I was doing the following exercise: if $ f: \mathbb{R}^n \setminus \{ 0 \} $ defined by $ f (x) = | x |^\alpha x_i $ for a fixed i, then $ f \in C^{0, \beta} (B_1 (0)) $, for $ 0<\alpha\leq 1$.
I tried to use the same idea as in the case where $ f (x) = |x|^\alpha $ (as for example in + Proving that $f(x) = \vert x \vert^{\alpha}$ is Holder continuous, inequality help). But I cannot simplify and adapt the expression properly. A tip would be welcome!
Here this is much easier since the function is Lipshitz. Since $∇|x| = \frac{x}{|x|}$, we can compute $$ ∇f(x) = \alpha \,|x|^{\alpha-2}\, x_i\, x + |x|^{\alpha} 1_i $$ where I used the notation $1_i = (\delta_{i,j})_{j=1..n}$ to denote the vector with coordinates $0$, except the $i$-th coordinate, which is $1$.
Therefore, $$ |∇f(x)| \leq \alpha\,|x|^{\alpha-1} |x_i| + |x|^{\alpha} ≤ (α +1) \, |x|^\alpha, $$ from which we deduce that $\|∇f\|_{L^\infty(B_1(0))} ≤ α+1$, so that $f$ is Lipshitz continuous $f∈C^{0,1}$ (and so Hölder continuous for any $\alpha\leq 1$ since $C^{0,1}⊂ C^{0,\alpha}$).