\| = 1, since \|\mathrm{Id}(x)\| = \|x\| \leq 1 \cdot \|x\| for all x \in X.[2] Multiplication operators provide another standard class of continuous linear operators. On the L^p space for $1 \leq p \leq \infty, the operator M_k: L^p(\Omega) \to L^p(\Omega) defined by (M_k f)(x) = k(x) f(x) for a measurable function k: \Omega \to \mathbb{C} is continuous if and only if k is essentially bounded, i.e., k \in L^\infty(\Omega), in which case the operator norm satisfies \|M_k\| = \|k\|_\infty.[25] Continuous linear functionals offer simple illustrations in specific spaces. For instance, on the space C[0,1] of continuous functions on [0,1] equipped with the supremum norm \|f\|_\infty = \sup_{t \in [0,1]} |f(t)|, the evaluation functional \phi_a: C[0,1] \to \mathbb{C} given by \phi_a(f) = f(a) for fixed a \in [0,1] is continuous with norm \|\phi_a\| = 1, as |f(a)| \leq \|f\|_\infty for all f \in C[0,1]. Integral operators serve as key examples of continuous linear functionals on L^2 spaces. Consider \phi: L^2([0,1]) \to \mathbb{C} defined by \phi(f) = \int_0^1 k(t) f(t) \, dt, where k \in L^2([0,1]). This functional is continuous by the Cauchy-Schwarz inequality, which yields |\phi(f)| \leq \|k\|_2 \|f\|_2, so \|\phi\| \leq \|k\|_2.[27] A historically significant example is Hilbert's integral operator from 1904, acting on L^2([0,1]) by (Tf)(s) = \int_0^1 K(s,t) f(t) \, dt with continuous kernel K: [0,1] \times [0,1] \to \mathbb{C}. Such an operator is bounded (hence continuous) with norm bounded by \sup_{s \in [0,1]} \int_0^1 |K(s,t)| \, dt \leq \max |K|.[28]Pathological cases
A prominent pathological case arises with unbounded linear operators, which are linear but not continuous in normed spaces. A standard example is the differentiation operator D: C^\infty[0,1] \to C[0,1], where C^\infty[0,1] is the space of infinitely differentiable functions on [0,1] equipped with the supremum norm \|f\|_\infty = \sup_{x \in [0,1]} |f(x)|, and C[0,1] is the space of continuous functions with the same norm. This operator, defined by Df = f', is linear but unbounded, as demonstrated by the sequence of functions f_n(x) = \sin(2\pi n x), where \|f_n\|_\infty = 1 but \|f_n'\|_\infty = 2\pi n \to \infty, so \sup_n \|Df_n\| / \|f_n\| = \infty.[2] The restriction of D to the dense subspace of polynomials is likewise unbounded in the supremum norm, although it is bounded (hence continuous) when restricted to any finite-dimensional subspace of polynomials of fixed degree.[29] In normed linear spaces, every continuous linear operator is bounded, and conversely, every bounded linear operator is continuous. However, this equivalence fails in more general topological vector spaces (TVS). Specifically, there exist bounded linear operators that are not continuous in non-metrizable TVS. For instance, let X be an infinite-dimensional Banach space equipped with its weak topology \sigma(X, X^*) and its norm topology. The identity operator \mathrm{id}: (X, \sigma(X, X^*)) \to (X, \|\cdot\|) is linear and bounded, since a set is bounded in the weak topology if and only if it is bounded in the norm topology, but it is discontinuous because the weak topology is strictly coarser than the norm topology.[30] Such pathologies highlight the role of the norm in ensuring the equivalence, which breaks in coarser or non-locally convex topologies like the weak topology on a Banach space. Discontinuous linear functionals provide another key pathology, existing only through the axiom of choice. On any infinite-dimensional normed space X, such as \ell^2, a Hamel basis (algebraic basis) \mathcal{B} can be constructed, allowing the definition of a linear functional \phi: X \to \mathbb{R} by arbitrarily assigning values \phi(b) for b \in \mathcal{B} (e.g., \phi(b) = 1 for all b) and extending linearly. This \phi is discontinuous because it lacks a bound on the unit ball, as the Hamel basis elements are linearly independent but the assignment ignores the norm structure. Without the axiom of choice, no such discontinuous functionals are known to exist on separable Banach spaces like \mathbb{R} over \mathbb{Q}.[31] These functionals are highly non-constructive and pathological, as they fail to be continuous despite linearity. In incomplete normed spaces, such as a dense proper subspace M of a Banach space X (e.g., polynomials in C[0,1] with supremum norm), continuous linear extensions of operators defined on further subspaces exhibit pathologies related to completeness. A bounded linear functional on a subspace of M extends continuously to M via Hahn-Banach, but multiple such extensions may exist, and the unique continuous extension to the full X requires passing to the completion, where non-uniqueness can arise if considering discontinuous alternatives; however, the continuous extension to X remains unique due to density. Without completeness, some bounded operators on M to an incomplete range may lack a continuous extension within the space itself.[2] Without completeness assumptions, theorems characterizing continuous operators, such as the open mapping theorem, can fail dramatically.[32]Advanced properties
Uniform boundedness principle
The uniform boundedness principle, also known as the Banach–Steinhaus theorem, asserts that if \{T_\alpha\} is a family of continuous linear operators from a Banach space X to a normed space Y, and if the family is pointwise bounded—meaning \sup_\alpha \|T_\alpha(x)\| < \infty for every x \in X—then the family is uniformly bounded, i.e., \sup_\alpha \|T_\alpha\| < \infty.[33] This result was first established in 1927 by Stefan Banach and Hugo Steinhaus in their work on the condensation of singularities, motivated by issues in Fourier series convergence.[33] The theorem highlights the interplay between pointwise and uniform behavior in operator families, leveraging the completeness of the domain space. A quantitative formulation of the principle states that if \sup_\alpha |T_\alpha(x)| \leq M_x < \infty for each x \in X, then the operator norms satisfy \sup_\alpha \|T_\alpha\| \leq \sup \{ M_x / \|x\| : x \neq 0 \} < \infty.[34] This bound ensures that the supremum of the norms is finite, preventing any single operator in the family from dominating the others excessively. The proof relies on the Baire category theorem applied to the complete metric space X. For each positive integer n, define the closed set F_n = \{ x \in X : \sup_\alpha \|T_\alpha(x)\| \leq n \}. Since the family is pointwise bounded, these sets cover X, so \bigcup_{n=1}^\infty F_n = X. By Baire's theorem, some F_n has nonempty interior, and scaling arguments show that the unit ball is contained in a multiple of this set, implying uniform boundedness on the unit ball and thus for the entire family.[34] One key application arises in the study of dual spaces, where the principle implies that weak*-bounded sets are norm-bounded. Combined with Tychonoff's theorem, this yields the result that closed convex balanced subsets of the dual space X^* that are bounded in the weak* topology are compact.[35] This compactness property, originally proved by Leonidas Alaoglu in 1940, underpins many results in reflexive spaces and optimization.[35]Closed graph theorem
The graph of a linear operator T: X \to Y between normed linear spaces X and Y is defined as the subset G(T) = \{ (x, Tx) \mid x \in X \} of the product space X \times Y, where X \times Y is equipped with the product norm \|(x, y)\| = \|x\| + \|y\|.[36] The operator T is said to have a closed graph if G(T) is a closed subset of X \times Y. The closed graph theorem, established by Stefan Banach in 1932, asserts that if X and Y are Banach spaces and T: X \to Y is a linear operator with closed graph, then T is continuous.[37] This result characterizes continuous linear operators among those with closed graphs in the Banach space setting. The proof leverages the completeness of X and Y to establish boundedness of T. Consider the closed sets F_n = \{ x \in X : \|Tx\| \leq n \|x\| \} for n \in \mathbb{N}; the closedness of the graph ensures each F_n is closed, as if x_k \to x and Tx_k \to y, then y = Tx and \|y\| \leq n \|x\|. These sets cover X, and by the Baire category theorem applied to the complete metric space X, one F_n has nonempty interior. Linearity then implies T is bounded on X, hence continuous.[38] This boundedness aligns with the equivalence of continuity and boundedness for linear operators between normed spaces. In the framework of Banach spaces, the closed graph theorem is logically equivalent to the open mapping theorem.[36] Without completeness of the domain space, the theorem fails; for instance, discontinuous linear operators with closed graphs exist on incomplete normed spaces, such as certain extensions via Hamel bases on dense subspaces like the rationals equipped with the Euclidean norm.[38]Spectral properties
The spectrum of a continuous linear operator T on a Banach space X is defined as the set\sigma(T) = \{ \lambda \in \mathbb{C} : \lambda I - T \text{ is not invertible in } \mathcal{B}(X) \},
where \mathcal{B}(X) denotes the algebra of bounded linear operators on X.[39] This set is nonempty and compact in the complex plane, and it is always closed.[39] The complement, known as the resolvent set \rho(T) = \mathbb{C} \setminus \sigma(T), consists of those \lambda for which \lambda I - T is bijective and has a bounded inverse.[39] The resolvent operator is given by R(\lambda, T) = (\lambda I - T)^{-1} for \lambda \in \rho(T), and it belongs to \mathcal{B}(X).[39] As a function of \lambda, the resolvent is analytic on \rho(T), enabling the extension of holomorphic functional calculus to bounded operators.[39] A key estimate bounds its norm by the distance to the spectrum:
\| R(\lambda, T) \| \leq \frac{1}{\mathrm{dist}(\lambda, \sigma(T))}
for all \lambda \in \mathbb{C}.[39] This inequality highlights how the resolvent's growth controls the location and structure of the spectrum. For compact operators on infinite-dimensional Banach spaces, the spectrum \sigma(T) excluding zero consists of at most countably many eigenvalues, forming a discrete set whose only possible accumulation point is zero.[39] In Hilbert spaces, normal continuous linear operators admit the spectral theorem, which decomposes them via a spectral measure: if T is normal, there exists a unique resolution of the identity E on \sigma(T) such that T = \int_{\sigma(T)} \lambda \, dE(\lambda).[40]