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For example, consider the space of continuous functions $C[a,b]$. Let $\{f_n\}$​ be a sequence of functions in this space. If $\{f_n\}$​ converges pointwise to $f$, does there exist a norm on $C[a,b]$ such that $\|f_n-f\|$ converges to 0?

If not, then does $\{f_n\}$​ converges to $f$ in a norm implies $\{f_n\}$ converges pointwise to $f$?

(I just started learning this part. Sorry for I can't provide some thoughts about the question.)

Ted Shifrin
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Yuyu
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Feb 16 '25 at 03:53
  • Your question is answered by this post. – Ted Shifrin Feb 16 '25 at 04:21
  • @TedShifrin not exactly. There is a subtlety that a topology on a space can be non-metrizable, but there can exist a metric $d$ such that $a_n\to a$ iff $d(a_n, a)\to 0$. The problem is pointwise convergence topology is about convergence of nets and not just about convergence of sequences of functions. – Jakobian Feb 16 '25 at 04:26
  • @Jakobian Ah, subtle point. Interesting. – Ted Shifrin Feb 16 '25 at 04:34
  • Second question sounds silly. Doesn't the usual sup norm do the job? – Kavi Rama Murthy Feb 16 '25 at 04:45
  • @geetha290krm What about an arbitrary norm? Certainly the $L^1$ norm does not. – Ted Shifrin Feb 16 '25 at 04:50
  • Arbitrary norm, no chance. My point is OP made no attempt at all to answer the (silly) second part and I have downvoted the question. – Kavi Rama Murthy Feb 16 '25 at 04:53
  • I know reference that there is no metric $d$ such that $f_n\to f$ pointwise iff $d(f_n, f)\to 0$, but the OP seems to be asking only for one direction of this implication (for norms, not metrics), and moreover, it seems that the norm should depend on the sequence $f_n\to f$. It would be nice to have this clarified, maybe I misunderstood something. – Jakobian Feb 16 '25 at 05:44

2 Answers2

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Assume $(f_n)$ is a sequence in $C[0,1]$ that converges to $f \in C[0,1]$ pointwise. Then we can construct a norm on $C[0,1]$ such that $f_n \to f$ under this norm. Indeed, fix a countable dense subset $\mathcal{D} =\{x_1, x_2, \ldots\}$ of $[0,1]$ and define

$$\|g\| = \sum_{n=1}^{\infty} \frac{1}{2^n(1+\sup_k |f_k(x_n)|)} |g(x_n)|. $$

It is not hard to check that this is indeed a norm on $C[0,1]$ such that $f_n \to f$ under this norm. Of course, this norm depends on $(f_n)$ and may (in fact, must) fail to capture the topology of pointwise convergence on $C[0,1]$.

Sangchul Lee
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  • https://math.meta.stackexchange.com/questions/33508/enforcement-of-quality-standards?cb=1 – Kavi Rama Murthy Feb 16 '25 at 07:18
  • @geetha290krm, I understand that we enforce quality standards in this community, and I agree that OP's post lacks some details. However, since they mentioned their level of understanding and explained that they couldn't find any good leads (which I think is a reasonable explanation for not providing their own thoughts), I believed this was a marginally acceptable case. If the majority thinks OP hasn't shown enough effort, I'll accept that and delete my answer. – Sangchul Lee Feb 16 '25 at 07:33
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    I think your answer is great, and if this question gets deleted, it'd be sad to see this answer deleted too. – Jakobian Feb 16 '25 at 19:54
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On the space $\mathbb R^I$ of all functions $f:I\to\mathbb R$, the pointwise convergence is the convergence with respect to the standard product topology, where all the open sets have the form $\{f:f(a_i)\in U_i\}$ where $a_1,...,a_n\in I$ and $U_i\subset\mathbb R$ are open.

This is not a normable space, when $I$ is uncountable, which is in your case, is not metrizable. See this post.

Liding Yao
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