Uniform Boundedness Principle — Let be a Banach space, a normed vector space and the space of all continuous linear operators from into . Suppose that is a collection of continuous linear operators from to If, for every ,
then
In the case that is not the trivial vector space, then the semi-inequality used in the supremum of the first term in this last chain of equalities (which has range over the closed unit ball) may be replaced by a proper equality (which has range over the closed unit sphere).
The completeness of enables the following short proof, using the Baire category theorem.
Corollary — If a sequence of bounded operators converges pointwise, that is, the limit of exists for all then these pointwise limits define a bounded linear operator
The above corollary does not claim that converges to in operator norm, that is, uniformly on bounded sets. However, since is bounded in operator norm, and the limit operator is continuous, a standard "" estimate shows that converges to uniformly on compact sets.
Proof
Essentially the same as that of the proof that a pointwise convergent sequence of equicontinuous functions on a compact set converges to a continuous function.
By uniform boundedness principle, let be a uniform upper bound on the operator norms.
Fix any compact . Then for any , finitely cover (use compactness) by a finite set of open balls of radius
Since pointwise on each of , for all large , for all .
Then by triangle inequality, we find for all large , .
Corollary — Any weakly bounded subset in a normed space is bounded.
Indeed, the elements of define a pointwise bounded family of continuous linear forms on the Banach space which is the continuous dual space of
By the uniform boundedness principle, the norms of elements of as functionals on that is, norms in the second dual are bounded.
But for every the norm in the second dual coincides with the norm in by a consequence of the Hahn–Banach theorem.
Let denote the continuous operators from to endowed with the operator norm.
If the collection is unbounded in then the uniform boundedness principle implies:
In fact, is dense in The complement of in is the countable union of closed sets
By the argument used in proving the theorem, each is nowhere dense, i.e. the subset is of first category.
Therefore is the complement of a subset of first category in a Baire space. By definition of a Baire space, such sets (called comeagre or residual sets) are dense.
Such reasoning leads to the principle of condensation of singularities, which can be formulated as follows:
Theorem — Let be a Banach space, a sequence of normed vector spaces, and for every let an unbounded family in Then the set
is a residual set, and thus dense in
Proof
The complement of is the countable union
of sets of first category. Therefore, its residual set is dense.
Let be the circle, and let be the Banach space of continuous functions on with the uniform norm. Using the uniform boundedness principle, one can show that there exists an element in for which the Fourier series does not converge pointwise.
For its Fourier series is defined by
and the N-th symmetric partial sum is
where is the -th Dirichlet kernel. Fix and consider the convergence of
The functional defined by
is bounded.
The norm of in the dual of is the norm of the signed measure namely
It can be verified that
So the collection is unbounded in the dual of
Therefore, by the uniform boundedness principle, for any the set of continuous functions whose Fourier series diverges at is dense in
More can be concluded by applying the principle of condensation of singularities.
Let be a dense sequence in
Define in the similar way as above. The principle of condensation of singularities then says that the set of continuous functions whose Fourier series diverges at each is dense in (however, the Fourier series of a continuous function converges to for almost every by Carleson's theorem).
Attempts to find classes of locally convex topological vector spaces on which the uniform boundedness principle holds eventually led to barrelled spaces.
That is, the least restrictive setting for the uniform boundedness principle is a barrelled space, where the following generalized version of the theorem holds (Bourbaki 1987, Theorem III.2.1):
A family of subsets of a topological vector space is said to be uniformly bounded in if there exists some bounded subset of such that
which happens if and only if
is a bounded subset of ;
if is a normed space then this happens if and only if there exists some real such that
In particular, if is a family of maps from to and if then the family is uniformly bounded in if and only if there exists some bounded subset of such that which happens if and only if is a bounded subset of
Proposition[2] — Let be a set of continuous linear operators between two topological vector spaces and and let be any bounded subset of
Then the family of sets is uniformly bounded in if any of the following conditions are satisfied:
is equicontinuous.
is a convexcompact Hausdorff subspace of and for every the orbit is a bounded subset of
Although the notion of a nonmeager set is used in the following version of the uniform bounded principle, the domain is not assumed to be a Baire space.
Every proper vector subspace of a TVS has an empty interior in [3] So in particular, every proper vector subspace that is closed is nowhere dense in and thus of the first category (meager) in (and the same is thus also true of all its subsets).
Consequently, any vector subspace of a TVS that is of the second category (nonmeager) in must be a dense subset of (since otherwise its closure in would a closed proper vector subspace of and thus of the first category).[3]
Let be balanced neighborhoods of the origin in satisfying It must be shown that there exists a neighborhood of the origin in such that for every
Let
which is a closed subset of (because it is an intersection of closed subsets) that for every also satisfies and
(as will be shown, the set is in fact a neighborhood of the origin in because the topological interior of in is not empty).
If then being bounded in implies that there exists some integer such that so if then
Since was arbitrary,
This proves that
Because is of the second category in the same must be true of at least one of the sets for some
The map defined by is a (surjective) homeomorphism, so the set is necessarily of the second category in
Because is closed and of the second category in its topological interior in is not empty.
Pick
Because the map defined by is a homeomorphism, the set
is a neighborhood of in which implies that the same is true of its superset
And so for every
This proves that is equicontinuous.
Q.E.D.
Proof that :
Because is equicontinuous, if is bounded in then is uniformly bounded in
In particular, for any because is a bounded subset of is a uniformly bounded subset of Thus
Q.E.D.
Theorem[3] — If is a sequence of continuous linear maps from an F-space into a Hausdorff topological vector space such that for every the limit
exists in then is a continuous linear map and the maps are equicontinuous.