Consider $m(\xi)=\frac{1}{\mu+|\xi|^{2\alpha}}$, where $\xi\in\mathbb{R}^n$, $\mu, \alpha>0$, I want to know that if $m(\xi)$ is a multiplier of $\mathcal{M_{1}^{\infty}}$,i.e., whether the associated convolution operator $Tf=\mathcal{F^{-1}}(m)*f$ is bounded from $L^1$ to $L^\infty$.

When $2\alpha>n$, $m(\xi)\in L^{1}$,so the Fourier transform is bounded (actually $C_{0}$) simply by Riemman-Lebesgue lemma. I want to know whether this is true for all $\alpha>0$. Since the function is radius, so I think there are some way to analyze it.

3more comments