Bounded Linear Operators

A linear operator $T : X \to Y$ between normed spaces is bounded if there exists $C \geq 0$ such that $\|Tx\|_Y \leq C\|x\|_X$ for all $x \in X$. The operator norm is

Proposition 3.1. A linear operator is bounded if and only if it is continuous.

Proposition 3.2. $T$ is bounded if and only if it maps bounded sets to bounded sets.

The space $\mathcal{B}(X, Y)$ of all bounded linear operators from $X$ to $Y$ is a Banach space when $Y$ is complete, with the operator norm.

Example 1. The identity operator $I : X \to X$ has $\|I\| = 1$.

Example 2. The zero operator $0 : X \to Y$ has $\|0\| = 0$.

Example 3. Let $T : \ell^2 \to \ell^2$ be defined by $T(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)$ (right shift). Then $\|T\| = 1$.

Example 4. The multiplication operator $(M_g f)(x) = g(x)f(x)$ on $L^2$ is bounded with $\|M_g\| = \|g\|_\infty$.

The dual space of $X$ is $X^* = \mathcal{B}(X, \mathbb{R})$ (or $\mathcal{B}(X, \mathbb{C})$), the space of all bounded linear functionals.

Theorem 3.3. $X^*$ is always a Banach space.

Theorem 3.4. $(\ell^1)^* \cong \ell^\infty$ via $\varphi(x) = \sum x_n y_n$ for $y \in \ell^\infty$.

Theorem 3.5. $(\ell^p)^* \cong \ell^q$ for $1 \leq p < \infty$ where $1/p + 1/q = 1$.

Theorem 3.6. $(c_0)^* \cong \ell^1$ where $c_0 = \{(x_n) : x_n \to 0\}$.

Let $M$ be a subspace of a normed space $X$. The annihilator of $M$ is

Let $N$ be a subspace of $X^*$. The pre-annihilator of $N$ is

Proposition 3.7. If $M \subseteq X$ is a subspace, then $M^\perp$ is a closed subspace of $X^*$. If $N \subseteq X^*$ is a subspace, then ${}^\perp N$ is a closed subspace of $X$.

The page states fifteen results; the proofs below are the ones that carry the subject, chosen because their arguments; linearity converts boundedness to continuity at a point, completeness passes to operator spaces, Hölder identifies duals; are the ones every later page reuses.

Proposition 3.1 (bounded iff continuous). A linear operator $T$ between normed spaces is bounded iff it is continuous.

Proof. ($\Rightarrow$) If $\|T\| \leq C$ then $\|Tx - Ty\| = \|T(x-y)\| \leq C\|x-y\|$, and $\epsilon$–$\delta$ continuity follows with $\delta = \epsilon/C$.

($\Leftarrow$) Continuity at $0$ gives $\delta > 0$ with $\|x\| \leq \delta \Rightarrow \|Tx\| \leq 1$. Every nonzero $x$ can be scaled to $x/(\|x\|/\delta) = \delta x/\|x\|$, which has norm $\delta$, so $\|Tx\| \leq \|x\|/\delta$; bounded with $C = 1/\delta$. Linearity enters in the scaling step: without it, continuity at a point says nothing about other points. $\blacksquare$

Theorem 3.3 ($\mathcal{B}(X,Y)$ is Banach). If $Y$ is complete, then $\mathcal{B}(X, Y)$ with the operator norm is a Banach space.

Proof. $\mathcal{B}(X,Y)$ is a vector space, and the operator norm is a norm: homogeneity and the triangle inequality come from the norm on $Y$, and $\|T\| = 0 \Rightarrow Tx = 0$ for all $\|x\| \leq 1 \Rightarrow T = 0$.

For completeness, let $(T_n)$ be Cauchy in operator norm. Then for each $x$, $\|T_nx - T_mx\| \leq \|T_n - T_m\|\,\|x\|$ shows $(T_nx)$ is Cauchy in $Y$, which is complete, so $T_nx \to Sx$ for some $Sx$. Define $T$ by $Tx = Sx$; linearity passes to the limit. The Cauchy condition gives, for each $\epsilon$ an $N$ with $\|T_n - T_m\| < \epsilon$ for $n, m \geq N$; letting $m \to \infty$ in $\|T_nx - T_mx\| \leq \epsilon\|x\|$; legitimate because $Y$‘s norm is continuous; gives $\|(T_n - T)x\| \leq \epsilon\|x\|$ for all $x$ and $n \geq N$. So $T - T_n \in \mathcal{B}(X,Y)$ with norm at most $\epsilon$: $T$ is bounded and $T_n \to T$ in operator norm. $\blacksquare$

The proof is the completeness transfer pattern; pointwise Cauchy, identify the limit, then upgrade the convergence using the operator-norm structure; and it is the same argument that makes the space of bounded bilinear forms, and every $\mathcal{B}(X,Y)$ variant, complete. Note that completeness of $Y$ is essential: $\mathcal{B}(X, Y)$ inherits Cauchy limits pointwise, and without $Y$‘s completeness the pointwise limits need not exist.

Theorem 3.5 ($( \ell^p)^* \cong \ell^q$). For $1 \leq p < \infty$ with $1/p + 1/q = 1$, every $\varphi \in (\ell^p)^*$ is $\varphi(x) = \sum_n x_n y_n$ for a unique $y \in \ell^q$, and the map $y \mapsto \varphi$ is an isometric isomorphism.

Proof. Let $e_n$ be the standard basis. Set $y_n = \varphi(e_n)$. For $x \in \ell^p$ define the truncations $x^{(N)} = (x_1, \dots, x_N, 0, \dots)$; by linearity $\varphi(x^{(N)}) = \sum_{n \leq N} x_n y_n$.

Bound on the tail coefficients. For fixed $n$, consider the vector $x(\alpha) = (\dots, 0, e^{i\alpha_n}|y_n|^{q-1}\operatorname{sgn}(y_n), 0, \dots)$ with the phase chosen so $|y_n|^q = e^{-i\alpha_n}y_n|y_n|^{q-1}$, supported at $n$ and scaled to make $\|x(\alpha)\|_p = |y_n|^{q-1}$. Then

when $y_n \neq 0$; for $p = 1$ the same computation with the sup norm gives $\sup_n |y_n| \leq \|\varphi\|$. So $y \in \ell^\infty$; for $q < \infty$, apply the same construction to the truncations: for $x^{(N)}$ scaled by $t = \|x^{(N)}\|_p^{1-p} \geq 0$,

uniformly in $N$, so $y \in \ell^q$ with $\|y\|_q \leq \|\varphi\|$.

Representation and isometry. Hölder’s inequality says the map $y \mapsto \varphi_y$, $\varphi_y(x) = \sum x_ny_n$, is bounded with $\|\varphi_y\| \leq \|y\|_q$, and equality is attained by the phase-aligned sequence; so $\|\varphi_y\| = \|y\|_q$. The truncations $x^{(N)} \to x$ in $\ell^p$ (dominated convergence, since $x \in \ell^p$), and $\varphi$ is continuous, so

which is the representation, with isometry from $\|y\|_q \leq \|\varphi\| \leq \|y\|_q$. Uniqueness: $\varphi(e_n) = y_n$ determines $y$. $\blacksquare$

The two estimates worth internalising: the phase-aligned truncation that bounds $\ell^q$-norms of the coefficients uniformly in $N$; choosing signs to make each term contribute maximally is the standard move; and Hölder’s inequality providing both the boundedness and the isometry. The companion results, that $(c_0)^* = \ell^1$ and $(L^p)^* = L^q$, follow the same proof with the approximating family changed: finitely supported sequences for $c_0$, simple functions for $L^p$. What has no such proof is $(L^\infty)^*$ and $(\ell^\infty)^*$: their duals are the ba spaces, strictly larger than $L^1$; a non-separability fact that marks the boundary of this theorem.

Proposition 3.7 (annihilators are closed) and the double-annihilator identity $(M^\perp)^\perp = \bar M$.

Proof of closedness. For $x \in M$, the map $\varphi \mapsto \varphi(x)$ is continuous on $X^*$, and $M^\perp$ is the intersection of its kernels; an intersection of closed sets is closed. The pre-annihilator is the same argument with roles swapped. $\blacksquare$

Proof of $(M^\perp)^\perp = \bar M$ for a subspace $M$ of a normed space $X$. $\bar M \subseteq (M^\perp)^\perp$: if $x \in \bar M$, take $m_k \in M$ with $m_k \to x$; for $\varphi \in M^\perp$, continuity gives $\varphi(x) = \lim \varphi(m_k) = 0$, so $x \in (M^\perp)^\perp$.

For the converse, suppose $x \in (M^\perp)^\perp$ but $x \notin \bar M$. Since $\bar M$ is closed and $x$ is at positive distance from it, the geometric form of Hahn–Banach separates: there is $\varphi \in X^*$ and $c$ with $\operatorname{Re}\varphi(x) > c \geq \operatorname{Re}\varphi(m)$ for all $m \in \bar M$. Scaling by a constant and using that $M$ is a subspace; so $m/ \|m\|$-style scaling keeps $\varphi(m)$ at whatever value it has; gives $\varphi|_M = 0$, i.e. $\varphi \in M^\perp$. But then $\varphi(x) = 0$ by $x \in (M^\perp)^\perp$, contradicting $\operatorname{Re}\varphi(x) > c \geq 0$ for the $c$ chosen positive. $\blacksquare$

This identity is why quotient-space duality works: $(X/M)^* \cong M^\perp$, and combined with $(M^\perp)^\perp = \bar M$ it says the dual of the quotient is exactly the functionals vanishing on $M$; the bridge between the quotient-space constructions of the abstract algebra track and the normed-space theory here.

Proposition 3.8. For a subspace $M \subseteq X$, ${}^\perp(M^\perp) = \overline{M}$ (the closure of $M$).

Proposition 3.9. For finite-dimensional subspaces $M \subseteq X$, $(X/M)^* \cong M^\perp$.

The double dual (or bidual) of $X$ is $X^{**} = (X^*)^*$. There is a natural embedding $J : X \hookrightarrow X^{**}$ defined by $(Jx)(\varphi) = \varphi(x)$ for $\varphi \in X^*$. The Hahn-Banach theorem guarantees that $J$ is an isometric embedding: $\|Jx\|_{X^{**}} = \|x\|_X$.

Definition. A normed space $X$ is reflexive if the canonical embedding $J : X \to X^{**}$ is surjective, i.e., $J(X) = X^{**}$.

Proposition 3.10. Every reflexive space is a Banach space.

Example. $\ell^p$ is reflexive for $1 < p < \infty$ since $(\ell^p)^{**} \cong (\ell^q)^* \cong \ell^p$.

Example. $L^p(\mu)$ is reflexive for $1 < p < \infty$.

Example. $\ell^1$, $\ell^\infty$, and $c_0$ are not reflexive. In particular, $(\ell^1)^{**} \cong (\ell^\infty)^* \supsetneq \ell^1$, and $(c_0)^{**} \cong \ell^\infty \supsetneq c_0$.

Theorem 3.11. A Banach space $X$ is reflexive if and only if its closed unit ball is weakly compact.

Theorem 3.12. If $X$ is reflexive, then every bounded sequence has a weakly convergent subsequence (Eberlein-Smulian theorem). In particular, every continuous linear functional achieves its norm on the closed unit ball.

Worked Example. Show that $c_0$ is not reflexive. Since $(c_0)^* \cong \ell^1$ by Theorem 3.6, and $(\ell^1)^* \cong \ell^\infty$ by Theorem 3.4, we have $(c_0)^{**} \cong \ell^\infty$. The canonical embedding $J : c_0 \hookrightarrow \ell^\infty$ is the inclusion map. Since $\ell^\infty$ contains bounded sequences that do not converge to zero (e.g., the constant sequence $(1, 1, 1, \ldots)$), $J$ is not surjective, so $c_0$ is not reflexive.

• Boundedness and continuity are equivalent for linear operators between normed spaces.
• The operator norm is submultiplicative: $\|ST\| \leq \|S\|\|T\|$ for composable operators.
• Reflexivity implies the Banach space property but not conversely.
• The double dual $X^{**}$ is always reflexive when $X$ is a Banach space.

• Assuming that every bounded linear functional on a subspace extends to the whole space without invoking the Hahn-Banach theorem. Extension requires the Hahn-Banach theorem and is not automatic.
• Confusing weak convergence with strong convergence. A sequence can converge weakly but not strongly (e.g., the standard basis in $\ell^2$).
• Forgetting that $\mathcal{B}(X,Y)$ is a Banach space only when $Y$ is complete. If $Y$ is not complete, the space of bounded operators need not be.
• Assuming reflexivity when only the canonical embedding is injective. Injectivity holds for all normed spaces by Hahn-Banach; reflexivity requires surjectivity.

• Quantum mechanics: Observables are modelled as self-adjoint bounded operators on Hilbert spaces.
• Numerical analysis: The spectral radius of an iteration matrix determines convergence of iterative methods.
• Partial differential equations: Bounded operators on Sobolev spaces encode weak formulations of PDEs.
• Signal processing: Bounded linear operators on $L^2$ spaces represent filters and transforms.

Theorem 3.13 (Open Mapping Theorem). If $X$ and $Y$ are Banach spaces and $T : X \to Y$ is a surjective bounded linear operator, then $T$ is an open map (it maps open sets to open sets).

Corollary 3.14 (Bounded Inverse Theorem). If $T : X \to Y$ is a bijective bounded linear operator between Banach spaces, then $T^{-1}$ is also bounded.

Theorem 3.15 (Closed Graph Theorem). A linear operator $T : X \to Y$ between Banach spaces is bounded if and only if its graph $\{(x, Tx) : x \in X\}$ is closed in $X \times Y$.

• Normed Spaces and Banach Spaces: Defines the domain and codomain spaces for bounded linear operators and establishes the dual space framework.

• The Fundamental Theorems: Presents the Hahn-Banach, open mapping, and closed graph theorems that govern the behaviour of bounded operators between Banach spaces.

• Compact Operators: Studies a subclass of bounded operators with finite-dimensional-like properties, crucial for spectral theory.

• Quantum Mechanics

• Graph Theory

• Classical Mechanics

• Electromagnetism

Bounded linear operators are the continuous linear maps between normed spaces, they do not blow up small inputs into large outputs. The operator norm measures the maximum stretching factor. The open mapping theorem says that surjective bounded operators between Banach spaces are automatically open maps, which implies the bounded inverse theorem: an invertible bounded operator has a bounded inverse. The closed graph theorem provides a practical test: a linear operator is bounded if and only if its graph is closed. These three results, open mapping, bounded inverse, and closed graph, are the fundamental theorems of functional analysis, guaranteeing that well-behaved operators behave as expected.

Problem. Let $T : C[0,1] \to C[0,1]$ be defined by $(Tf)(x) = xf'(x)$. Show that $T$ is unbounded.

• Applications
• Compact Operators
• Historical Context
• Inner Product Spaces and Hilbert Spaces