Uniform Boundedness Principle - Uniform Boundedness Principle

Uniform Boundedness Principle

The precise statement of the result is:

Theorem. Let X be a Banach space and Y be a normed vector space. Suppose that F is a collection of continuous linear operators from X to Y. The uniform boundedness principle states that if for all x in X we have

then

The completeness of X enables the following short proof, using the Baire category theorem:

Proof.

Define the closed sets X_n with n = 1, 2, 3, … by

By hypothesis, the union of all the X_n is X. Since X is a Baire space, one of the X_n, say X_m, has an interior point (X_n are closed sets), i.e., there exists a δ > 0 and a y in X such that all x ∈ X with ||x − y|| < δ are elements of X_m. Now, choose an arbitrary z in X with ||z|| < δ. Then, y and y + z are elements of X_m and hence, for every operator T in the family F, ||T(z)|| ≤ ||T(y + z)|| + ||T(y)|| ≤ m + m = 2m. Since z is arbitrary in the ball of radius δ, it follows that ||T|| ≤ 2m / δ for all T in F, which proves the theorem.

A direct consequence is:

Corollary. If a sequence of bounded operators (T_n) converges pointwise, that is, lim T_n(x) exists for all x in X, then these pointwise limits define a bounded operator T.

Note it is not claimed above that T_n converges to T in operator norm, i.e. uniformly on bounded sets. (However, since {T_n} is bounded in operator norm, and the limit operator T is continuous, a standard "3-ε" estimate shows that T_n converges to T uniformly on compact sets.)

Another corollary is that any weakly bounded subset S in a normed space Y is bounded; indeed, the elements of S define a pointwise bounded family of continuous linear forms on the Banach space X = Y∗, continuous dual of Y. By the uniform boundedness principle, the norms of elements of S, as functionals on X, that is, norms in the second dual Y∗∗, are bounded. But for every s ∈ S, the norm in the second dual coincides with the norm in Y, by a consequence of the Hahn–Banach theorem.

Let L(X, Y) denote the continuous operators from X to Y, with the operator norm. If the collection F is unbounded in L(X, Y), then by the uniform boundedness principle, the set

is not empty. In fact, it is dense in X. The complement of R in X is the countable union of closed sets ∪X_n. By the argument used in proving the theorem, each X_n is nowhere dense, i.e. the subset ∪X_n is of first category. Therefore R is the complement of a subset of first category in a Baire space. By definition of a Baire space, such sets (called residual sets) are dense. Such reasoning leads to the principle of condensation of singularities, which can be formulated as follows:

Theorem. Let X be a Banach space, {Y_n} a sequence of normed vector spaces, and F_n a unbounded family in L(X, Y_n). Then the set

is dense in X.

Proof.

The complement of R is the countable union

of sets of first category. Therefore its residual set R is dense.