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Consider the set $\ K=C[0,1]$, the set of continuous functions on $\ [0,1]$, with the supremum norm. Let $\ A= \{f: f$ is power series with infinite radius of convergence and is bounded by $1\}$. I am asked to prove or disprove whether A is compact or not.

I tried noting a counterexample to show that the set isn't closed. $$ \sum^{\infty}_{k=0}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}=\sin(x) $$

This is a power series which converges to $\sin(x)$ which isn't in the set. Does this work?

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    $\begingroup$ What makes you think $ \sin(x) $ isn't in the set? $\endgroup$ – Starfall Sep 24 '18 at 3:41
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    $\begingroup$ Why wouldn’t sine be in that set? $\endgroup$ – Joel Sep 24 '18 at 3:41
  • $\begingroup$ Isn’t what you gave a power series? Isn’t sine bounded by $1$? $\endgroup$ – Clayton Sep 24 '18 at 3:42
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Let $f_n(x) = x^n, n=1,2,\dots $ Then each $f_n\in A.$ If $A$ were compact, then some subsequence $f_{n_k}$ would converge uniformly to some $f\in A.$ This would imply $f_{n_k}\to f$ pointwise on $[0,1].$ But the full sequence $f_n,$ hence the subsequence $f_{n_k},$ converges pointwise to the function that equals $0$ on $[0,1)$ and equals $1$ at $1.$ That function isn't even in $K,$ much less $A,$ contradiction. Thus $A$ is not compact.

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Hint: Stone-Weierstrass theorem.

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  • $\begingroup$ Can I say that by the Weierstrass approximation theorem we can find a sequence of polynomials, or power series, which converges in norm to a continuous function which is not in the set A? $\endgroup$ – Gaby Alfonso Sep 24 '18 at 3:49
  • $\begingroup$ @GabeBoyAnalysis Yes. $\endgroup$ – Starfall Sep 24 '18 at 3:51

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