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.

  • $\begingroup$ This function is not in $L^{1}$ for small $\alpha$, so how would you like to define this operator? maybe an operator over $L^{\text{something}}_{loc}$? In particular, as long as the function is not in $L^{1}$, the zero'th Fourier coefficient will be $\infty$. $\endgroup$
    – Asaf
    Dec 26, 2013 at 19:46
  • $\begingroup$ @Asaf, the fourier transform of $L^{1}$ function can be continuously extended to tempered distributions, which is the dual space of Schwartz function with rapidly decreasing, and my function should be seen as a tempered distribution. $\endgroup$
    – Tomas
    Dec 26, 2013 at 23:12
  • 5
    $\begingroup$ If it were, then so would be $M_t(\xi)=\frac {e^{-t|\xi|^2}}{\mu+|\xi|^{2\alpha}}$ for all $t>0$, and, moreover, the norms would be uniformly bounded (one more convolution with scaled $e^{-|x|^2}$ on the space side). Now the Fourier transform is classical and it becomes obvious that the answer is "No" (value $[\mathcal F^{-1}M_t](0)$ is huge for small $t$). $\endgroup$
    – fedja
    Dec 31, 2013 at 1:24
  • $\begingroup$ @fedja, thank you very much for your comments. But I don't quite understand your first sentence, why should the Fourier transform of $M_{t}(\xi)$ be uniformly bounded? It seems that your argument applies for all $\alpha$, but when $2\alpha>n$, $\mathcal{F}^{-1}(m)$ is indeed bounded, do I miss something? $\endgroup$
    – Tomas
    Dec 31, 2013 at 1:45
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    $\begingroup$ Fedja's argument is correct. As he notes parenthetically, the Fourier transform of $M_t$ is the convolution of the (distributional) Fourier transform of $m$ with a gaussian approximation to the identity, so if the latter is bounded, the former is also. The difference when $2\alpha > n$ is that ${\mathcal F}^{-1} M_t(0)$ remains bounded as $t \to 0$ in this case. $\endgroup$
    – Terry Tao
    Dec 31, 2013 at 16:42


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