Uniform Boundedness Theorem

Also known as uniform boundedness principle and Banach–Steinhaus theorem.

Uniform Boundedness Theorem

If T\mathcal{T} is a set of bounded linear operators from a Banach space XX into a normed space YY such that {Tx:TT}\braces{\norm{Tx} : T \in \mathcal{T}} is a bounded set for every xXx \in X, then {T:TT}\braces{\norm{T} : T \in \mathcal{T}} is a bounded set.

Proof: For each nNn \in \NN the set

An=TT{xX:Txn}A_n = \bigcap_{T \in \mathcal{T}} \braces{x \in X : \norm{Tx} \le n}

is closed, since it is the intersection of the preimages of the closed interval [0,n][0,n] under the continuous maps xTxx \mapsto \norm{Tx}. Given any xXx \in X, the set {Tx:TT}\braces{\norm{Tx} : T \in \mathcal{T}} is bounded by assumption. This means that there exists a nNn \in \NN such that Txn\norm{Tx} \le n for all TTT \in \mathcal{T}. In other words, xAnx \in A_n. Thus we have show that An=X\bigcup A_n = X. XXX Apart from the trivial case X=X = \emptyset, the union An\bigcup A_n has nonempty interior. Now, utilizing the completeness of XX, the Baire Category Theorem implies that there exists a mNm \in \NN such that AmA_m has nonempty interior. It follows that AmA_m contains an open ball B(y,ϵ)B(y,\epsilon).

To show that {T:TT}\braces{\norm{T} : T \in \mathcal{T}} is bounded, let zXz \in X with z1\norm{z} \le 1. Then y+ϵzB(y,ϵ)y+\epsilon z \in B(y,\epsilon). Using the reverse triangle inequality and the linearity of TT, we find

ϵTzTy+T(y+ϵz)2m.\epsilon \norm{Tz} \le \norm{Ty} + \norm{T(y + \epsilon z)} \le 2m.

This proves T2m/ϵ\norm{T} \le 2m/\epsilon for all TTT \in \mathcal{T}. \square\enspace

In particular, for a sequence of operators (Tn)(T_n), if there are pointwise bounds cxc_x such that

TnxcxnN,xX,\norm{T_n x} \le c_x \quad \forall n \in \NN, \forall x \in X,

the theorem implies the existence of bound cc such that

TncnN.\norm{T_n} \le c \quad \forall n \in \NN.

If XX is not complete, this may be false.

  1. Munkres, J. (2014). Topology (2nd ed.). Pearson Education Limited.