Convergence of a sequence of real convex analytic functions

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?

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

• Did you forget to divide by $n?$
– zhw.
Aug 27 '20 at 17:30
• @zhw Yes, thank you! Haha, I checked the limit graphically, but didn't notice the $y$-axis range was changing :D Aug 27 '20 at 17:32
• Thanks! So I guess, the additional assumption that could make it work would be to add the constraint of bounded derivatives. Aug 27 '20 at 19:28
• In this and my example, the derivatives are bounded.
– zhw.
Aug 27 '20 at 19:32
• The derivative in 0 is actually growing with n in your example. I meant bounded with a constant K independent on n Aug 27 '20 at 20:02

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 $$\mathbb R.$$)

• Great and simple example! Aug 27 '20 at 19:29