Bounded Linear Operators & Operator Norms

Linearity reduces global continuity to one quantitative question: can the map amplify a unit input without bound? The operator norm answers it and turns linear maps themselves into a normed space.

Bounded, continuous, and Lipschitz are equivalent #

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

• $T$ is continuous everywhere;
• $T$ is continuous at $0$;
• there is $C\ge0$ with $\|Tx\|\le C\|x\|$ for every $x$;
• $T$ is globally Lipschitz.

The smallest such $C$ is

$$\|T\|=\sup_{\|x\|\le1}\|Tx\| =\sup_{x\ne0}\frac{\|Tx\|}{\|x\|}.$$

Linearity turns a local bound around zero into a global one by scaling. “Bounded operator” means bounded sets map to bounded sets; it does not mean the entire range is bounded.

Standard examples #

• A matrix $A:\mathbb K^n\to\mathbb K^m$ is bounded under any norms because the spaces are finite-dimensional. Under Euclidean norms, $\|A\|_2$ is the largest singular value.
• Multiplication $M_g:L^p\to L^p$, $M_gf=gf$, has $\|M_g\|=\|g\|_\infty$ under the standard hypotheses.
• The Volterra operator $(Vf)(x)=\int_0^x f(t)dt$ is bounded on several common function spaces.
• Evaluation $f\mapsto f(x_0)$ is bounded on $C(K)$ with the sup norm, but is not even well-defined on $L^p$ equivalence classes.
• Differentiation $C^1[0,1]\to C[0,1]$ is bounded under the $C^1$ norm, but unbounded when the domain inherits only the sup norm.

These examples show that boundedness belongs to a map together with its exact domain and codomain norms.

Algebra and the operator space #

Bounded operators form $\mathcal B(X,Y)$. They satisfy

$$\|S+T\|\le\|S\|+\|T\|, \qquad \|ST\|\le\|S\|\|T\|.$$

If $Y$ is Banach, then $\mathcal B(X,Y)$ is Banach in the operator norm; the domain $X$ need not be complete. Operator-norm convergence $T_n\to T$ implies uniform convergence on the unit ball and preserves boundedness. Pointwise convergence $T_nx\to Tx$ is weaker and requires uniform boundedness to recover global control.

The kernel of a bounded operator is closed. Its range need not be closed. Closed range is a substantive stability property because it prevents outputs from being arbitrarily approximable without being attainable.

Neumann-series inversion #

If $X$ is Banach and $A\in\mathcal B(X)$ satisfies $\|A\|<1$, then

$$(I-A)^{-1}=\sum_{n=0}^{\infty}A^n.$$

The series converges in operator norm because $\sum\|A^n\|\le\sum\|A\|^n<\infty$. Multiplying partial sums gives

$$(I-A)\sum_{n=0}^{N}A^n=I-A^{N+1}\to I.$$

Moreover,

$$\|(I-A)^{-1}\|\le\frac1{1-\|A\|}.$$

This is the operator form of a geometric series and the local engine behind resolvents, perturbation estimates, and second-kind integral equations.

Invertibility and conditioning #

A bounded bijection need not have a bounded inverse unless completeness hypotheses permit the bounded inverse theorem. Once both $T$ and $T^{-1}$ are bounded, the condition number

$$\kappa(T)=\|T\|\,\|T^{-1}\|$$

bounds first-order relative amplification. A small residual does not imply a small solution error when $\|T^{-1}\|$ is large.

Operator checklist #

• Name domain, codomain, and both norms.
• Prove linearity and a global norm bound separately.
• Distinguish operator-norm convergence from strong/pointwise convergence.
• Check whether the range is closed before interpreting approximate solutions.
• For inversion by Neumann series, verify $\|A\|<1$ in the actual operator norm; spectral radius below one alone requires an additional argument.