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Suppose that we have a sequence of functions $\{f_{n}\}$ that converges uniformly to a function $f$ on any $(a,b)\subset \mathbb{R}$. If $K\subset \mathbb{R}$ is compact prove that $\{f_{n}\}$ converges uniformly on $K$.

I know that for something to be uniformly convergent we would have $$|f_{n}(x) - f(x)| \leq~ \epsilon~~~~~ \forall~ x \in E$$

According to Rudin's "Principles of Mathematical Analysis" Theorem 7.12 States that "If $\{f_{n}\}$ is a sequence of continuous functions on $E$, and if $f_{n} \rightarrow f$ uniformly on $E$, then $f$ is continuous on $E$.

In this case I would substitute $\mathbb{R}$ for $E$ and make $K$ a subset of $\mathbb{R}$

Now the following theorem 7.13 states that "Suppose $K$ is compact", and

  • 1) $\{f_{n}\}$ is a sequence of continuous functions on $K$,

  • 2) $\{f_{n}\}$ converges pointwise to a continuous function $f$ on $K$

  • 3) $f_{n}(x)\geq f_{n+1}(x)$ for all $x\in K$, $n = 1, 2, 3\ldots$

Then $f_{n}\rightarrow f$ uniformly on $K$

I'm wondering if I can not only use these two theorems and, if it works it does converge uniformly. Thanks for the help, as always it's very much appreciated.

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    $\begingroup$ Can you use that any compact $K$ is contained in some $(a,b)$? $\endgroup$ – Orest Bucicovschi May 6 '15 at 0:06
  • $\begingroup$ @orangeskid can we let (a,b) $\in$ K? $\endgroup$ – Chris Millett May 6 '15 at 0:07
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    $\begingroup$ The hypothesis " on any $(a,b)$ " is so strong, you might as well use it and finish off the problem. Any compact $K$ is contained in some $(a,b)$. $\endgroup$ – Orest Bucicovschi May 6 '15 at 0:11
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Since $K$ is compact, then is bounded, then $K\subset(a,b)$ for some $a,b\in\mathbb{R}$, Then $$\{f_n\}\to f,\,\mbox{ in } K\subset(a,b)$$ it is all you need.

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