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This is a question on the convergence of a sequence of real, convex, analytic functions (it does not get better than that!):

Let $(f_n)_{n\in \mathbb N}$ be a sequence of convex analytic functions on $\mathbb R$.

Suppose that $f_n(x) \to f(x)$ as $n \to \infty$ for all $x \in \mathbb R$ (or in $\mathbb R^+$).

Is $f(x)$ analytic?

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No—not even necessarily differentiable! The function $f_n(x) = \frac1n\log(1+e^{nx})$ is convex and analytic on $\Bbb R$, but $$ \lim_{n\to\infty} \frac1n\log(1+e^{nx}) = \begin{cases} 0, &\text{if } x\le 0, \\ x, &\text{if } x\ge0. \end{cases} $$

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  • $\begingroup$ Did you forget to divide by $n?$ $\endgroup$
    – zhw.
    Aug 27 '20 at 17:30
  • $\begingroup$ @zhw Yes, thank you! Haha, I checked the limit graphically, but didn't notice the $y$-axis range was changing :D $\endgroup$ Aug 27 '20 at 17:32
  • $\begingroup$ Thanks! So I guess, the additional assumption that could make it work would be to add the constraint of bounded derivatives. $\endgroup$
    – LaplaceFan
    Aug 27 '20 at 19:28
  • $\begingroup$ In this and my example, the derivatives are bounded. $\endgroup$
    – zhw.
    Aug 27 '20 at 19:32
  • $\begingroup$ The derivative in 0 is actually growing with n in your example. I meant bounded with a constant K independent on n $\endgroup$
    – LaplaceFan
    Aug 27 '20 at 20:02
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Counterexample: Define $f_n(x) = (x^2+1/n)^{1/2}.$ Then each $f_n$ is analytic and convex on $\mathbb R.$ Clearly $f_n(x)\to |x|$ pointwise everywhere. (A little more work shows $f_n(x)\to |x|$ uniformly on $\mathbb R.$)

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  • $\begingroup$ Great and simple example! $\endgroup$
    – LaplaceFan
    Aug 27 '20 at 19:29

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