Schwartz Space and the Plancherel Theorem

The series converges since its terms are bounded by $2^{-j-1}$. Symmetry and translation invariance follow termwise; $\min (1,a+b)\le \min (1,a)+\min (1,b)$ proves the triangle inequality. Vanishing distance forces $q_0(f-g)=0$, hence $f=g$. To make $d(f,0)<\epsilon$, choose $J$ with ${\sum }_{j>J}{2}^{-j-1}<\epsilon \mathrm{/}2$ and require $q_J(f)<\epsilon \mathrm{/}2$. Conversely, to ensure $q_J(f)<\eta$, put ${\eta }^{\mathrm{\prime }}=\min (\eta \mathrm{/}2,1\mathrm{/}2)$ and require $d(f,0)<2^{-J-1}{\eta }^{\mathrm{\prime }}$. Then $\min (1,q_J(f))<{\eta }^{\mathrm{\prime }}<1$, as required. Every finite seminorm neighbourhood contains such a $q_J$ ball and each $q_J$ ball is a finite seminorm neighbourhood, proving equality of topologies.

Let $(f_k)$ be $d$-Cauchy. The converse estimate in step 1.1 makes it Cauchy in every $q_J$. Each ${\mathrm{\partial }}^{\beta }{f}_k$ therefore has a unique uniform complex limit $g_{\beta }$, by applying [F1] to its real and imaginary parts, and this limit is continuous by [F2]. On any fixed coordinate segment $x+t{e}_j$, $a\le t\le b$ with $a<b$, [F4] gives derivative ${\mathrm{\partial }}^{\beta +e_j}{f}_k$. Apply [F3] componentwise to these restricted functions: their values converge at $a$ and their derivatives converge uniformly. Consequently ${\mathrm{\partial }}_j{g}_{\beta }=g_{\beta +e_j}$ along every segment. Induction on the length of an ordered derivative now gives $f=g_0\in C^{\mathrm{\infty }}$ and ${\mathrm{\partial }}^{\beta }f=g_{\beta }$. Limits are unique specified values, so forming this family requires no choice.

Fix $\alpha ,\beta$ and $\epsilon >0$. The seminorm Cauchy property gives $N$ with $\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}^{\beta }(f_k-f_l)(x)\mathrm{\mid }<\epsilon \mathrm{/}2$ for every $x$ and $k,l\ge N$. For fixed $x,k$, let $l\to \mathrm{\infty }$ using step 2.1. Then $p_{\alpha \beta }(f_k-f)\le \epsilon \mathrm{/}2<\epsilon$. Taking $k=N$ also shows $p_{\alpha \beta }(f)\le p_{\alpha \beta }(f_N)+\epsilon \mathrm{/}2<\mathrm{\infty }$. Thus $f\in \mathcal{S}$ and $f_k\to f$ in every seminorm and hence in $d$. Combined with local convexity, this proves the claimed Fréchet property.

Put $W(x)={\prod }_{j<n}(1+x_j^2)$ and $A={\sum }_{ϵ\in \{0,1{\}}^n}{p}_{\alpha +2ϵ,\beta }(f)$. Expanding the product and taking absolute values gives $W(x)\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}^{\beta }f(x)\mathrm{\mid }\le A$. Also $\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}^{\beta }f(x)\mathrm{\mid }\le B:=p_{\alpha \beta }(f)\le A$. Each factor $(1+t^2{)}^{-1}$ has integral at most $6$: its integral on $[-1,1]$ is at most $2$, while on $2^k\le \mathrm{\mid }t\mathrm{\mid }<2^{k+1}$ it contributes at most $2\cdot 2^k{2}^{-2k}$; the sum for $k\ge 0$ is $4$. All functions are continuous and hence Borel measurable. For $n=1$ this proves $\int W^{-1}\le 6$. For $n>1$, [F2] identifies integration of this Borel function against ${\lambda }_n$ with integration against ${\lambda }_{n-1}\times {\lambda }_1$; induction on $n$ and [F1] therefore give $\int W^{-1}\le 6^n$.

For $g=x^{\alpha }{\mathrm{\partial }}^{\beta }f$, step 1.1 gives $\mathrm{\parallel }g{\mathrm{\parallel }}_1\le 6^{n}A$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le B$. If $1<p<\mathrm{\infty }$, integrating $\mathrm{\mid }g{\mathrm{\mid }}^p\le B^{p-1}\mathrm{\mid }g\mathrm{\mid }$ gives $\mathrm{\parallel }g{\mathrm{\parallel }}_p\le B^{1-1\mathrm{/}p}(6^{n}A{)}^{1\mathrm{/}p}\le 6^{n}A$. When $B=0$, $g=0$ and this conclusion holds directly. This treats both endpoint spaces and every intermediate exponent.

Define $a(t)=e^{-1\mathrm{/}t}$ for $t>0$ and $a(t)=0$ for $t\le 0$. If $s=1\mathrm{/}t$, induction gives $a^{(k)}(t)=P_k(s)e^{-s}$ on $t>0$, with $P_0=1$ and $P_{k+1}(s)=s^2(P_k(s)-P_k^{\mathrm{\prime }}(s))$. By [F1], both $P_k(s)e^{-s}$ and $s{P}_k(s)e^{-s}$ tend to zero as $s\to \mathrm{\infty }$. Extending each derivative by zero to $t\le 0$ is therefore continuous; its difference quotient at zero also tends to zero. Induction proves $a\in C^{\mathrm{\infty }}(\mathbb{R})$ with every derivative zero at zero.

Set $\sigma (t)=a(t)\mathrm{/}(a(t)+a(1-t))$. For $t\le 0$, the second summand in the denominator is positive; for $t\ge 1$, the first is positive; for $0<t<1$, both are positive. Thus the quotient is smooth, $0\le \sigma \le 1$, and $\sigma =0$ on $t\le 0$, $\sigma =1$ on $t\ge 1$. Define $\chi (x)=\sigma ((4-\mathrm{\mid }x{\mathrm{\mid }}^2)\mathrm{/}3)$. Repeated [F2] proves smoothness, and the two constant regions give the claimed unit and zero regions, including their boundaries. Its closed support lies in the closed radius-two ball and is compact by [F3].

Differentiating $\chi (x\mathrm{/}R)$ once in coordinate $j$ gives $R^{-1}({\mathrm{\partial }}_j\chi )(x\mathrm{/}R)$. Iterating this identity in the prescribed multi-index order gives the asserted factor, including $\beta =0$. No selection was made in any construction.

Apply [F2] successively in each coordinate and use [F3] to regroup derivatives; for complex functions apply the real rule to the four real products. This gives ${\mathrm{\partial }}^{\beta }(uv)={\sum }_{\gamma \le \beta }\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right)({\mathrm{\partial }}^{\gamma }u)({\mathrm{\partial }}^{\beta -\gamma }v)$, where $\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right)={\prod }_{j<n}\left(\genfrac{}{}{0px}{}{{\beta }_j}{{\gamma }_j}\right)$. Also, on $\mathrm{\mid }x\mathrm{\mid }\ge R$, $\mathrm{\mid }x\mathrm{\mid }\le {\sum }_{j<n}\mathrm{\mid }{x}_j\mathrm{\mid }$ gives $\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}^{\delta }f(x)\mathrm{\mid }\le R^{-1}{\sum }_{j<n}{p}_{\alpha +e_j,\delta }(f)\mathrm{.}$ Indeed multiply the left side by $\mathrm{\mid }x\mathrm{\mid }$ and bound each $\mathrm{\mid }{x}_j{x}^{\alpha }{\mathrm{\partial }}^{\delta }f\mathrm{\mid }$ by its seminorm.

For $u={\chi }_R-1$, [F1] makes the undifferentiated product-rule term vanish on $\mathrm{\mid }x\mathrm{\mid }\le R$ and have coefficient at most one elsewhere. For $\gamma \mathrm{\ne }0$, ${\mathrm{\partial }}^{\gamma }{\chi }_R$ is supported on $R\le \mathrm{\mid }x\mathrm{\mid }\le 2R$, with supremum $R^{-\mathrm{\mid }\gamma \mathrm{\mid }}{C}_{\gamma }$, $C_{\gamma }=\mathrm{\parallel }{\mathrm{\partial }}^{\gamma }\chi {\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$. Boundedness follows from continuity on its compact support. Hence for $R\ge 1$ the formula in step 1.1 gives $p_{\alpha \beta }(({\chi }_R-1)f)\le R^{-1}{\sum }_{j<n}{p}_{\alpha +e_j,\beta }(f)+{\sum }_{0\mathrm{\ne }\gamma \le \beta }\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right)C_{\gamma }{R}^{-1-\mathrm{\mid }\gamma \mathrm{\mid }}{\sum }_{j<n}{p}_{\alpha +e_j,\beta -\gamma }(f)\mathrm{.}$ This tends to zero. The product is smooth with compact support inside the dilated support of $\chi$; each weighted derivative is bounded on that compact set, so ${\chi }_{R}f\in C_c^{\mathrm{\infty }}\subseteq \mathcal{S}$. Taking the explicit integers $R=1,2,\dots$ proves density.

By [F1], $p_{\alpha \beta }({\mathrm{\partial }}^{\delta }f)=p_{\alpha ,\beta +\delta }(f)$. For a monomial multiplier $x^{\delta }$, its product-rule term indexed by $\gamma \le \beta ,\delta$ is $\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right)\delta !\mathrm{/}(\delta -\gamma )!$ times $x^{\delta -\gamma }{\mathrm{\partial }}^{\beta -\gamma }f$, giving the bound by the corresponding finite sum of $p_{\alpha +\delta -\gamma ,\beta -\gamma }(f)$. A polynomial is a finite sum of these monomials. For $T_{a}f(x)=f(x-a)$, put $y=x-a$ and expand $(y+a{)}^{\alpha }$; then $p_{\alpha \beta }(T_{a}f)\le {\sum }_{\gamma \le \alpha }\left(\genfrac{}{}{0px}{}{\alpha }{\gamma }\right)\mathrm{\mid }{a}^{\alpha -\gamma }\mathrm{\mid }{p}_{\gamma \beta }(f)$.

For $M_{b}f(x)=e^{2\pi ib\cdot x}f(x)$, [F1] and [F2] give $p_{\alpha \beta }(M_{b}f)\le {\sum }_{\gamma \le \beta }\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right)\mathrm{\mid }(2\pi ib{)}^{\gamma }\mathrm{\mid }{p}_{\alpha ,\beta -\gamma }(f)$. Reflection $Rf(x)=f(-x)$ has $p_{\alpha \beta }(Rf)=p_{\alpha \beta }(f)$, and conjugation has the same identity, since coordinate derivatives commute with real and imaginary parts. These formulas also prove the asserted linearity or antilinearity.

All bounds in steps 1.1 and 1.2 are finite seminorm sums; requiring their finitely many input seminorms to be sufficiently small proves continuity at zero directly from the topology, and linearity or antilinearity translates this to every point. Product Leibniz further gives $p_{\alpha \beta }(fg)\le {\sum }_{\gamma \le \beta }\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right)p_{\alpha \gamma }(f)p_{0,\beta -\gamma }(g)$, proving closure. At $(f_0,g_0)$ write $fg-f_0{g}_0=(f-f_0)g_0+f_0(g-g_0)+(f-f_0)(g-g_0)$ and apply this bound to all three terms. Each of the finitely many errors tends to zero with the relevant input seminorms, proving joint continuity and bilinearity.

The interval FTC for $e^{it}$ gives $\mathrm{\mid }{e}^{it}-1\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }$. Thus the difference quotient in frequency coordinate $j$ is dominated in absolute value by $2\pi \mathrm{\mid }{x}_{j}f(x)\mathrm{\mid }$, integrable by [F2]. By [F4] its limit is $\mathcal{F}(-2\pi i{x}_{j}f)$. The convergence for arbitrary real increments follows either directly from the dominated estimate by truncating the majorant to a finite box and using uniform convergence there, or by the sequential criterion under the stated countable choice. The derivative is continuous by [F3]. Repeating for each weighted function, which remains Schwartz by [F1], proves all ordered derivatives and the second formula.

Fix the other coordinates and integrate in coordinate $j$. For $u=f$ restricted to this line and $v=e^{-2\pi ix\cdot \xi }$, both $u^{\mathrm{\prime }}v$ and $u{v}^{\mathrm{\prime }}$ are integrable on the line: multiply $u,u^{\mathrm{\prime }}$ by $1+x_j^2$ and use their bounded Schwartz seminorms. Also $uv\to 0$ at both ends, since $x_{j}u$ is bounded. [F6] therefore gives the derivative identity in that coordinate. Integrating over the other coordinates is legitimate by [F2] and [F5], since the full integrals of $\mathrm{\mid }{\mathrm{\partial }}_{j}f\mathrm{\mid }$ and $\mathrm{\mid }f\mathrm{\mid }$ are finite. Iteration using [F1] proves the first formula, including zero components of $\xi$ without division by them.

Combine the two identities to obtain ${\xi }^{\alpha }{\mathrm{\partial }}^{\beta }\widehat{f}=(2\pi i{)}^{-\mathrm{\mid }\alpha \mathrm{\mid }}\mathcal{F}({\mathrm{\partial }}^{\alpha }((-2\pi ix{)}^{\beta }f))\mathrm{.}$ By [F3], its supremum is at most $(2\pi {)}^{-\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }((-2\pi ix{)}^{\beta }f){\mathrm{\parallel }}_1$. By [F2] and the explicit operation bounds in [F1], this is a finite linear combination of input Schwartz seminorms. Thus every output seminorm is finite, and the finite-neighbourhood definition proves continuity.

By [F1] and [F2], both $f$ and $\widehat{f}$ are integrable, and the displayed integral is absolutely convergent since the exponential has modulus one. Fix $x$. Smoothness implies continuity, so for every $\epsilon >0$ some $\delta >0$ gives $\mathrm{\mid }f(x-y)-f(x)\mathrm{\mid }<\epsilon$ for $\mathrm{\mid }y\mathrm{\mid }<\delta$. Averaging over any ball of radius $0<r<\delta$ bounds its mean oscillation by $\epsilon$. Thus $x$ is a Lebesgue point with specified value $f(x)$.

Apply [F3] at this arbitrary point. This proves the formula everywhere, inheriting exactly the countable-choice assumption of these three suppliers. The proof never exchanges an undamped double Fourier integral.

Evaluate [F1] at $-x$. Its right-hand side is $\mathcal{F}(\widehat{f})(x)$, so ${\mathcal{F}}^{2}f=Rf$. Also $R^2=I$ directly. By associativity, $\mathcal{F}R=\mathcal{F}{\mathcal{F}}^2={\mathcal{F}}^2\mathcal{F}=R\mathcal{F}$. All compositions are defined by [F2] and [F3].

Consequently $(R\mathcal{F})\mathcal{F}=R^2=I$ and $\mathcal{F}(R\mathcal{F})=R{\mathcal{F}}^2=I$. These two identities prove both injectivity and surjectivity and the asserted inverse. Both the map and its inverse are continuous by [F2], [F3] and composition, establishing the topological automorphism.

By [F1]–[F3], $h={\mathcal{F}}^{-1}(\widehat{f}\widehat{g})$ is Schwartz and integrable. Also the convolution integral exists for every $x$, bounded absolutely by $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }g{\mathrm{\parallel }}_1$. It is continuous: for any $x_k\to x$, its integrands converge pointwise by continuity of $f$ and are dominated by $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }g\mathrm{\mid }$; [F7] gives convergence of the integrals. The sequential continuity criterion is valid under countable choice. By [F4] the integrable convolution class has transform $\widehat{f}\widehat{g}$, so [F6] identifies it with $h$ almost everywhere. Two continuous functions equal almost everywhere are equal everywhere, since a nonzero difference persists on a ball containing a box of positive measure. Hence the actual convolution is $h\in \mathcal{S}$.

The product $fg$ is Schwartz by [F2]. By [F1] and [F3], $f,g,\widehat{f}$ are integrable, so [F5] applies. Its continuous inverse representative of $f$ is $f$ itself by [F1]. Thus its identity gives the second displayed formula everywhere; step 1.1 and [F4] give the first.

By [F1], $\overline{g(x)}=\int \overline{\widehat{g}(\xi )}{e}^{-2\pi ix\cdot \xi }d\xi$. The integrand after multiplying by $f(x)$ is jointly measurable and has absolute double integral $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }\widehat{g}{\mathrm{\parallel }}_1<\mathrm{\infty }$, by [F1], [F2] and product integration. Hence [F3] gives $\int f\bar{g}=\int \overline{\widehat{g}(\xi )}[\int f(x)e^{-2\pi ix\cdot \xi }dx]d\xi =\int \widehat{f}\bar{\widehat{g}}$. This also proves absolute integrability of the final product; the original product is integrable since $g$ is bounded and $f$ integrable.

Taking $g=f$ gives equality of the nonnegative square integrals, finite by [F2] for the input and by step 1.1 for its transform. Taking nonnegative square roots proves the norm identity. The displayed pairing is linear in its first entry and conjugate-linear in its second directly from integration and conjugation.

For $u\in C_c^{\mathrm{\infty }}$, every derivative vanishes off its compact support: outside the support, $u$ is zero on a neighbourhood. Thus $x^{\alpha }{\mathrm{\partial }}^{\beta }u$ is continuous with compact support, hence bounded, for all $\alpha ,\beta$. The empty-support case is the zero function. Therefore $C_c^{\mathrm{\infty }}\subseteq \mathcal{S}$.

If $u\in \mathcal{S}$, then [F2] gives $u\in L^1$, while $\mathrm{\mid }u(x)\mathrm{\mid }\le p_{00}(u)$ by the defining seminorm. Hence ${\int }_{{\mathbb{R}}^n}\mathrm{\mid }u{\mathrm{\mid }}^2\le p_{00}(u){\int }_{{\mathbb{R}}^n}\mathrm{\mid }u\mathrm{\mid }<\mathrm{\infty }\mathrm{.}$ Thus every Schwartz function determines an $L^2$ class.

Given $f\in L^2$ and $\epsilon >0$, apply [F1] with $p=2$ to obtain $u\in C_c^{\mathrm{\infty }}$ with $\mathrm{\parallel }f-u{\mathrm{\parallel }}_2<\epsilon$. Step 1.1 puts this same $u$ in Schwartz space, and step 1.2 confirms that its class belongs to $L^2$. Every norm ball about $f$ therefore meets the Schwartz classes, proving density with precisely the countable-choice assumption of the smooth-density supplier.

View Schwartz functions as a normed subspace of $L^2$: a continuous function vanishing a.e. vanishes everywhere, since any nonzero value persists on a positive-volume box. Thus the association with its class is injective. By [F1], Fourier is a bounded linear isometry on this dense subspace, and [F5] makes the target Banach. [F4] and [F2] give a unique bounded linear extension ${\mathcal{F}}_2$. For each fixed $f$, countable choice selects a sequence $u_k\in \mathcal{S}$ with $\mathrm{\parallel }{u}_k-f{\mathrm{\parallel }}_2<1\mathrm{/}(k+1)$ for $k\in \mathbb{N}$; its Fourier images converge to ${\mathcal{F}}_{2}f$. The isometry on the subspace gives $\mathrm{\parallel }{\mathcal{F}}_{2}f{\mathrm{\parallel }}_2={\lim }_k\mathrm{\parallel }{\widehat{u}}_k{\mathrm{\parallel }}_2={\lim }_k\mathrm{\parallel }{u}_k{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$. Independence of the sequence follows also from $\mathrm{\parallel }{\widehat{u}}_k-{\widehat{v}}_k{\mathrm{\parallel }}_2=\mathrm{\parallel }{u}_k-v_k{\mathrm{\parallel }}_2\to 0$.

For any $g\in L^2$, [F2] and countable choice give $v_k\in \mathcal{S}$ tending to $g$. By [F3], define the uniquely determined $u_k={\mathcal{F}}^{-1}{v}_k\in \mathcal{S}$. By [F1], $\mathrm{\parallel }{u}_k-u_l{\mathrm{\parallel }}_2=\mathrm{\parallel }{v}_k-v_l{\mathrm{\parallel }}_2$, so [F5] gives a limit $u\in L^2$. Continuity of the extension in step 1.1 gives ${\mathcal{F}}_{2}u={\lim }_k{v}_k=g$. Thus the extension is surjective.

For approximants $u_k\to f$, $v_k\to g$, Cauchy–Schwarz in [F5] bounds $\mathrm{\mid }⟨u_k,v_k⟩-⟨f,g⟩\mathrm{\mid }$ by $\mathrm{\parallel }{u}_k-f{\mathrm{\parallel }}_2\mathrm{\parallel }{v}_k{\mathrm{\parallel }}_2+\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }{v}_k-g{\mathrm{\parallel }}_2\to 0$, since a convergent sequence is norm bounded. Apply the same estimate to their transform images and pass to the limit in [F1]. This proves pairing preservation. Steps 1.1 and 2.1 give the remaining unitary properties. Countable choice was used only in the cited interfaces and to select countable approximation sequences for each fixed input; no simultaneous arbitrary-index selection or Hilbert basis is needed.

Null-equivalent finite representatives give null-equivalent products; this applies to changes of $m$ as well as $f$, so domain and value are well-defined. The domain is a vector subspace. With $E_N=\{\mathrm{\mid }m\mathrm{\mid }\le N\}$ and $f_N={\mathbf{1}}_{E_N}f$, one has $\mathrm{\parallel }m{f}_N{\mathrm{\parallel }}_2\le N\mathrm{\parallel }f{\mathrm{\parallel }}_2$, so $f_N\in D(M)$. Finiteness of $m$ gives $E_N\uparrow {\mathbb{R}}^n$, and [F2] applied to ${\mathbf{1}}_{E_N^c}\mathrm{\mid }f{\mathrm{\mid }}^2\le \mathrm{\mid }f{\mathrm{\mid }}^2$ gives $f_N\to f$ in norm. This proves density. If $h,h^{\mathrm{\prime }}$ both satisfy the adjoint identity for $g$, then $⟨f,h-h^{\mathrm{\prime }}⟩=0$ on this dense domain; continuity in [F1] extends it to every $f\in H$, including $h-h^{\mathrm{\prime }}$, forcing $h=h^{\mathrm{\prime }}$.

By [F5], $V_t{V}_s=V_{t+s}$, $V_0=I$, $V_{-t}=V_t^{-1}$ and $⟨V_{t}f,V_{t}g⟩=⟨f,g⟩$. For fixed $f$, $\mathrm{\mid }{e}^{itm}-1{\mathrm{\mid }}^2\mathrm{\mid }f{\mathrm{\mid }}^2\to 0$ pointwise as $t\to 0$, with majorant $4\mathrm{\mid }f{\mathrm{\mid }}^2$. [F2] gives strong continuity at zero; the isometry and group law give it at every $t$. Countable choice permits the sequential criterion for these real-parameter norm limits.

Real-valuedness of $m$ and [F1] give $⟨Mf,g⟩=⟨f,Mg⟩$ for $f,g\in D(M)$, with both integrals absolutely convergent. Thus $D(M)\subseteq D(M^{\ast })$ with the same value. Conversely let $M^{\ast }g=h$. On $E_N$, $w_N={\mathbf{1}}_{E_N}(mg-h)$ is in $H$, and $m{w}_N\in H$ because $\mathrm{\mid }m\mathrm{\mid }\le N$ there, so $w_N\in D(M)$. Inserting $f=w_N$ into the adjoint identity gives $0=⟨M{w}_N,g⟩-⟨w_N,h⟩={\int }_{E_N}\mathrm{\mid }mg-h{\mathrm{\mid }}^2$. By [F4], $mg=h$ a.e. on each $E_N$. Their countable union is the whole space, so $mg=h$ a.e. globally; in particular $mg\in H$. Thus $D(M^{\ast })=D(M)$ and the operators agree.

If $mf\in H$, [F6] gives pointwise $(e^{itm}-1)f\mathrm{/}t\to imf$ and the squared error is at most $4\mathrm{\mid }mf{\mathrm{\mid }}^2$. [F2] proves norm convergence to $iMf$. Conversely, if the norm derivative exists, the quotients at $t=1\mathrm{/}(N+1)$ have bounded norms for $N\in \mathbb{N}$. Their squared moduli tend pointwise to $\mathrm{\mid }mf{\mathrm{\mid }}^2$ by [F6]. [F3] gives $\int \mathrm{\mid }mf{\mathrm{\mid }}^2\le {\mathrm{lim\hspace{0.17em}inf}}_N\mathrm{\parallel }(V_{1\mathrm{/}(N+1)}f-f)\mathrm{/}(1\mathrm{/}(N+1)){\mathrm{\parallel }}_2^2<\mathrm{\infty }$. Thus $f\in D(M)$, and the forward part identifies the derivative. This proves both directions, including points where $m=0$.

Since $U$ and $U^{-1}$ preserve norms and pairings, $U^{-1}D(M)$ is dense. For $g,h\in H$, the assertion $⟨Pf,g⟩=⟨f,h⟩$ for every $f\in D(P)$ is equivalent, by writing $v=Uf$, to $⟨Mv,Ug⟩=⟨v,Uh⟩$ for every $v\in D(M)$. By step 2.1 this holds exactly when $Ug\in D(M)$ and $Uh=MUg$. Therefore $D(P^{\ast })=D(P)$ and $P^{\ast }=P$. Conjugating the group identities and norm limits of steps 1.2 and 2.2 by $U$ proves the asserted unitary group, continuity, and both directions of the transported derivative-domain criterion. No spectral theorem or choice of a basis is used.

Fix a finite measurable representative of $f$ and set $g_j=f{\mathbf{1}}_{\{\mathrm{\mid }x\mathrm{\mid }\le j,\mathrm{\mid }f(x)\mathrm{\mid }\le j\}}$, $j\ge 1$. Then $g_j$ is bounded and compactly supported, and $\mathrm{\mid }{g}_j-f{\mathrm{\mid }}^p\le \mathrm{\mid }f{\mathrm{\mid }}^p$ tends pointwise to zero for $p=1,2$. [F1] therefore gives $\mathrm{\parallel }{g}_j-f{\mathrm{\parallel }}_p\to 0$ in both norms. Let $\chi$ be [F3]'s cutoff and put $\rho =\chi \mathrm{/}\int \chi$. Its integral is finite since it is bounded and supported in a finite-volume ball, and positive since it equals one on a ball containing a positive-volume box. Thus $\rho$ is a specified real smooth compactly supported kernel of mass one.

For fixed $j$, [F2] gives ${\rho }_{1\mathrm{/}k}\ast g_j\to g_j$ as $k\to \mathrm{\infty }$ in both norms. Let $k_j$ be the least positive integer for which both errors are below $1\mathrm{/}j$, and define $f_j={\rho }_{1\mathrm{/}{k}_j}\ast g_j$. The qualifying set is nonempty since both convergences hold, and the least-integer rule needs no further choice. [F2] makes $f_j$ smooth with compact support (contained in the radius $j+2\mathrm{/}{k}_j$ ball). Finally $\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_p\le 1\mathrm{/}j+\mathrm{\parallel }{g}_j-f{\mathrm{\parallel }}_p\to 0$ for both $p=1,2$. The same sequence works, with countable choice inherited only from the Euclidean mollification interface.