For fixed $d\geq 1$ and $\beta\in (1,2]$, consider the two following 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 since the defining condition in the second bullet point above is symmetric in $x$ and $y$, we can apply the triangle inequality to 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 $n\to\infty$.