[proofplan] The proof is a weighted $L^2$ estimate on the Fourier side. For $u \in H^s(\mathbb{R}^n)$, the weighted Fourier representative $\langle \xi \rangle^s \hat{u}(\xi)$ lies in $L^2(\mathbb{R}^n)$. The multiplier hypothesis says that multiplication by $\langle \xi \rangle^{-m}a(\xi)$ is bounded pointwise by $M_a$, so the weighted transform $\langle \xi \rangle^{s-m}a(\xi)\hat{u}(\xi)$ lies in $L^2$. The Fourier-side definition of Sobolev spaces then produces the element $Au \in H^{s-m}(\mathbb{R}^n)$, and the same estimate gives the operator norm bound. [/proofplan]

[step:Represent the input by its weighted Fourier transform] Fix $s \in \mathbb{R}$ and let $u \in H^s(\mathbb{R}^n)$. Let $\mathcal{B}(\mathbb{R}^n)$ denote the Borel $\sigma$-algebra on $\mathbb{R}^n$. By the Fourier-side definition of $H^s(\mathbb{R}^n)$, the [Fourier transform](/page/Fourier%20Transform) $\hat{u}$ has a measurable representative, still denoted $\hat{u}: \mathbb{R}^n \to \mathbb{C}$, and the function $v: \mathbb{R}^n \to \mathbb{C}$ defined by $v(\xi) = \langle \xi \rangle^s \hat{u}(\xi)$ belongs to $L^2(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n), \mathcal{L}^n)$. Moreover, $\|u\|_{H^s(\mathbb{R}^n)} = \|v\|_{L^2(\mathbb{R}^n)}$. [/step]

[step:Use the multiplier bound to place the output in the target weighted space]Define the measurable function \begin{align*} w: \mathbb{R}^n \to \mathbb{C},\quad \xi \mapsto a(\xi)\hat{u}(\xi). \end{align*} For $\mathcal{L}^n$-a.e. $\xi \in \mathbb{R}^n$, \begin{align*} \langle \xi \rangle^{s-m} w(\xi) = \langle \xi \rangle^{-m} a(\xi) v(\xi). \end{align*} Taking absolute values and using the essential supremum defining $M_a$ gives \begin{align*} |\langle \xi \rangle^{s-m} w(\xi)| \le M_a |v(\xi)|. \end{align*} Since $v \in L^2(\mathbb{R}^n)$, squaring and integrating with respect to $\mathcal{L}^n$ gives \begin{align*} \int_{\mathbb{R}^n} |\langle \xi \rangle^{s-m} w(\xi)|^2\, d\mathcal{L}^n(\xi) \le M_a^2 \int_{\mathbb{R}^n} |v(\xi)|^2\, d\mathcal{L}^n(\xi). \end{align*} Thus $\langle \xi \rangle^{s-m}w(\xi) \in L^2(\mathbb{R}^n)$.[/step]

[guided]The goal is to prove that $a\hat{u}$ is the Fourier transform of an element of $H^{s-m}(\mathbb{R}^n)$. By the Fourier-side definition of that [Sobolev space](/page/Sobolev%20Space), this means we must prove that the weighted function $\langle \xi \rangle^{s-m}a(\xi)\hat{u}(\xi)$ is square-integrable with respect to $\mathcal{L}^n$. We already know that \begin{align*} v: \mathbb{R}^n \to \mathbb{C},\quad \xi \mapsto \langle \xi \rangle^s \hat{u}(\xi) \end{align*} belongs to $L^2(\mathbb{R}^n)$ because $u \in H^s(\mathbb{R}^n)$. Define \begin{align*} w: \mathbb{R}^n \to \mathbb{C},\quad \xi \mapsto a(\xi)\hat{u}(\xi). \end{align*} Then the target Sobolev weight gives \begin{align*} \langle \xi \rangle^{s-m}w(\xi) = \langle \xi \rangle^{s-m}a(\xi)\hat{u}(\xi). \end{align*} We rewrite this expression so that the known $H^s$ weight appears: \begin{align*} \langle \xi \rangle^{s-m}a(\xi)\hat{u}(\xi) = \langle \xi \rangle^{-m}a(\xi)\bigl(\langle \xi \rangle^s\hat{u}(\xi)\bigr). \end{align*} The expression in parentheses is exactly $v(\xi)$. The multiplier hypothesis says that $|\langle \xi \rangle^{-m}a(\xi)| \le M_a$ for $\mathcal{L}^n$-a.e. $\xi \in \mathbb{R}^n$. Therefore \begin{align*} |\langle \xi \rangle^{s-m}w(\xi)| \le M_a |v(\xi)| \end{align*} for $\mathcal{L}^n$-a.e. $\xi \in \mathbb{R}^n$. Squaring this pointwise inequality and integrating with respect to [Lebesgue measure](/page/Lebesgue%20Measure) gives \begin{align*} \int_{\mathbb{R}^n} |\langle \xi \rangle^{s-m}w(\xi)|^2\, d\mathcal{L}^n(\xi) \le M_a^2 \int_{\mathbb{R}^n} |v(\xi)|^2\, d\mathcal{L}^n(\xi). \end{align*} The right-hand side is finite because $v \in L^2(\mathbb{R}^n)$. Hence $\langle \xi \rangle^{s-m}w(\xi) \in L^2(\mathbb{R}^n)$, which is exactly the weighted Fourier condition needed for membership in $H^{s-m}(\mathbb{R}^n)$.[/guided]

[step:Define the output by the Fourier-side Sobolev characterization] Since $w$ is measurable and $\langle \xi \rangle^{s-m}w(\xi) \in L^2(\mathbb{R}^n)$, the Fourier-side definition of $H^{s-m}(\mathbb{R}^n)$ gives a unique element $Au \in H^{s-m}(\mathbb{R}^n)$ whose Fourier transform is represented by $w$. Thus $\widehat{Au}(\xi) = a(\xi)\hat{u}(\xi)$ for $\mathcal{L}^n$-a.e. $\xi \in \mathbb{R}^n$. This definition is independent of the chosen representative of $\hat{u}$. If $\hat{u}_1$ and $\hat{u}_2$ are two measurable representatives of the Fourier transform of $u$, then $\hat{u}_1 = \hat{u}_2$ for $\mathcal{L}^n$-a.e. $\xi \in \mathbb{R}^n$, and hence $a\hat{u}_1 = a\hat{u}_2$ for $\mathcal{L}^n$-a.e. $\xi \in \mathbb{R}^n$. [/step]

[step:Estimate the Sobolev norm of the output] Using the Fourier-side norm in $H^{s-m}(\mathbb{R}^n)$ and the definition of $w$, we obtain $\|Au\|_{H^{s-m}(\mathbb{R}^n)}^2 = \int_{\mathbb{R}^n} \langle \xi \rangle^{2(s-m)} |a(\xi)\hat{u}(\xi)|^2\, d\mathcal{L}^n(\xi)$. By the multiplier bound $|a(\xi)| \le M_a \langle \xi \rangle^m$ for $\mathcal{L}^n$-a.e. $\xi \in \mathbb{R}^n$, the integrand satisfies $\langle \xi \rangle^{2(s-m)} |a(\xi)\hat{u}(\xi)|^2 \le M_a^2 \langle \xi \rangle^{2s}|\hat{u}(\xi)|^2$. Integrating this pointwise inequality with respect to $\mathcal{L}^n$ gives $\|Au\|_{H^{s-m}(\mathbb{R}^n)}^2 \le M_a^2 \int_{\mathbb{R}^n} \langle \xi \rangle^{2s}|\hat{u}(\xi)|^2\, d\mathcal{L}^n(\xi)$. Therefore $\|Au\|_{H^{s-m}(\mathbb{R}^n)} \le M_a \|u\|_{H^s(\mathbb{R}^n)}$. [/step]

[step:Verify linearity and conclude boundedness] Let $u_1, u_2 \in H^s(\mathbb{R}^n)$ and let $\alpha, \beta \in \mathbb{C}$. Choose measurable Fourier representatives $\hat{u}_1$ and $\hat{u}_2$. The Fourier transform of $\alpha u_1 + \beta u_2$ is represented by $\alpha \hat{u}_1 + \beta \hat{u}_2$, so the definition of $A$ gives \begin{align*} \widehat{A(\alpha u_1 + \beta u_2)}(\xi) = a(\xi)(\alpha \hat{u}_1(\xi) + \beta \hat{u}_2(\xi)). \end{align*} Distributing multiplication in $\mathbb{C}$ gives \begin{align*} \widehat{A(\alpha u_1 + \beta u_2)}(\xi) = \alpha \widehat{Au_1}(\xi) + \beta \widehat{Au_2}(\xi) \end{align*} for $\mathcal{L}^n$-a.e. $\xi \in \mathbb{R}^n$. Uniqueness of Fourier representatives in $H^{s-m}(\mathbb{R}^n)$ implies \begin{align*} A(\alpha u_1 + \beta u_2) = \alpha Au_1 + \beta Au_2. \end{align*} Thus $A$ is linear. The estimate proved above holds for every $u \in H^s(\mathbb{R}^n)$, so $A$ is bounded and \begin{align*} \|A\|_{\mathcal{L}(H^s(\mathbb{R}^n), H^{s-m}(\mathbb{R}^n))} \le M_a. \end{align*} This is the desired Fourier multiplier mapping theorem. [/step]