[Math] Pointwise convergence to infinity implies uniformly convergence

general-topologyreal-analysisuniform-convergence

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.

Best Answer

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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