# Norm for pointwise convergence

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?

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.

• 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.) Apr 17, 2011 at 19:27
• @Glen: A basic open set is bounded only in "finitely many directions".
– t.b.
Apr 17, 2011 at 19:34
• Doesn't "countably infinite" imply "at most countable"? Perhaps you meant uncountable? Apr 17, 2011 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, 2011 at 21:15

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 $\ell^1$.

hardmath'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:

http://www.ams.org/journals/proc/1951-002-01/S0002-9939-1951-0040647-7/S0002-9939-1951-0040647-7.pdf

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.

• 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. Apr 17, 2011 at 16:19
• @Byron Schmuland: Right, I realized my mistake after posting. Apr 17, 2011 at 16:20
• Perhaps you are thinking of Dini's theorem (en.wikipedia.org/wiki/Dini%27s_theorem), which says that continuous functions on a compact set converging monotonically pointwise must converge uniformly. Apr 17, 2011 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"!
– user940
Apr 17, 2011 at 16:31