Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following result was proved in a previous post: Bounded functionals on Banach spaces.

Let $(X, \|.\|)$ be a Banach space such that

  • $X \subset C([0,1]) $
  • For every $r\in \mathbb{Q}\cap[0,1], f\mapsto f(r)$ defines a bounded linear functional on $X$.

There exists a $C>0$ such that, for all $f\in X$,
$$\sup_{x\in[0,1]} |f(x)| \leq C\|f\|.$$


Does anyone know an example of space $X$ where this result is interesting?

Indeed I feel that the example above could be a very nice application of the Banach-Steinhaus theorem, but the examples of spaces $X$ I thought of were too simple:

  1. One could easily prove the result without the Banach-Steinhaus theorem.
  2. The assumption on boundedness would be easy to prove for all $r$ in $[0,1]$.

If someone has an example of space $X$ satisfing the first point, even without the second one, I am already interested.

share|cite|improve this question
The problem when we work with a specific example is that the norm is already given. We know that it will involve a norm greater than the supremum norm (up to an universal constant). For example, if $X$ is the space of $C^1$ functions on $[0,1]$ endowed with the norm $\lVert f\rVert:=|f(0)|+\sup_{0\leqslant x\leqslant 1}|f'(x)|$ and we are in category 1. and 2.. – Davide Giraudo Dec 16 '12 at 18:09
Yes @DavideGiraudo, that's indeed the type of problem I have. So that I feel that I can't really convince someone new to functional analysis that the Banach-Steinhaus theorem is usefull with this result. – Sebastien B Dec 16 '12 at 20:34
Well, at least it gives an interesting result per se, that is, if a have a Banach space which consists of continuous functions on $[0,1]$, and the evaluations at rationals are bounded linear functionals then the norm is, up to an universal constant, greater than the supremum norm (so we have to take "strong norms", that is, not norms likes the integral or things like that). – Davide Giraudo Dec 28 '12 at 22:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.