Schwartz Space and the Plancherel Theorem — Examples

Differentiation in coordinate $j$ sends $Q(x)e^{-\pi t\mathrm{\mid }x{\mathrm{\mid }}^2}$ to $({\mathrm{\partial }}_{j}Q(x)-2\pi t{x}_{j}Q(x))e^{-\pi t\mathrm{\mid }x{\mathrm{\mid }}^2}$ by [F1]. Starting at $Q=P$, this recurrence proves that every ordered derivative is a polynomial times the same Gaussian and is continuous. Multiplication by any $x^{\alpha }$ leaves this form unchanged.

For any polynomial $Q$ of degree at most $d$, the sum of the absolute coefficients gives a constant $C$ with $\mathrm{\mid }Q(x)\mathrm{\mid }\le C(1+\mathrm{\mid }x\mathrm{\mid }{)}^d$. For $\mathrm{\mid }x\mathrm{\mid }\ge 1$, choose an integer $N$ with $2N\ge d$; then $(1+\mathrm{\mid }x\mathrm{\mid }{)}^d{e}^{-\pi t\mathrm{\mid }x{\mathrm{\mid }}^2}\le 2^d(\mathrm{\mid }x{\mathrm{\mid }}^2{)}^N{e}^{-\pi t\mathrm{\mid }x{\mathrm{\mid }}^2}\to 0$ by [F2]. On $\mathrm{\mid }x\mathrm{\mid }\le 1$ the bound is at most $C{2}^d$. Thus every weighted derivative in step 1.1 has finite supremum, which is precisely the Schwartz condition. If $P=0$, all derivatives and bounds are zero directly.

The denominator $1+x^2$ is strictly positive. Inductively $f^{(k)}(x)=Q_k(x)\mathrm{/}(1+x^2{)}^{k+1}$, where $Q_0=1$ and $Q_{k+1}=(1+x^2)Q_k^{\mathrm{\prime }}-2(k+1)x{Q}_k$; these derivatives are continuous, proving smoothness. On $[0,1]$, $f\le 1$, and for $x\ge 1$, $f(x)\le x^{-2}$. Thus for $R\ge 1$, ${\int }_0^{R}f\le 1+{\int }_1^R{x}^{-2}dx=2-1\mathrm{/}R\le 2$. The nonnegative improper integral exists and is finite; evenness gives the identical bound on the negative half-line. [F1] identifies the two improper integrals with the Lebesgue integrals, so $\mathrm{\parallel }f{\mathrm{\parallel }}_1\le 4$.

For $\mathrm{\mid }x\mathrm{\mid }\ge 1$, $x^{4}f(x)=x^4\mathrm{/}(1+x^2)\ge x^2\mathrm{/}2\to \mathrm{\infty }$ as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$. Therefore $p_{4,0}(f)=\mathrm{\infty }$, violating the defining Schwartz condition despite step 1.1. Countable choice is used only through the improper-to-Lebesgue interface, not for the explicit smoothness or failed seminorm.

Taking $P=1,t=1$ in [F1] gives $g\in \mathcal{S}$. Positivity of the real exponential gives $g(x)>0$ for every $x$, so [F2] gives $\mathrm{supp}g=\overline{{\mathbb{R}}^n}={\mathbb{R}}^n$.

The increasing open balls $B(0,k)$, $k=1,2,\dots$, cover this support and have no finite subcover: a finite union lies in the largest of those balls and misses a point further along the first coordinate axis. Thus the support is not compact, while step 1.1 supplies the Schwartz hypothesis.

Put $a=\sqrt{\pi }x+(2\sqrt{\pi }{)}^{-1}d\mathrm{/}dx$ and $u_m=(a^{†}{)}^m{h}_0$. Differentiation gives $a{h}_0=0$ and $a{a}^{†}-a^{†}a=I$, since $(d\mathrm{/}dx)(xf)-x{f}^{\mathrm{\prime }}=f$. Induction gives $a{u}_m=m{u}_{m-1}$ for $m\ge 1$, and $a^{†}a{u}_m=m{u}_m$ for $m\ge 0$. If $u_m=Q_m{h}_0$, then $Q_{m+1}=2\sqrt{\pi }x{Q}_m-(2\sqrt{\pi }{)}^{-1}{Q}_m^{\mathrm{\prime }}$. Thus $Q_m$ is real of degree $m$ with leading coefficient $(2\sqrt{\pi }{)}^m$, and [F1] makes every $u_m$ Schwartz.

For polynomial Gaussians $v,w$, [F4] gives $⟨a^{†}v,w⟩=⟨v,aw⟩$: derivative products are polynomial Gaussians and integrable, and their endpoint products vanish. Consequently $N=a^{†}a$ is symmetric on these functions, since $⟨Nv,w⟩=⟨av,aw⟩=⟨v,Nw⟩$. Step 1.1 implies $(m-l)⟨u_m,u_l⟩=0$. Also $\mathrm{\parallel }{u}_m{\mathrm{\parallel }}_2^2=⟨a^{†}{u}_{m-1},u_m⟩=m\mathrm{\parallel }{u}_{m-1}{\mathrm{\parallel }}_2^2$. The base norm is $\mathrm{\parallel }{h}_0{\mathrm{\parallel }}_2^2=\sqrt{2}\int e^{-2\pi x^2}dx=1$ by [F2]. Hence $\mathrm{\parallel }{u}_m{\mathrm{\parallel }}_2^2=m!$ and the normalized functions are orthonormal, including $m=0$.

The derivative identities [F3] give $\mathcal{F}(a^{†}v)=(i\mathrm{/}(2\sqrt{\pi }))(\widehat{v}{)}^{\mathrm{\prime }}-i\sqrt{\pi }\xi \widehat{v}=-i{a}^{†}\widehat{v}$. Since [F2] gives ${\widehat{h}}_0=h_0$, induction yields ${\widehat{u}}_m=(-i{)}^m{u}_m$ and the asserted normalized identity. All operations are on Schwartz functions, so this also holds for their Plancherel classes.

Suppose $f\in L^2$ is orthogonal to all $h_m$. By the nonzero real leading coefficients in step 1.1, triangular induction expresses each monomial $x^r$ as a real linear combination of $Q_0,\dots ,Q_r$. Therefore $\int f(x)h_0(x)x^{r}dx=0$ for every $r$; these integrals exist by [F5], since $x^r{h}_0\in L^2$. Put $v=f{h}_0\in L^1$ by [F5]. For fixed real $\xi$, the exponential Taylor partial sums are bounded by $e^{2\pi \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}$. The majorant $\mathrm{\mid }f\mathrm{\mid }{h}_0{e}^{2\pi \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}$ is integrable by [F5]: its second factor has finite square integral, because $-2\pi x^2+4\pi \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }\le -\pi x^2+4\pi {\xi }^2$. Thus [F7] integrates the exponential series termwise, all terms being the zero moments. It gives $\widehat{v}(\xi )=0$ for every $\xi$. By [F6], $v=0$ a.e.; positivity of $h_0$ gives $f=0$ a.e.

For arbitrary $f\in L^2$, put $s_N={\sum }_{m=0}^N⟨f,h_m⟩h_m$. Finite orthogonality in step 2.1 gives $\mathrm{\parallel }f-s_N{\mathrm{\parallel }}_2^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2-{\sum }_{m=0}^N\mathrm{\mid }⟨f,h_m⟩{\mathrm{\mid }}^2\ge 0$. Hence the coefficient-square partial sums are bounded increasing and converge; their tails give $\mathrm{\parallel }{s}_M-s_N{\mathrm{\parallel }}_2^2\to 0$. Completeness in [F5] supplies $s\in L^2$ with $s_N\to s$. Pairing continuity shows $⟨f-s,h_m⟩=0$ for every $m$, so step 2.3 gives $f=s$. This proves the promised expansion, not merely orthogonality. All sequences are specified; countable choice is inherited from the complex integral and completeness interfaces.

Set $f={\mathbf{1}}_{[-1\mathrm{/}2,1\mathrm{/}2]}$. Then $\mathrm{\parallel }f{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2=1$, so $f\in L^1\cap L^2$. For $\xi \mathrm{\ne }0$, [F1] gives $\widehat{f}(\xi )=[e^{-2\pi ix\xi }\mathrm{/}(-2\pi i\xi ){]}_{-1\mathrm{/}2}^{1\mathrm{/}2}=(e^{-\pi i\xi }-e^{\pi i\xi })\mathrm{/}(-2\pi i\xi )=\sin (\pi \xi )\mathrm{/}(\pi \xi )$. For $\xi =0$ the defining integral is the interval length, one.

By [F2] and [F3], the square modulus of this explicitly computed transform has integral $\mathrm{\parallel }{\mathcal{F}}_{2}f{\mathrm{\parallel }}_2^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2=1$. The sinc quotient is real, so its squared modulus is its square, yielding the statement. This supplies integrability of the square as well as its value.

The series defining $\theta (t)$ converges: for $k\ge 1$, $e^{-\pi t{k}^2}\le e^{-\pi tk}$ and $0<e^{-\pi t}<1$, so its positive and negative tails are bounded by geometric series; the zero term is one. The same proof applies to $1\mathrm{/}t>0$. By [F1], $g_t(x)=e^{-\pi t{x}^2}$ is Schwartz, and [F2] gives its transform with factor $t^{-1\mathrm{/}2}$.

Apply [F3] at $x=0$ to $g_t$: ${\sum }_k{g}_t(k)={\sum }_k{\widehat{g}}_t(k)=t^{-1\mathrm{/}2}{\sum }_k{e}^{-\pi k^2\mathrm{/}t}$. By step 1.1 these are the two absolutely convergent theta series, proving the identity. At $t=1$ both sides agree termwise. The parameter is only real and positive; this example makes no complex modular-form assertion.

Put $g(y)=e^{-2\pi ib\cdot (y+a)}f(y+a)$. By [F6] it is Schwartz, and [F3] gives $\widehat{g}(\eta )=e^{2\pi ia\cdot \eta }\widehat{f}(\eta +b)$. Translation substitution and unit modulus therefore identify $\mathrm{\parallel }g{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$, $\mathrm{\parallel }\mathrm{\mid }y\mathrm{\mid }g{\mathrm{\parallel }}_2=\mathrm{\parallel }\mathrm{\mid }x-a\mathrm{\mid }f{\mathrm{\parallel }}_2$, and $\mathrm{\parallel }\mathrm{\mid }\eta \mathrm{\mid }\widehat{g}{\mathrm{\parallel }}_2=\mathrm{\parallel }\mathrm{\mid }\xi -b\mathrm{\mid }\widehat{f}{\mathrm{\parallel }}_2$. It suffices to prove the zero-centre assertion for $g$.

For each coordinate $j$, apply [F5] along that line to $u=x_{j}g$ and $v=\bar{g}$. The endpoint product $x_j\mathrm{\mid }g{\mathrm{\mid }}^2$ tends to zero at both ends by rapid decay, and both differentiated products are line-integrable. Their full-space integrability follows from [F6] and [F4], so [F7] permits integrating the identity in the other coordinates. It yields $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=-2\mathrm{Re}\int x_{j}g\overline{{\mathrm{\partial }}_{j}g}$. Sum over $j$ and define tuples $A=(x_{j}g{)}_j$, $B=({\mathrm{\partial }}_{j}g{)}_j$. Then $n\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=-2\mathrm{Re}⟨A,B⟩\le 2\mathrm{\mid }⟨A,B⟩\mathrm{\mid }\le 2\mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }\mathrm{.}$ By [F1] and [F2], $\mathrm{\parallel }B{\mathrm{\parallel }}^2=4{\pi }^2\int \mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{\mid }\widehat{g}(\xi ){\mathrm{\mid }}^{2}d\xi$, while $\mathrm{\parallel }A\mathrm{\parallel }=\mathrm{\parallel }\mathrm{\mid }x\mathrm{\mid }g{\mathrm{\parallel }}_2$. This proves the inequality with the claimed constant.

Suppose $g\mathrm{\ne }0$ and equality holds. Step 2.1 gives $n\mathrm{\parallel }g{\mathrm{\parallel }}_2^2>0$, so both tuple norms are nonzero. Equality in [F4] makes $B=dA$ for one complex scalar $d$. Equality in the real-part bound, together with $⟨A,B⟩=\bar{d}\mathrm{\parallel }A{\mathrm{\parallel }}^2$, forces $d$ to be a negative real number. Write $d=-\lambda$, $\lambda >0$. The identities ${\mathrm{\partial }}_{j}g=-\lambda x_{j}g$ hold a.e., hence everywhere by continuity and the positive measure of nondegenerate boxes. Thus every partial derivative of $G(x)=e^{\lambda \mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}2}g(x)$ is zero. Applying the interval FTC [F5] along successive coordinate segments shows $G(x)=G(0)=c$, so $g(x)=c{e}^{-\lambda \mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}2}$. Nonzero $g$ forces $c\mathrm{\ne }0$. Undoing step 1.1 gives exactly the displayed form for $f$.

Conversely let $g(x)=c{e}^{-\lambda \mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}2}$ with $\lambda >0$, $c\mathrm{\ne }0$. By [F8] it is Schwartz, and direct differentiation gives $B=-\lambda A$. Thus both inequalities in step 2.1 are equalities, giving equality in the uncertainty inequality. By step 1.1 the translated and modulated functions have the same equality property. If $f=0$, both sides are zero directly. The common scalar across all coordinates is essential to the radial equality assertion proved here.

The multiplier $m(\xi )=2\pi \xi$ is real, finite and measurable. Apply [F1] with the specified unitary $U={\mathcal{F}}_2$ from [F2]. The domain condition $mUf\in L^2$ is equivalent to $\xi Uf\in L^2$ since $2\pi \mathrm{\ne }0$. Hence [F1] gives exactly the stated self-adjoint operator and the strongly continuous group $e^{itP}={\mathcal{F}}_2^{-1}{M}_{e^{2\pi it\xi }}{\mathcal{F}}_2$, including both directions of its derivative-domain criterion.

If $f\in \mathcal{S}$, [F3] gives $\mathcal{F}(-i{f}^{\mathrm{\prime }})=2\pi \xi \widehat{f}$. The left side is a Schwartz function and hence is in $L^2$; by [F6] and [F2], this proves $f\in D(P)$ and $Pf=-i{f}^{\mathrm{\prime }}$. Also [F4] with $a=-t$ gives $\mathcal{F}(f(\cdot +t))=e^{2\pi it\xi }\widehat{f}$. By [F6] and step 1.1 it follows that $e^{itP}f=f(\cdot +t)$ for Schwartz $f$, with the plus sign appropriate to $e^{itP}$.

For any $f\in L^2$, use [F5] and countable choice to take $f_j\in \mathcal{S}$ tending to $f$. Both $e^{itP}$ and $f\mapsto f(\cdot +t)$ are isometries, respectively by step 1.1 and [F7]. Therefore the norm of their difference on $f$ is at most $2\mathrm{\parallel }f-f_j{\mathrm{\parallel }}_2$, since it is zero on $f_j$ by step 2.1. Let $j\to \mathrm{\infty }$. This proves the group identity on every class. The exact domain and derivative assertion remain those proved in step 1.1; no unspecified self-adjoint extension or general functional calculus is used.