1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Try $f_n(x)=\frac1n\sqrt{1+n^2x^2}$ on $(-1,1)$. $\endgroup$ – Did Dec 14 '18 at 15:30
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.