operator norm
The operator norm answers a single question about a linear map: what is the worst-case stretch? Feed the map every vector of length one and watch how long the outputs are; the operator norm is the supremum of those output lengths. It is the maximum amplification factor, a single number summarizing how powerfully the operator can magnify.
For a bounded linear operator T : X -> Y between normed spaces, the operator norm is ||T|| = sup over ||x|| <= 1 of ||T x||, equivalently the supremum of ||T x||/||x|| over nonzero x, and equivalently the smallest constant M such that ||T x|| <= M ||x|| for all x. From the definition one reads off the fundamental estimate ||T x|| <= ||T|| ||x||, used everywhere, and the submultiplicativity ||S T|| <= ||S|| ||T|| for composable operators.
This norm makes the set of bounded operators from X to Y into a normed vector space itself; when the target Y is complete (a Banach space), so is that operator space. Note the supremum need not be attained — there may be no unit vector that achieves the maximum stretch — which is one of the subtle ways infinite dimensions differ from the finite-dimensional case, where the maximum is always realized.
On l^2 let T multiply the n-th coordinate by 1 - 1/n: (T x)_n = (1 - 1/n) x_n. Then ||T x|| <= ||x|| always, and using the single basis vector e_n gives ||T e_n|| = 1 - 1/n -> 1, so ||T|| = 1. Yet no unit vector reaches output length 1 — the supremum is approached but never attained.
An operator with norm 1 whose supremum is never achieved by any unit vector.
For a matrix A acting on Euclidean space, the operator norm equals the largest singular value of A — the longest semi-axis of the ellipsoid that A makes from the unit sphere. This geometric picture is the right intuition to carry into infinite dimensions.