Suppose $K\in L^2(\mathbb R)$ is symmetric s.t. $$ \text{ess sup}_{t\in \mathbb R/\{0\}} \frac{|1-\hat K(t)|}{|t|^{\beta}}\leq A, $$ where $\hat K$ is the Fourier transform.
Theorem 1.5 in "Introduction to Nonparametric Estimation" by B.Tsybakov claim that the condition above is equivalent to $$ \exists t_0, A_0<\infty, \text{ess sup}_{0<t\leq|t_0|} \frac{|1-\hat K(t)|}{|t|^{\beta}}\leq A_0. $$ I'm not sure why this is true. Can Fourier transform of $K$ be bounded by a polynomial at infinity?
Remark: Exercise 1.6 in the same book adds the assumption that $K\in L^{\infty}(R)$, which makes the equivalence quite obvious. I wonder if the equivalence still holds without such an assumption.
Assuming "symmetric" means "even" it's clearly false. Because $\hat K$ can be any symmetric $L^2$ function...