# Proof of a weighted norm inequality.

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.

• Sorry, might be a stupid question, what's the definition of kernel here? Commented Mar 24, 2022 at 1:00
• 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. Commented Mar 24, 2022 at 1:09
• You are right about the remaining. The first inequality is just Holder's, and the second follows from Fubini's. Commented Mar 24, 2022 at 1:18
• @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. Commented Mar 24, 2022 at 8:25
• Try to compute $T_{m\lambda}(f) = \mathcal{F}^{-1} (\widehat{T_{m\lambda}(f)}) = \mathcal{F}^{-1}(m(\lambda\xi) \hat{f})$. Commented Mar 24, 2022 at 15:15