Norm for pointwise convergence

Solution 1:

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.

Solution 2:

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

Solution 3:

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.