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Let $(f_n)$ be sequence of continuous, real valued functions on a compact subset $\Omega \subset \mathbb{R}^n$. Assume that $(f_n(x))$ converges pointwise to infinity. Are there any conditions ensure a subset $A \subset \Omega$ that $(f_n)$ converges uniformly to infinity on $A$?

I found Dini's Theorem and Egorov's Theorem, but both of them don't have infinity limit.

Thank you.

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  • $\begingroup$ A sufficient condition for uniform convergence: the $f_n$ are positive and have the same Lipschitz constant. The proof is very straightforward. $\endgroup$ – Gabriel Romon Feb 28 '16 at 18:54
  • $\begingroup$ A subset? That's easy: Any finite subset of $\Omega.$ And in some cases, that's all you can say. $\endgroup$ – zhw. Feb 28 '16 at 21:32
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For $x\in \mathbb R,$ define $f_n(x) = n^2|x-1/n|.$ Then each $f_n$ is continuous everywhere, and $f_n \to \infty$ pointwise on $\mathbb R.$

Set $K = \{1/k: k \in \mathbb N \} \cup \{0\}.$ Then $K$ is compact. The only subsets of $K$ where $f_n \to \infty$ uniformly are the finite subsets. Proof: $f_n(1/n) = 0, n \in \mathbb N.$

As for Egorov's theorem, it certainly holds for pointwise convergence to $\infty.$ Just replace $f_n$ by $\arctan f_n.$ The latter sequence converges pointwise to the constant $\pi/2.$ Apply the usual Egorov to get uniform convergence to $\pi/2$ on a "large" subset of $K,$ then apply $\tan$ to get the desired conclusion.

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  • $\begingroup$ Thank you. We can still use D_S's arguments above to get a uniformly convergent subsequence, can't we? How do you think about that? And may I ask you how you came up with that example? $\endgroup$ – user1210321 Feb 29 '16 at 4:43
  • $\begingroup$ No, there is no subsequence that converges uniformly on $K$. $\endgroup$ – zhw. Feb 29 '16 at 5:23
  • $\begingroup$ I added to my answer with a comment on Egorov's theorem. $\endgroup$ – zhw. Feb 29 '16 at 16:40

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