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 $\mathbb{R}$.
Uniform converge and strict convexity
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analysis
1 Answers
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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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0@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