I posted the following question on StackExchange a few months ago (https://math.stackexchange.com/questions/2898620/notions-of-beta-h%C3%B6lder-smoothness-when-beta-in-1-2-are-they-equivalent), but didn't receive any replies. As this is still unresolved, I'm reposting this here in the hope that someone in the MathOverflow community might be able to help me out. Thanks!

For fixed $d\geq 1$ and $\beta\in (1,2]$, consider the following two classes of functions:

- Let $\mathcal{H}^\beta$ denote the collection of all $C^1$ functions $\phi:\mathbb{R}^d\to\mathbb{R}$ for which there exists $L>0$ such that $\|\nabla\phi(y)-\nabla\phi(x)\|_2\leq L\|y-x\|_2^{\beta-1}$ for all $x,y\in\mathbb{R}^d$.

[*This is sometimes referred to as a $\beta$-Hölder class.*] - Let $\tilde{\mathcal{H}}^\beta$ denote the collection of all $C^1$ functions $\phi:\mathbb{R}^d\to\mathbb{R}$ for which there exists $\tilde{L}>0$ such that $\lvert\phi(y)-\phi(x)-\nabla\phi(x)^\top(y-x)\rvert\leq\tilde{L}\|y-x\|_2^\beta$ for all $x,y\in\mathbb{R}^d$.

Is it true that $\mathcal{H}^\beta=\tilde{\mathcal{H}}^\beta$? I believe the answer is yes if either $d=1$ or $\beta=2$, but I can't see how to tackle the general case. Here are some preliminary remarks and observations:

- It is not hard to see that $\mathcal{H}^\beta\subseteq\tilde{\mathcal{H}}^\beta$: indeed, fix $x,y\in\mathbb{R}^d$ and apply the chain rule and mean value theorem to the function $\Phi:[0,1]\to\mathbb{R}$ defined by $\Phi(t):=\phi(x+t(y-x))$.
*The case $d=1$*: if $\phi\in\tilde{\mathcal{H}}^\beta$, observe that if we interchange $x$ and $y$ in the defining condition in the second bullet point above and apply the triangle inequality, we can deduce that $\lvert\{\phi'(y)-\phi'(x)\}(y-x)\rvert\leq 2\tilde{L}\lvert y-x\rvert^\beta$, which implies that $\phi\in\mathcal{H}^\beta$, as required. However, this argument does not generalise to higher dimensions ...*The case $\beta=2$*: fix $\phi\in\tilde{\mathcal{H}}^\beta$ and suppose first that $\phi$ is twice differentiable on $\mathbb{R}^d$. Then in view of Taylor's theorem with the Peano form of the remainder (cf. Exercise 9.30(b) in Rudin's Principles of Mathematical Analysis), the defining condition for $\tilde{\mathcal{H}}^2$ implies that the second derivative of $\phi$ is bounded in operator norm by $2\tilde{L}$. But then it follows from the mean value inequality (Theorem 5.15 in Rudin) that $\|\nabla\phi(y)-\nabla\phi(x)\|_2\leq 2\tilde{L}\|y-x\|_2$ for all $x,y$, as required.

To handle the case where $\phi$ is not twice differentiable, define $\phi_n:=\phi\ast g_{1/n}$ for each $n\in\mathbb{N}$, where we let $g$ be a smooth bump function supported on the unit ball and define $g_\varepsilon(x):=\varepsilon^{-d}g(x/\varepsilon)$ for $x\in\mathbb{R}^d$ and $\varepsilon>0$. It is easy to verify that each $\phi_n$ is smooth and belongs to $\tilde{\mathcal{H}}^2$ (with the same $\tilde{L}$ as before). Since $\nabla\phi_n\to\nabla\phi$ pointwise, we can obtain the desired conclusion by sending $n\to\infty$.