Bounded Linear Operators - Core Definitions

Linear operators between normed spaces form the natural morphisms in functional analysis, generalizing linear transformations from finite-dimensional linear algebra to infinite dimensions.

DefinitionBounded Linear Operator

Let $X$ and $Y$ be normed spaces. A linear map $T : X \to Y$ is bounded if there exists a constant $C \geq 0$ such that $\|T(x)\|_Y \leq C \|x\|_X$ for all $x \in X$ .

The operator norm of $T$ is defined by $\|T\| = \sup_{\|x\| \leq 1} \|T(x)\| = \sup_{x \neq 0} \frac{\|T(x)\|}{\|x\|} = \inf\{C : \|T(x)\| \leq C\|x\| \text{ for all } x\}$

The terminology "bounded" might seem unusual—it means that $T$ maps bounded sets to bounded sets, not that $T$ itself is bounded as a function. In finite dimensions, all linear operators are bounded, but this fails dramatically in infinite dimensions.

TheoremBoundedness and Continuity

For a linear operator $T : X \to Y$ between normed spaces, the following are equivalent:

$T$ is bounded
$T$ is continuous
$T$ is continuous at $0$

Proof

$(1) \Rightarrow (2)$ : If $x_n \to x$ , then $\|T(x_n) - T(x)\| = \|T(x_n - x)\| \leq \|T\| \|x_n - x\| \to 0$ .

$(2) \Rightarrow (3)$ : Obvious.

$(3) \Rightarrow (1)$ : If $T$ is not bounded, there exists $(x_n)$ with $\|x_n\| = 1$ but $\|T(x_n)\| \to \infty$ . Then $y_n = x_n/\|T(x_n)\| \to 0$ but $\|T(y_n)\| = 1$ , contradicting continuity at $0$ .

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ExampleBounded and Unbounded Operators

Bounded: Multiplication operator $M_\phi : L^2[0,1] \to L^2[0,1]$ by $(M_\phi f)(x) = \phi(x)f(x)$ where $\phi \in L^\infty[0,1]$ is bounded with $\|M_\phi\| = \|\phi\|_\infty$
Unbounded: The differentiation operator $D : C^1[0,1] \to C[0,1]$ with $D(f) = f'$ is unbounded. Consider $f_n(x) = \sin(nx)$ with $\|f_n\|_\infty = 1$ but $\|D(f_n)\|_\infty = n \to \infty$
Bounded: Any finite rank operator $T(x) = \sum_{i=1}^n \phi_i(x) y_i$ where $\phi_i \in X^*$ and $y_i \in Y$

DefinitionSpace of Bounded Operators

Let $\mathcal{B}(X, Y)$ denote the space of all bounded linear operators from $X$ to $Y$ . When $X = Y$ , we write $\mathcal{B}(X) = \mathcal{B}(X, X)$ .

With the operator norm, $\mathcal{B}(X, Y)$ is a normed space. If $Y$ is a Banach space, then so is $\mathcal{B}(X, Y)$ .

Remark

The operator norm makes $\mathcal{B}(X, Y)$ into a Banach space whenever $Y$ is complete, even if $X$ is not. This completeness is crucial for many fixed point arguments and iterative methods in analysis.

The study of bounded operators forms the foundation for spectral theory, operator algebras, and the theory of linear partial differential equations.