Functions $f: \mathbb{R}^n \to \mathbb{R}$ such that $|f(x) -f(y)| \le C \prod_{i=1}^n |x_i - y_i|^{\alpha_i}$

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The standard definition of a Holder continuous function between metric spaces $X,Y$ is a function $f: X \to Y$ such that there exist $C>0$ and $0 < \alpha \le 1$ such that $$ d_Y(f(x),f(y)) \le C (d_X(x,y))^\alpha \text{ for all }x,y \in X. $$ When specialized to $X= \mathbb{R}^n$ and $Y = \mathbb{R}$ or $\mathbb{C}$ this reads $$ |f(x)-f(y)| \le C |x-y|^\alpha \text{ for all }x,y \in \mathbb{R}^n. $$ It is a simple matter to find examples of such functions, with the easiest case being when $\alpha =1$, as in this case any differentiable function with bounded derivative will obey the $\alpha =1$ (Lipschitz) estimate.

I was recently looking at Classical Fourier Analysis by Grafakos. In the study of pointwise convergence of Fourier series he proves a result (Corollary 3.3.8) that says that if we have a function on the $n-$torus, $f: \mathbb{T}^n \to \mathbb{C}$, and a point $y \in \mathbb{T}^n$ such that $$ |f(x) - f(y)| \le C \prod_{i=1}^n |x_i - y_i|^{\alpha_j} \text{ for all }x \in \mathbb{T}^m $$ then the Fourier partial sums converge at $y$. This suggests that the collection of functions satisfying this condition is of some interest.

It's not hard to build a function that satisfies this condition at a point. To make things simpler let's swap to $\mathbb{R}^n$ from the torus and restrict to real-valued functions. Then we can define $f(x) = \prod_{i=1}^n |x_i|^{\alpha_i}$ to get a trivial example satisfying the condition at $y=0$. I then started thinking about this condition as a general one, satisfied for all pairs of points $x,y$, i.e. I want a function $f: \mathbb{R}^n \to \mathbb{R}$ such that there exist $C>0$ and $0 < \alpha_i \le 1$ such that $$ |f(x) - f(y)| \le C \prod_{j=1}^n |x_i - y_i|^{\alpha_i} \text{ for all }x,y \in \mathbb{R}^n. $$
This is a sort of strange variant of the standard Holder continuity condition. Note that to avoid triviality we should assume that $\sum_i \alpha_i \le 1$ as otherwise $$ |f(x)-f(y)| \le C |x-y|^{1+\epsilon} $$ for $\epsilon = \sum_{i} \alpha_i -1 >0$, and so $f$ is constant.

My problem is that I can't seem to come up with any example that satisfies this condition for all $x,y$. Initially I guessed functions of the form $f(x) = \prod_{i=1}^n f_i(x_i)$ would work, but this doesn't fly. For example, with $n=2$ we get $$ f_1(x_1)f_2(x_2) - f_1(y_1) f_2(y_2) = (f_1(x_1) - f_1(y_1))f_2(x_2) + f_1(y_1)(f_2(x_2) - f_2(y_2)) $$ and so if $f_1,f_2$ are bounded and Holder continuous (with exponents $\alpha_1,\alpha_2$) we get $$ |f(x) - f(y)| \le C |x_1 - y_1|^{\alpha_1} + C |x_2 - y_2|^{\alpha_2}. $$ This is not what we want.

Do such functions exist? If so, is there a sufficient condition for generating them (something like the bounded derivative condition above)? Any ideas would be greatly appreciated!

EDIT: @Dap made the observation I had overlooked. If $x_i = y_i$ then $f(x)=f(y)$ and so $f$ must be constant on every hypersurface parallel to the coordinate axes. This shows that $f$ is constant.