Is $f_{n}$ is analytic on $(a, b)$ and $f_{n} \rightarrow f$ uniformly on $(a, b)$ then is $f$ analytic on $(a, b)$?

Intuitively, I think that the answer is no. I know that the statement holds for integrability and continuity; however, I don't think it's necessary for analyticity. Am I correct?

  • 2
    $\begingroup$ Try $f_n(x)=\frac1n\sqrt{1+n^2x^2}$ on $(-1,1)$. $\endgroup$ – Did Dec 14 '18 at 15:30

Yes, you are correct. Just consider$$\begin{array}{rccc}f_n\colon&(-1,1)&\longrightarrow&\mathbb R\\&x&\mapsto&\sqrt{x^2+\frac1{n^2}}.\end{array}$$The sequence $(f_n)_{n\in\mathbb N}$ is a sequence of analytic functions that converges uniformly to the absolute value functions, which isn't differentiable.

  • $\begingroup$ Nice counterexample $\endgroup$ – joseph Dec 14 '18 at 15:33
  • $\begingroup$ @joseph If my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ – José Carlos Santos Dec 14 '18 at 16:59

Stone-Weierstrass theorem indicates that any continuous function $f:[a,b]\to\Bbb R$ is a uniform limit of a sequence of polynomials $p_n$. Polynomials are obviously analytic but $f$ need not be differentiable.


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