In mathematics, the **uniform boundedness principle** or **Banach–Steinhaus theorem** is one of the fundamental results in functional analysis.
Together with the Hahn–Banach theorem and the open mapping theorem, it is considered one of the cornerstones of the field.
In its basic form, it asserts that for a family of continuous linear operators (and thus bounded operators) whose domain is a Banach space, pointwise boundedness is equivalent to uniform boundedness in operator norm.

The theorem was first published in 1927 by Stefan Banach and Hugo Steinhaus, but it was also proven independently by Hans Hahn.

**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

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

Let X be a Banach space. Suppose that for every

For every integer let

Each set is a closed set and by the assumption,

By the Baire category theorem for the non-empty complete metric space there exists some such that has non-empty interior; that is, there exist and such that

Let with and Then:

Taking the supremum over in the unit ball of and over it follows that

There are also simple proofs not using the Baire theorem (Sokal 2011).

**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.

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

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

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).

In a topological vector space (TVS) "bounded subset" refers specifically to the notion of a von Neumann bounded subset. If happens to also be a normed or seminormed space, say with (semi)norm then a subset is (von Neumann) bounded if and only if it is *norm bounded*, which by definition means

Main article: Barrelled space |

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):

**Theorem** — Given a barrelled space and a locally convex space then any family of pointwise bounded continuous linear mappings from to is equicontinuous (and even uniformly equicontinuous).

Alternatively, the statement also holds whenever is a Baire space and is a locally convex space.^{[1]}

Main article: Uniformly bounded sets (topological vector space) |

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:

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.

**Theorem ^{[2]}** — Let be a set of continuous linear operators between two topological vector spaces and (not necessarily Hausdorff or locally convex).
For every denote the orbit of by

and let denote the set of all whose orbit is a bounded subset of
If is of the second category (that is, nonmeager) in then and is equicontinuous.

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]}

*Proof that is equicontinuous:*

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.

The following theorem establishes conditions for the pointwise limit of a sequence of continuous linear maps to be itself continuous.

**Theorem ^{[4]}** — Suppose that is a sequence of continuous linear maps between two topological vector spaces and

- If the set of all for which is a Cauchy sequence in is of the second category in then
- If the set of all at which the limit exists in is of the second category in and if is a complete metrizable topological vector space (such as a Fréchet space or an F-space), then and is a continuous linear map.

**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.

If in addition the domain is a Banach space and the codomain is a normed space then

Dieudonné (1970) proves a weaker form of this theorem with Fréchet spaces rather than the usual Banach spaces.

**Theorem ^{[2]}** — Let be a set of continuous linear operators from a complete metrizable topological vector space (such as a Fréchet space or an F-space) into a Hausdorff topological vector space
If for every the orbit

is a bounded subset of then is equicontinuous.

So in particular, if is also a normed space and if

then is equicontinuous.