# Uniform converge and strict convexity

Suppose $f_n$ converges uniformly to a function $f$. Both the sequence and the function are continuous. Moreover, $f$ is strictly convex on $(0,\delta)$ for some arbitrarily small $\delta$. Is it the case that for some $N$, $f_n$ is strictly convex on "almost everywhere" $(0,\delta)$ for $n>N$ if each $f_n$ is convex fucntion on $\mathbb{R}$.

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Consider defining $f_n$ on $(0,1)$ via $f_n(x)=x^2+{1\over n}\sin (1/x)$ and $f$ via $f(x)=x^2$. Then for any $x$ we have $|f(x)-f_n(x)|\le{1\over n}$, thus $f_n$ converges to $f$ uniformly on $(0,1)$. Given $\delta>0$, the function $f$ is strictly convex $(0,\delta)$. But for each $n$, $f'_{\kern-3pt n}(x)=2x-{1\over nx^2}\cos(1/x)$ takes on both positive and negative values in any interval $(0,\epsilon)$, $\epsilon>0$. From this it follows that $f_n$ is not even convex on $(0,\delta)$.
David, What if I add the condition that $f_n$ is convex everywhere for every $n$. – Stuck_pls_help Apr 24 '12 at 0:44
@Stuck_pls_help I don't think the result has to hold, even then. Loosely speaking: take a continuous, strictly convex, bounded function $f$ defined on $(0,\delta)$ and let $f_n$ be the function whose graph is comprised of $n$ line segments of equal length joining points on the graph of $f$. – David Mitra Apr 24 '12 at 0:53
Thanks, @David Mitra. I'm wondering if imposing a smoothness condition on $f_n$ might remedy this. Sorry for all the questions. – Stuck_pls_help Apr 24 '12 at 2:12