Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I'm learning about the weak and weak* topologies on a normed vector space following the book of Brezis. He limits his discussion to case where $E$ is a Banach space, and my question is most simply stated as, "Why?". I can't find an example of a theorem where the completeness of $E$ is a necessary hypothesis. Most of the basic results on weak and weak* topologies proceed by applications of Hahn-Banach, which holds for a much wider class of spaces than just Banach spaces.

Are there any examples of reasonably elementary (i.e. of relevance to a first-year graduate student who does not anticipate having heavy contact with functional analysis in the future) facts about weak or weak* topologies that are true for Banach spaces but not all normed vector spaces?

EDIT: I should add that, as Yemon Choi points out, dual spaces of normed vector spaces are complete, so the weak-star topology will never be defined on a space that isn't Banach. With regards to the weak-star topology, then, my question should refer to aspects of the weak-star topology on a space $E^*$ where the original $E$ is not Banach.

share|cite|improve this question
Of course, the dual of a normed space is already complete, so you can cross out the weak-star part of your question... – user16299 Jan 18 '12 at 0:53
Of course. That answers part of it, I guess... – NKS Jan 18 '12 at 1:55
Although, thinking about it more, I don't think that on principle the weak-star question is frivolous. Its definition still makes reference to the original space E, so unless there's a reason why the weak-star topology couldn't possibly "remember" anything about the completeness of E, it seems possible to me that there might be situations in which it behaves differently based on properties of the original space E. – NKS Jan 18 '12 at 2:00
Just a general remark: results in functional analysis textbooks are almost never the "best possible": the hypotheses are often far stronger than what you'd need just to establish the conclusion. The most general hypotheses for any given conclusion are often very hard to figure out, and vary enough from one result to another that there is little interest in identifying the class of spaces that makes any one given thing happen. (This was a minor industry in the 50s/60s, though. The fact that so little of it filtered down into textbooks tells you all you need to know, as a student) ... – leslie townes Jan 18 '12 at 23:31
.. because of this, trying to "optimize" the statements of textbook results--- the most natural way to get into many topics--- is an unusual way to learn functional analysis. Instead of teaching you the fundamentals, it leads you to very technical questions that aren't of interest (or much use) to nonspecialists. If you "don't anticipate heavy contact with functional analysis in the future", you can probably assume that all spaces you would ever want to use functional analysis on are complete. (Having said that: your question is interesting, and I'm sorry I don't have a good answer.) – leslie townes Jan 18 '12 at 23:34
up vote 6 down vote accepted

I don't know whether the property I discuss is advanced for the first year student. This example shows that weak$^*$ boundness in $X^*$ imply uniform boundness provided $X$ is complete (see uniform boundness principle). But this is not true in general

Consider the space $c_{00}(\mathbb{N})$ of finitely supported sequences with uniform norm. This is not a complete space, and its completion is $c_0(\mathbb{N})$. Consider family of functionals $\{f_n:n\in\mathbb{N}\}$ defined by equality $$ f_n :c_{00}(\mathbb{N})\to\mathbb{C}:x\mapsto\sum\limits_{k=0}^nkx(k) $$ It is easy to check that

Each functional $f_n$ is bounded and $\Vert f_n\Vert=\frac{1}{2}n(n+1)$. Hence $$ \sup\{f_n:n\in\mathbb{N}\}=+\infty $$

For each $x\in c_{00}(\mathbb{N})$ the sum $\sum\limits_{k=0}^\infty k|x(k)|$ is finite. Hence we can obtain $$ \sup\{|f_n(x)|:n\in\mathbb{N}\}\leq\sum\limits_{k=0}^\infty k|x(k)|<\infty $$ Thus we constructed a family of bounded linear operators (in particular - functionals) which are pointwise bounded but not uniformly bounded. The raison d'etre of such a family is that our space $c_{00}(\mathbb{N})$ is not complete.

On the other hand uniform boundness principle guarantees boundness of any family of functionals in $X^*$ provided the space $X$ is complete.

share|cite|improve this answer
This is exactly the sort of thing I was looking for. Thanks! – NKS Jan 19 '12 at 0:09
The completion of $c_{00}$ is not $l_\infty$, but rather the space $c_0$ of sequences converging to zero. – Martin Jan 25 '13 at 8:42
@Martin You are right. I'll fix it – Norbert Jan 25 '13 at 9:01

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.