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

$\|T\| = \sup\{\|Tx\|_Y : \|x\|_X \leq 1\} = \sup\{\|Tx\|_Y : \|x\|_X = 1\}$

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

$M^\perp = \{\varphi \in X^* : \varphi(x) = 0 \text{ for all } x \in M\}$

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

${}^\perp N = \{x \in X : \varphi(x) = 0 \text{ for all } \varphi \in N\}$

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 $X$.

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 $X \times Y$.

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