# Compactness of $\ A= \{f: f$ is power series with infinite radius of convergence and is bounded by $1\}$

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?

• What makes you think $\sin(x)$ isn't in the set? – Starfall Sep 24 '18 at 3:41
• Why wouldn’t sine be in that set? – Joel Sep 24 '18 at 3:41
• Isn’t what you gave a power series? Isn’t sine bounded by $1$? – Clayton Sep 24 '18 at 3:42

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.