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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?

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2 Answers 2

up vote 23 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.

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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
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@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.

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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 (en.wikipedia.org/wiki/Dini%27s_theorem), which says that continuous functions on a compact set converging monotonically pointwise must converge uniformly. –  Nate Eldredge Apr 17 '11 at 16:22
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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

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