If $f$ is in $\mathcal{C}^{1.5}$, is it true that there exists a $C$ such that $|f(x)-f(y)|\leq C |x-y|^{1.5}$

Question

For $f$ to be in $\mathcal{C}^{1.5}$ means that $f$ is continuously differentiable and that its derivative $f'$ is Holder-continuous with exponent $0.5$.

Answer 1

If a continuous function satisfies a Holder condition of order $\alpha>1$ in an interval, then it is a constant there, so the answer must be negative.

Answer 2

We can deduce as follows: $$0\leq \frac{|f(x)-f(y)|}{|x-y|}\leq C|x-y|^{0.5}$$ for any $x\ne y.$

Let $y\to x$ we have $f^{\prime}(x)$ exists and $f^{\prime}(x)=0$.

Since $x$ is arbitrary, we must have $f$ is a constant.