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

Does there exist a norm on the space of all real-valued functions on the real line (or on an open set? a compact set?) such that convergence in this norm is equivalent to pointwise convergence?

share|cite|improve this question
up vote 26 down vote accepted

No. First note that if $X$ is any set, the space of all functions $f: X \to \mathbb{R}$ with the topology of pointwise convergence is simply the space $\prod_{X} \mathbb{R}$ with the product topology. It is well-known that this space is metrizable if and only if $X$ is at most countable, see e.g. J. Dugundji, Topology, IX.7.1, p.189. However, if $X$ is countably infinite, then it is an easy exercise to show that there is no norm inducing the topology of pointwise convergence (see e.g. Rudin, Functional analysis, Theorem 1.39, p.28, and also Exercise 7 on p.37).

To sum it all up, the space you're interested in is

  • (completely) normable if and only if $X$ is finite;
  • (completely) metrizable if and only if $X$ is countable;
  • not even metrizable in all other situations.

I leave the fact that the norm and metric can be chosen to be complete as an exercise.

share|cite|improve this answer
Hi @Theo: could you give a hint or one line justification for why no norm can exist? Is it in contradiction with the triangle inequality? (I haven't done the exercise.) – Glen Wheeler Apr 17 '11 at 19:27
@Glen: A basic open set is bounded only in "finitely many directions". – t.b. Apr 17 '11 at 19:34
Doesn't "countably infinite" imply "at most countable"? Perhaps you meant uncountable? – Jesse Madnick Apr 17 '11 at 21:14
@Jesse: For me countable means finite or countably infinite, so I said countably infinite to exclude the finite case. A finite product certainly is normable, a countably infinite product is metrizable but it isn't normable and finally, an uncountable product is not even metrizable. – t.b. Apr 17 '11 at 21:15

Edit: [Realized the sup-norm doesn't even work for continuous functions. Revised example.] An example shows that even for continuous functions on a compact set, pointwise convergence doesn't imply convergence in the sup-norm. Consider $f_k$ defined as $f_k(1/k) = 1$ and piecewise linearly interpolated to $f_k(0) = f_k(2/k) = f_k(1) = 0$, for $k = 2,3,... $. Pointwise the sequence ${f_k}$ converges to zero on $[0,1]$, but this is not true obviously for convergence in the sup-norm.

share|cite|improve this answer
Thank you, but I think this does not answer my question. I did not restrict my question to continuous functions. I am interested if there is another norm for a larger class of functions, or for a proof that such a norm does not exists. – the L Apr 17 '11 at 16:19
@Byron Schmuland: Right, I realized my mistake after posting. – hardmath Apr 17 '11 at 16:20
Perhaps you are thinking of Dini's theorem (, which says that continuous functions on a compact set converging monotonically pointwise must converge uniformly. – Nate Eldredge Apr 17 '11 at 16:22
Don't feel bad, I also have a wrong answer (now deleted). When a probabilist hears "pointwise convergence", he usually thinks "almost sure pointwise convergence"! – Byron Schmuland Apr 17 '11 at 16:31

The first answer does not reply exactly to the question: indeed, the topology of the pointwise convergence is not metrizable; however, there could be another (metrizable or not) topology with the same convergent sequences! This phenomen happens with the Schur property of $\ell^1$.

harmath's example answers completely the question. Interestingly, the following article shows also that there is no metric having the same set of convergent sequences as the pointwise convergent sequences:

share|cite|improve this answer

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.