A nontopological proof of the uniform boundedness principle

Uniform boundedness principle is a well known and powerful result in functional analysis. This theorem was proved in a paper of Banach and Steinhaus (1927) (Banach is a student of Steinhaus). Some book says their original manuscript was lost during the war, so we do not know how they proved it. Nevertheless, the referee of the paper, Saks, suggested proving it by Baire category theorem, which is a result in general topology. This way of proving is shown in most of the texts and in undergraduate course on functional analysis.

It is believed that Banach and Steinhaus did not prove this by Baire category theorem, but by a so called “gliding hump argument”, which is found in many papers at that time. It is probably first appeared in the work of Lebesgue (1905). Note that “glide” means moving with a smooth quiet continuous motion, and “hump” means a rounded raised mass.

This post shows how to prove this result by the gliding hump technique.

First recall

\text{\bf{Uniform boundedness principle:}}
Let \{T_{\alpha}\}_{\alpha\in A} be a family of bounded linear operators from a Banach space X to a normed space Y. If the family is pointwisely bounded, then it is uniformly bounded. In other words, if \sup\{\|T_{\alpha}x\|:\alpha\in A\}<\infty for all x\in X, then \displaystyle\sup\{\|T_{\alpha}\|:\alpha\in A\}=\sup_{\alpha\in A}\sup_{\|x\|=1}\|T_{\alpha}x\|<\infty.

\text{\it{Proof.}} Suppose the assumption is true but the contrary that \sup\{\|T_{\alpha}\|:\alpha\in A\}=\infty. We want to construct sequences \{T_n\} in the family and \{x_n\}\subset X such that for all n,

(1) \|x_n\|=4^{-n} and

(2) \|T_{n}x\|>n, where \displaystyle x=\sum_{1}^{\infty}x_n.
Indeed, first choose x_1 and T_1 in the family such that \|x_1\|=4^{-1} and \|T_1\|>12.
Suppose x_1,\cdots,x_{n-1} and T_1,\cdots,T_{n-1} are chosen. Set \displaystyle M_{n-1}=\sup_{\alpha\in A}\|T_{\alpha}(x_1+\cdots +x_{n-1})\| \text{\bf{---(1)}} .

Choose T_n in the family so that \|T_n\|\geq 3\cdot 4^n(M_{n-1}+n)\text{\bf{---(2)}} .

Since \displaystyle \|T_n\|>\frac{2}{3}\|T_n\|, choose x_n\in X such that \|x_n\|=4^{-n} and \displaystyle \frac{\|T_n x_n\|}{\|x_n\|}>\frac{2}{3}\|T_n\|\text{\bf{---(3)}}.

Hence by (2) and (3)

\displaystyle \|T_n x_n\|>\frac{2}{3}\|T_n\|\|x_n\|>\frac{2}{3}(3\cdot 4^n)(M_{n-1}+n)4^{-n}=2(M_{n-1}+n)\text{\bf{---(4)}} ,

and by (3)

\displaystyle \|T_n(x_{n+1}+\cdots)\|\leq\|T_n\|\sum_{k=n+1}^{\infty}\|x_k\|

\displaystyle=\|T_n\|\cdot \frac{1}{3}\cdot 4^{-n}=\frac{1}{3}\|T_n\|\|x_n\|<\frac{1}{3}\frac{3}{2}\|T_n x_n\|=\frac{1}{2}\|T_n x_n\|\text{\bf{---(5)}} .

Let \displaystyle x=\sum_{n=1}^{\infty}x_n (exists in X as the sequence of partial sums is Cauchy and X is complete). Then by (1), (4) and (5)

\displaystyle \|T_n x\|\geq\|T_n x_n\|-\|T_n(x_1+\cdots+x_{n-1})\|-\|T_n{x_{n+1}+\cdots}\|>

\displaystyle\frac{1}{2}\|T_n x_n\|-\|T_n(x_1+\cdots+x_{n-1})\|>(M_{n-1}+n)-M_{n-1}=n,

contradicting the hypothesis, so the proof is complete.

\text{\bf{Remark:}} Why do people call this kind of argument “gliding hump”? Give me some time (or leave to readers) to figure out it.

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2 Responses to A nontopological proof of the uniform boundedness principle

  1. KKK says:

    Nice post.
    I have a question: do you require each T_\alpha in your family to be bounded?

  2. Hon Leung says:

    Yes. Corrected.

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