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I am trying to think of a topology (possibly metric, as I am more used to think about things in metric spaces) on possibly unbounded functions (on $\mathbb{R}$) such that

1) convergence in that defined topology implies pointwise convergence and

2) the limit of continuous functions is continuous itself.

I know that a sequence of functions converges to a function $f$ under the metric derived from the uniform norm if and only if converges to $f$ uniformly. And I know the existence of the Uniform limit theorem. (link ) Also convergence in uniform norm implies pointwise convergence. The problem is that the sup norm/ uniform norm is defined on bounded functions, whereas, I want to think about general functions.

Could I use a metric, $d'=\dfrac{d}{1+d}$ where d is the uniform metric, and 1 when d is infinity? The "normalization" makes sure that I do not get infinite values while getting the distance between functions. And it seems to be that given how this new metric is defined, it would satisfy 1) and 2). Do you guys think I am on the right track?

Also, is there a more obvious metric that I am missing?

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  • $\begingroup$ This will not work, I think. The way to go is to exhaust the real line with countably many compact sets which are increasing, define a bounded metric on each of these (e. g. with the trick you described) and the sum all these up with appropriate weights. $\endgroup$
    – Dirk
    Apr 15, 2016 at 7:05
  • $\begingroup$ Yes since $$\frac{d}{1+d} \leq \varepsilon \mbox{ implies } d\leq 2\varepsilon$$ for $\varepsilon \leq 0.5 $ $\endgroup$
    – user235708
    Apr 15, 2016 at 7:11
  • $\begingroup$ @Dirk, thanks for your reply. If I do this method, even if I use the uniform metric, the values would be bounded on each compact set, so for each such bounded value $v_k$ I can calculate $\sum{2^{-k}v_k}$, to find the final metric I guess. This still makes my life difficult. :( Could you tell me your intuition/ hunch of why you think the previous thing would not work? $\endgroup$
    – Juanito
    Apr 15, 2016 at 7:11
  • $\begingroup$ @MotylaNogaTomkaMazura I am sorry for being slow on this. My intuition was that given $\dfrac{d}{1+d}$ is monotonic in $d$, things would work out. Could you kindly elaborate on your counter example? $\endgroup$
    – Juanito
    Apr 15, 2016 at 7:13
  • $\begingroup$ $\dfrac{d}{1+d}\leq \epsilon$ means $d< \dfrac{\epsilon}{1-\epsilon}$, so, as $\epsilon$ gets smaller, so does $d$, going to zero in the limit. This was my intuition of why things might work. $\endgroup$
    – Juanito
    Apr 15, 2016 at 7:19

2 Answers 2

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The topology you are looking for is, I think, the topology of uniform convergence on compact sets. It is the topology defined on $\mathcal{C} (\mathbb{R}, \mathbb{R})$ such that:

$$f_n \to_{n \to + \infty} f \ \Leftrightarrow \ \lim_{n \to + \infty} \max \{|f_n (x) - f(x)| \ : \ x \in K\} \ \ \forall \text{ compact } K.$$

It is actually sufficient to check convergence on an exhaustive family of compact sets. On $\mathbb{R}$, take for instance $K_M := [-M, M]$. Hence:

$$f_n \to_{n \to + \infty} f \ \Leftrightarrow \ \lim_{n \to + \infty} \max \{|f_n (x) - f(x)| \ : \ x \in K_M \} \ \ \forall M \geq 0.$$

Finally, if you want a single metric, put $\|f\|_{\infty, K} := \max \{|f_n (x) - f(x)| \ : \ x \in K \}$. Then a possible metric is:

$$d(f,g) := \sum_{M \geq 0}2^{-M} \frac{\|f-g\|_{\infty, K_M}}{1+\|f-g\|_{\infty, K_M}},$$

although in all examples I have seen, it has been more convenient to work with the definition (first equation) than with this explicit metric.

This construction works for functions on more general spaces (we need only $\sigma$-compactness, so any open set or any manifold is fair game), for higher regularity...

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  • $\begingroup$ The metric $d$ is equivalent to the one in my comment to my answer. Why do all books prefer your formula? $\endgroup$
    – Jochen
    Apr 16, 2016 at 13:05
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Here is a metric which describes uniform convergence on $\mathbb R$: $D(f,g)=\sup\lbrace \min\lbrace|f(x)-g(x)|,1\rbrace: x\in\mathbb R\rbrace$.

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  • $\begingroup$ That's too strong, I think. For instance, $\varepsilon x$ does not converges to $0$ as $\varepsilon$ goes to $0$. $\endgroup$
    – D. Thomine
    Apr 15, 2016 at 7:32
  • $\begingroup$ @Jochen Does the metric suggested in my question not work? Could you kindly explain why that would be the case? $\endgroup$
    – Juanito
    Apr 15, 2016 at 7:36
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    $\begingroup$ @ Juanito: I think that a good test case is to compute the distance between $\varepsilon x$ and $0$. For your metric, it is not even defined (you need $d$ to be finite so that $d'$ is well-defined), although you could arguably extend it to $d = \infty$. With Jochen's answer, which essentially exchanges the $\sup$ and the formula for $d'$, there is no such problem. $\endgroup$
    – D. Thomine
    Apr 15, 2016 at 7:47
  • $\begingroup$ A metric which describes uniform convergence on all compact sets is $\delta(f,g)=\sup\lbrace \min\lbrace \|f-g\|_{[-n,n]}, 1/n\rbrace : n\in\mathbb N\rbrace$ where $\|h\|_K=\sup\lbrace |h(x)|:x\in K\rbrace$. $\endgroup$
    – Jochen
    Apr 15, 2016 at 10:14

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