Bounded Linear Operators and the Operator Norm

Continuity equals boundedness

Let X and Y be normed spaces. A linear map T: X -> Y is a bounded linear operator if there is a constant C with ||Tx|| ≤ C ||x|| for all x. The word bounded does not mean the range is a bounded set — it means T does not stretch lengths by more than a fixed factor. The key theorem is that for linear maps, bounded and continuous are equivalent conditions.

Theorem. For a linear map T: X -> Y between normed spaces,
the following are equivalent:
   (a) T is continuous at every point;
   (b) T is continuous at 0;
   (c) T is bounded: ∃ C with ||Tx|| ≤ C ||x|| for all x.

(a) ⇒ (b): trivial, 0 is a point.

(b) ⇒ (c): Continuity at 0 with ε = 1 gives δ > 0 such that
    ||u|| ≤ δ  ⇒  ||Tu|| ≤ 1.
Take any x ≠ 0 and set u = δ x / ||x||, so ||u|| = δ. Then
    ||T u|| = (δ / ||x||) ||T x|| ≤ 1
  ⇒  ||T x|| ≤ (1/δ) ||x||.
So C = 1/δ works (x = 0 is trivial).

(c) ⇒ (a): For any x, a, linearity gives
    ||T x - T a|| = ||T(x - a)|| ≤ C ||x - a||.
Given ε > 0, choose δ = ε / C; then ||x - a|| < δ forces
    ||T x - T a|| < ε.
So T is (in fact Lipschitz, hence uniformly) continuous.  ∎

The homogeneity trick u = δx/||x|| converts one ε at the origin into a global bound — this is why one point of continuity suffices.

The operator norm

The smallest valid constant C is the operator norm: ||T|| = sup{ ||Tx|| : ||x|| ≤ 1 }. Equivalently ||T|| = sup_{x ≠ 0} ||Tx|| / ||x||. By construction ||Tx|| ≤ ||T|| ||x|| for every x, and ||T|| is the best such constant. The operator norm is a genuine norm on the set of bounded operators B(X, Y), and it is submultiplicative under composition: ||ST|| ≤ ||S|| ||T||.

Worked value. Let X = C[0,1] with sup norm, and define the
integration functional  T f = ∫_0^1 f(t) dt.

Upper bound:  |T f| = |∫_0^1 f| ≤ ∫_0^1 |f(t)| dt
                    ≤ ∫_0^1 ||f||_∞ dt = ||f||_∞.
So ||T|| ≤ 1.

Attained:  take f ≡ 1, then ||f||_∞ = 1 and T f = 1.
So ||T|| ≥ |T f| / ||f||_∞ = 1.

Therefore  ||T|| = 1.

Submultiplicativity, sanity check. If S, T are bounded and
   ||T x|| ≤ ||T|| ||x||,  ||S y|| ≤ ||S|| ||y||,  then
   ||S T x|| ≤ ||S|| ||T x|| ≤ ||S|| ||T|| ||x||,
so ||S T|| ≤ ||S|| ||T||.

To pin an operator norm you do two things: bound it above for all x, then exhibit one x that nearly attains the bound.

A crowning fact: if the target Y is a Banach space, then B(X, Y) with the operator norm is itself complete. In particular the dual space X* = B(X, scalars) is always a Banach space, even when X is not. We return to this dual in guide 4.