I am reading the book Fourier Analysis by Javier Duandikoetxea and I am stuck in the proof of a lemma that it is the key part to prove the Hörmander's multiplier theorem. This is the lemma:

Let $m\in L_a^2(\mathbb{R}^n),a>\frac{n}{2}$ and $\lambda>0$. We define the operator:
\begin{align*} T_{\lambda m}:L^2(\mathbb{R}^n)&\rightarrow L^2(\mathbb{R}^n) \\ f&\mapsto T_{\lambda m}(f) \end{align*} where $\widehat{T_{\lambda m}(f)}(\xi)=m(\lambda\xi)\hat{f}(\xi)$. Then $\forall~u:\mathbb{R}^n\to\mathbb{C}$: $$\int_{\mathbb{R}^n}|T_{\lambda m}(f)(\xi)|^2u(\xi)d\xi\leq C\int_{\mathbb{R}^n}|f(\xi)|^2\mathcal{M}u(\xi)d\xi$$ where $C>0$ is a constant which is independent from $u$ and $\lambda$ and $\mathcal{M}$ is the Hardy-Littlewood maximal operator which is defined as: $$\mathcal{M}(f)(x)=\sup_{x\in Q}\frac{1}{|Q|}\int_{Q}|f(\xi)|d\xi$$ where $Q$ is any cube containing $x$, and: $$L_a^2(\mathbb{R}^n)=\lbrace g:L^2(\mathbb{R}^n)\to L^2(\mathbb{R}^n)\mid (1+|\xi|^2)^{a/2}\hat{g}(\xi)\in L^2(\mathbb{R}^n)\rbrace$$

Proof: If $K =\hat{m}$ in then by our hypothesis, $(1 + |x|^2 )^{a/2}K(x) = R(x)\in L^2(\mathbb{R}^n)$ , and the kernel of $T_{\lambda m}$ is $\lambda^{-n}K(\lambda^{-1}x)$. Hence: $$\int_{\mathbb{R}^n}|T_{\lambda m}(f)(\xi)|^2u(\xi)d\xi=\int_{\mathbb{R}^n}\left|\int_{\mathbb{R}^n}\frac{\lambda^{-n} R(\lambda^{-1}(\xi-x))}{(1+\left|\lambda^{-1}(\xi-x)\right|)^{a/2}}f(x)dx\right|^2u(\xi)d\xi\leq$$ $$\leq||m||_{L_a^2}^2\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\frac{\lambda^{-n}|f(x)|^2}{(1+\left|\lambda^{-1}(\xi-x)\right|)^{a/2}}dx u(\xi)d\xi\leq C_a ||m||_{L_a^2}^2\int_{\mathbb{R}^n}|f(x)|^2Mu(x)dx $$

Honestly I am quite lost with the proof, since it skips many steps and it is hard for me to read it. It seems obvious that the first inequality should involve Cauchy-Scwarz and that the second inequality should involve Fubinni, but I am not able to write all the steps, especially the first equality has zero sense for me, since I can't get any of these symbols in my calculations. Sorry if I can't give many thoughts, but I don't have anything better. One of the things that annoys me is that I can't see what is the point of knowing what the kernel of $T_\lambda$ is.

Any hints or help will be thanked.

  • $\begingroup$ Sorry, might be a stupid question, what's the definition of kernel here? $\endgroup$
    – mathdoge
    Mar 24, 2022 at 1:00
  • $\begingroup$ I think that $T_{m\lambda}(f)$ equals to the convolution of the kernel with $f$. It follows from taking inverse transform to $m(\lambda \xi) \hat{f}$. Inverse transform of a product equals to convolution of inverse transforms. The first equality follows from this fact. $\endgroup$
    – mathdoge
    Mar 24, 2022 at 1:09
  • $\begingroup$ You are right about the remaining. The first inequality is just Holder's, and the second follows from Fubini's. $\endgroup$
    – mathdoge
    Mar 24, 2022 at 1:18
  • $\begingroup$ @mathdoge can you elaborate more on that? I don't undestand why is the convolution of the kernel and no the convolution of the original function. Also, I don't know what is the definition of the kernel here, I guess it has to be the usual one, but I am not able to prove that this is indeed the kernel. $\endgroup$
    – Marcos
    Mar 24, 2022 at 8:25
  • 1
    $\begingroup$ Try to compute $T_{m\lambda}(f) = \mathcal{F}^{-1} (\widehat{T_{m\lambda}(f)}) = \mathcal{F}^{-1}(m(\lambda\xi) \hat{f})$. $\endgroup$
    – mathdoge
    Mar 24, 2022 at 15:15


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