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

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?

• Try $f_n(x)=\frac1n\sqrt{1+n^2x^2}$ on $(-1,1)$. – 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.
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.