Is there an elementary example of a function f, such that $|f(x+t)+f(x-t)-2f(x)|/|t|^a\le C$, where $a>1$, such that $f$ is not $C^1$?
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2$\begingroup$ Maybe you mean $|t|^a$ , not $|x|^a$ , in the denominator? $\endgroup$– Pietro MajerCommented Jan 30, 2013 at 9:33
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1$\begingroup$ If you only want to have a fixed $x$ (as it seems since you accepted Xandi's answer) you can take any odd function $f$ with $f(0)=0$. $\endgroup$– Jochen WengenrothCommented Jan 30, 2013 at 15:53
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1$\begingroup$ (so, any odd function) $\endgroup$– Pietro MajerCommented Jan 30, 2013 at 16:28
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1 Answer
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Yes, i think.
Pick an integer $n>1$ and let $f$ be the function given by $f(x) = x^n\sin(x^{-n})$ for $x>0$ and $f(x)=0$ for $x\leq 0$. It is continuous, and smooth except at $x=0$. It is not $C^1$ at zero, because the derivative for $x>0$ is $nx^{n-1}\sin(x^{-n}) - nx^{-1}\cos(x^{-n})$, which enters in a state of flailing tantrum close to $0$.
On the other hand, we have $|f(0+t)+f(0-t)-2f(0)| = |f(t)|$, so every $\alpha < n$ is good.
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$\begingroup$ Are you sure? It seems to me that $f$ is not even in the Zygmund class $(a=1)$. In fact, it fails to satisfy the condition for any $a > 1 - 1/(n+1)$. $\endgroup$ Commented Jan 30, 2013 at 15:46
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1$\begingroup$ Are you sure that there is a constant $C$ that works in this example for all other $x$? $\endgroup$ Commented Jan 30, 2013 at 15:47
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$\begingroup$ Indeed, if a>1−1/(n+1), there are no positive numbers C,ϵ such that the inequality holds for all |x|< ϵ and |t|<ϵ. $\endgroup$ Commented Jan 30, 2013 at 16:36