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Question

Related to a nice problem I met yesterday, a question arises:

Suppose $\{f_n\}$ is a sequence of mappings from a connected complete metric space $X$ to a metric space $Y$. Given $f\colon X\to Y$ such that $f_n(x_n)\to f(x)$ whenever a real sequence $x_n\to x$. Must $\{f_n\}$ converge uniformly?

According to Canez's hint, it's true that $f_n\to f$ pointwise and $f$ is continuous (modifying my ugly answer, replace $+1/N_k$ with $+0$) even if $X$ is an arbitrary metric space (without assuming that $X$ is connected or complete).

If it's too abstract, we can consider a concrete one, such as $X=Y=\mathbb R$ or $X=Y=\mathbb C$.

Explanations

If $X$ is compact, it's true that $f_n\to f$ uniformly.

Prove by contradiction. Suppose $\exists\epsilon_0>0,\forall N,\exists n\ge N$ such that $$\sup_{x\in X}d_Y(f_n(x),f(x))>\epsilon_0$$

We can choose a increasing sequence $\{m_k\}\subset\mathbb N$, and a sequence $\{x_{m_k}\}\subset X$, such that $d_Y(f_{m_k}(x_{m_k}),f(x_{m_k}))>\epsilon_0$ for all $k$. Since $X$ is compact, there's a subsequence $\{n_k\}$ of $\{m_k\}$ such that $x_{n_k}\to x$ for some $x\in X$, then we can show that $f_{n_k}(x_{n_k})\to f(x)$ by constructing a sequence with duplicated $x_{n_k}$'s, just like this. On the other hand, since $f$ is continuous, we have $f(x_{n_k})\to f(x)$ contradicting to $d_Y(f_n(x),f(x))>\epsilon_0$.

In contrary to the preceding statement, if $X$ isn't connected (or at least, with finite connected components), even if $X$ is complete, $f_n\to f$ needn't be uniform, such as $X=\mathbb Z$.

Generalization

I don't know whether $X$ could be replaced with an arbitrary connected metric space, or even $X$, $Y$ are just Hausdorff spaces. I don't know whether the preceding compact metric space $X$ could be replaced with a compact Hausdorff space. It's out of my ability and I can't discuss more.

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  • $\begingroup$ For $X=Y=\Bbb R$, the sequence $(f_n)$ where $f_n=\chi_{[n,n+1]}$ would be a counterexample. (Note in the Explanations section, you might only have a subsequence of $(f_n)$ that satisfies your $\sup$ condition. But the argument can still be carried through.) $\endgroup$ Feb 23, 2013 at 5:16
  • $\begingroup$ @DavidMitra You're right. Could you pleased post an answer and I'll accept yours? $\endgroup$
    – Yai0Phah
    Feb 23, 2013 at 5:49
  • $\begingroup$ Continuous convergence is equivalent to continuity of $f$ and uniform convergence $f_n \to f$ on compact subsets. See here: math.stackexchange.com/q/182223 $\endgroup$
    – Martin
    Feb 23, 2013 at 12:31

1 Answer 1

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Note in the Explanations section, you might only have a subsequence of $(f_n)$ that satisfies your sup condition. But the argument still carries through. The result holds for compact sets.

For $X=Y=R$, the sequence $(f_n)$ where $f_n=\chi_{[n,n+1]}$ would be a counterexample.

The result also fails for general bounded sets, as the sequence $(x^n)$ on $(0,1)$ witnesses.

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