The uniform boundedness principle

Let X be Banach, let Y be a normed space, and let a family of bounded linear maps Tᵢ:X→Y be indexed by an arbitrary set. If for every x the values ∥Tᵢx∥ are bounded uniformly in i, then the operator norms are bounded uniformly in i. Completeness is required of X; Y need not be complete for this conclusion. Each individual map must already be bounded. A family of discontinuous linear maps cannot be smuggled into the statement merely by observing some finite values at selected test vectors.

The distinction is between one bound for each fixed x and one bound multiplying ∥x∥ for all x. Testing finitely many vectors cannot establish pointwise boundedness on an infinite-dimensional space. Conversely, the conclusion does not promise that an operator of maximum norm exists in the family, or that the operators converge. It supplies a finite common bound. The same theorem can be applied to sequences, nets or uncountable families once the pointwise hypothesis is genuinely verified over the entire Banach domain.

For each positive integer m, collect the vectors whose images under every operator have norm at most m. Each such set is closed, because every operator is continuous, and the pointwise hypothesis says these sets cover X. Baire’s theorem gives one set containing a nonempty open ball. Comparing the images of its center and a nearby point bounds the image of their difference. Linearity then rescales this local difference bound to the entire unit ball, producing the uniform operator estimate. The proof uses a countable cover even when the operator index set is uncountable.

One immediate consequence concerns pointwise limits. If bounded operators Tₙ from Banach X into Banach Y converge at every vector, their pointwise values are bounded; uniform boundedness gives a common operator bound. Taking limits proves that the pointwise limit is linear and bounded. The convergence need not be in operator norm. The target’s completeness is useful when one starts only with pointwise Cauchy images; if actual pointwise limits in Y are already given, the bounded-limit argument itself needs no additional completeness of Y.

A weakly bounded subset of a normed space is norm bounded. To see this, treat each vector x in the subset as its evaluation map on X*. The dual X* is Banach even when X is incomplete, and Hahn–Banach identifies the norm of this evaluation map with ∥x∥. Pointwise boundedness of these evaluation maps is exactly weak boundedness, so the theorem applies. In particular, a weakly convergent sequence is norm bounded, because every scalar convergent sequence of evaluations is bounded.

The complete-domain assumption cannot simply be deleted from the original operator statement. On finitely supported sequences with the supremum norm, the maps x↦nxₙ have norms n, yet for each input only finitely many values are nonzero. Completion adds vectors with infinitely many small coordinates, exposing the growth that every finite-support vector missed. This example also explains why an operator family can look harmless on a dense subspace and still fail the pointwise boundedness needed on its completion. Density alone does not replace a uniform norm estimate.