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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 $(0,\delta)$.

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    David, What if I add the condition that $f_n$ is convex everywhere for every $n$.2012-04-24
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    @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 $f$.2012-04-24
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    Thanks, @David Mitra. I'm wondering if imposing a smoothness condition on $f_n$ might remedy this. Sorry for all the questions.2012-04-24
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    Actually, I only need strict convexity almost everywhere. Is this even possible? Thanks again, @David.2012-04-24
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    @Stuck_pls_help I don't think it's good form to change your question entirely when editing, but should instead add the revised question as an addendum. As it stands now, my answer does not address the edited question. As for the latest version of your question, I think the example I described in my last comment shows it is not always the case.2012-04-24