Let $p>0$ be any real positive. Does there exist a function $f(x)$ which is $o(|x|^p)$ in $x=0$ yet not $O(|x|^{p+\varepsilon})$ for any $\varepsilon>0$ ?
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$f(0) = 0$ and on $[1/2^n, 1/2^n-1[$, $f(x)=x^{p+1/\sqrt{n}}$
The idea is to have something like $f(x) = x^{p+\alpha}$ with alpha smaller than epsilon for any $\varepsilon$ when $x \to 0$.
Or $f(x) = x^{p-log(x)}$
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Understood the idea but everything I tried made it not o(|x|^p) – ORBOT Inc. Nov 02 '13 at 11:24
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Like your example: f(x)/|x|^p = |x|^|x|^2->1 – ORBOT Inc. Nov 02 '13 at 11:25
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ok, the previous one worked, I thought that p+x^2 was cleaner but didnt check it, reverting to first answer – Thomas Nov 02 '13 at 11:27
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What about: f(x)=|x|^(p+1/sqrt(-ln(|x|))) ? Checking it now – ORBOT Inc. Nov 02 '13 at 11:29
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yeah, this is a cleaner version of the [1/2^n...] thing – Thomas Nov 02 '13 at 11:30
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Cool, that works. – ORBOT Inc. Nov 02 '13 at 11:32