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Suppose a function is Hölder continuous with exponent $\alpha_1 >0$. Then, for $0 < \alpha_2 < \alpha_1$, is it true that the Hölder continuity still holds with exponent $\alpha_2$?

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If your function is bounded then yes: Take $0 then

$ \frac{|f(x)-f(y)|}{|x-y|^a}=\left( \frac{|f(x)-f(y)|}{|x-y|^b} \right)^{\frac{a}{b}}|f(x)-f(y)|^{1-\frac{a}{b}} \leq 2^{1-\frac{a}{b}} \sup |f|^{1-\frac{a}{b}} \left( \frac{|f(x)-f(y)|}{|x-y|^b} \right)^{\frac{a}{b}} $ This says that $C^{0,b} \subset C^{0,a}$

If $f$ is not bounded then not always: $f(x)=x$ is Lipschitz but not Hölder for any other order in $[0,\infty)$ because if it were $|x|^{1-a}\leq c$ for some constant $c$ which is false

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