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Let $f_n:[0,1]\to\mathbb{R},\ n\in\mathbb{N}$ be a sequence of convex analytic functions.

Consider the sequence of derivatives $(f_n')_{n\in\mathbb{N}}$. Suppose that $$f'_n(x)\xrightarrow[n\to\infty]{}g(x)\in\mathbb{R}\ \text{for all}\ x\in\mathbb{Q\cap[0,1]}.$$

Is it true that $(f_n')_{n\in\mathbb{N}}$ converges uniformly on $\mathbb{Q}\cap[0,1]$?

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up vote 2 down vote accepted

Not necessarily. Let $f_n(x)=\dfrac{x^{n+1}}{n+1}$. Each $f_n$ is analytic and convex, and $f'_n(x)=x^n$. The sequence $f'_n$ converges pointwise in $[0,1]$ to the function defined by $g(x)=0$ if $0\le x<1$, $g(1)=1$. However, $f'_n$ does not converge uniformly to $g$ on $\mathbb{Q}\cap[0,1]$.

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OK , thank you! – qwertyuio Jan 30 '13 at 18:07

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