Let $f$ be a $\alpha$ Hölder continuous function. I need an estimate from above for the expression $\left|f(t_{i_{1}})-f(t_{i_2})\right|+\left|f(t_{i_2})-f(t_{i_{3}})\right|$ Is it possible to estimate this expression by $\left|t_{i_1}-t_{i_3}\right|$. $t_{i_1}\leq t_{i_2}\leq t_{i_3}$?
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$\leq L |t_{i_1}-t_{i_2}|^\alpha + |t_{i_2}-t_{i_3}|^\alpha$. But this seems to be too easy... Can you be more precise, please? – Siminore Nov 14 '16 at 12:34
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Yes, this is too easy. I have a sum of increments of a Hölder Continuous function. And I need an nice upper bound for that sum. Applying the Hölder property to each summand provides an bad upper bound. – ithusiasm Nov 14 '16 at 12:36
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Note that for $a < b$ and $0 < \alpha \le 1$ the function $x \mapsto (b - x)^\alpha + (x - a)^\alpha$ is concave on $[a, b]$ and therefore has a unique maximum. Since $f'(x) = -\alpha (b - x)^{\alpha - 1} + \alpha (x - a)^{\alpha - 1}$ is zero for $x = \frac{a + b}{2} \in [a, b]$, the maximum is attained at this point and $f\left(\frac{a + b}{2}\right) = 2^{1 - \alpha} (b - a)^{\alpha}$.
Applied to your problem, this means $$|f(t_{i_1}) - f(t_{i_2})| + |f(t_{i_2}) - f(t_{i_3})| \le C \left((t_{i_1} - t_{i_2})^\alpha + (t_{i_2} - t_{i_3})^\alpha\right) \le C 2^{1 - \alpha} (t_{i_3} - t_{i_1})^{\alpha}$$ for $t_{i_1} \le t_{i_2} \le t_{i_3}$.
Dominik
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This depends on how well the approximation needs to be. You can apply this repeatedly to get a bound of the form $C(2^{(n - 2)(1-\alpha)}(t_{i_n} - t_{i_1})^\alpha$. But unless $\alpha = 1$, the constant in front of $t_{i_n} - t_{i_1}$ will increase with $n$, since not every Hölder continuous function is absolutely continuous. C.f. http://math.stackexchange.com/a/972681/259493 – Dominik Nov 14 '16 at 13:21