# Holder continuity for $\alpha$ but not $\alpha+\varepsilon$?

Is it possible to find a function which is, say, α− Holder continuous, for 0<α≤1, but not α+ε- Holder continuous for arbitrary ε>0? I was thinking that perhaps $|x|^\alpha$ could work? It's easy to prove that function is α−Holder continuous but I'm not sure how to prove it isn't when you make α slightly larger.

## 1 Answer

You are right, $f(x) = |x|^{\alpha}$ works. You can show that it is not Holder continuous with exponent $\alpha + \epsilon$ by taking $x = s$ and $y = 0$ in the definition of Holder continuity: $$\frac{|f(x) - f(y)|}{|x -y|^{\alpha + \epsilon}} = s^{-\epsilon},$$ which tends to $\infty$ as $s \to 0$.

Notice, however, that $f(x)$ is Holder continuous for exponents less than $\alpha$ only if you look at a bounded interval. On $\mathbb R$, it is not, because $$\frac{|f(s) - f(0)|}{|s -0|^{\alpha - \epsilon}} = s^{\epsilon},$$ which tends to $\infty$ as $x \to \infty$.

• Ah, of course. Thank you! – Hugh Mungus Jan 28 '17 at 18:18
• You're welcome! – Roberto Rastapopoulos Jan 28 '17 at 18:27