Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 7 / 4

Interpret $H_{-\infty }({\mathbb{T}}^n)={\bigcup }_{s\in \mathbb{R}}{H}^s({\mathbb{T}}^n)$, the periodic distributions of finite negative Sobolev space order. Smooth multiplication is bounded on every periodic Sobolev space. One direct proof uses the rapid decay of the Fourier coefficients of $V$: the inequalityreduces the weighted convolution formula for $\widehat{Vf}$ to an ${\ell }^1\ast {\ell }^2$ estimate. Thus $\Vert Vf{\Vert }_{H^s}\le C_{V,s}\Vert f{\Vert }_{H^s}$ for every real $s$.

Suppose initially $f\in H^s$. The equation gives $\Delta f=u-Vf\in H^s$, since a smooth function belongs to every periodic Sobolev space. For $k\ne 0$, the Fourier multiplier of $\Delta$ is $|k{|}^2$, andThe zero coefficient is already finite. Hence $f\in H^{s+2}$. Iterating the argument gives $f\in H^{s+2j}$ for every $j\ge 0$. Applying the Sobolev embedding theorem also to each distributional partial derivative produces a smooth representative. Therefore $\boxed{f\in C^{\infty }({\mathbb{T}}^n)}$. This is an explicit elliptic regularity bootstrap, valid for complex or real $V$ and without a sign assumption on it.

Every entire holomorphic function has a Taylor series $f(z)={\sum }_{n\ge 0}{a}_n{z}^n$, uniformly convergent on each circle. Angular orthogonality givesIntegrate in polar coordinates and use Tonelli theorem for the nonnegative sum:The last integral equals $n!/2$ by substituting $u=r^2$ and repeated integration by parts.

Conversely, for any $(b_n)\in {\ell }^2$, set $a_n=b_n/\sqrt{n!}$. On $|z|\le R$, the Cauchy-Schwarz inequality givesApplied to the tails, the same bound proves uniform convergence on every disk. The series therefore defines an entire holomorphic function, and the norm identity puts it in the Bargmann-Fock space $\mathcal{F}$. The coefficient map $f\mapsto (\sqrt{n!}{a}_n)$ is consequently an onto linear isometry $\mathcal{F}\to {\ell }^2$; polarization also identifies the inner products. Completeness of ${\ell }^2$ proves that $\mathcal{F}$ is a Hilbert space. Its standard coordinate orthonormal basis corresponds toThe norm identity shows both their unit norms and completeness; it also gives $|f(z)|\le e^{|z{|}^2/2}\Vert f{\Vert }_{\mathcal{F}}$.

To distinguish the two normalizations, denote the physicists' Hermite polynomial by $h_n(x)=(-1{)}^n{e}^{x^2}(d/dx{)}^n{e}^{-x^2}$. Define the normalized Hermite functions byIntroduce the differential operators $a=(x+d/dx)/\sqrt{2}$, $a^{\ast }=(x-d/dx)/\sqrt{2}$ on the Schwartz space. Direct multiplication gives $[a,a^{\ast }]=1$ andThe Gaussian $H_0={\pi }^{-1/4}{e}^{-x^2/2}$ satisfies $a{H}_0=0$, and repeated differentiation gives $H_n=(a^{\ast }{)}^n{H}_0/\sqrt{n!}$. The commutator identity implies $a{H}_n=\sqrt{n}{H}_{n-1}$ and $a^{\ast }{H}_n=\sqrt{n+1}{H}_{n+1}$. ThereforeFor normalization, integration by parts makes $a^{\ast }$ the adjoint operator of $a$ on these rapidly decreasing functions, so $\Vert (a^{\ast }{)}^n{H}_0{\Vert }^2=n!\Vert H_0{\Vert }^2=n!$. Distinct eigenvalues of the symmetric quantum harmonic oscillator give orthogonality.

The generating identity follows directly from the Rodrigues definition of the Hermite polynomials:The Taylor expansion of the entire Gaussian justifies the middle equality for every complex $w$. Substitute $w=z/\sqrt{2}$ and the Bargmann-Fock space basis $e_n(z)=z^n/\sqrt{n!}$ to obtainThe convergence is locally uniform in $z$ by the same entire generating series.

For $0<t<1$, $A_x(\sqrt{t}z)$ belongs to the Bargmann-Fock space: after writing $z=u+iv$, its squared modulus times $e^{-|z{|}^2}$ has strictly negative quadratic coefficients $-(1+t)u^2-(1-t)v^2$. Taking the Bargmann-Fock space inner product of the two generating functions and using the orthonormal basis proved in part (b) givesThe coefficient series converges absolutely by the Cauchy-Schwarz inequality. Substitution factors the right side intoUse the Gaussian integral $\int e^{-a{u}^2+bu}du=\sqrt{\pi /a}{e}^{b^2/(4a)}$ for $a>0$ and complex $b$; the complex case follows from the real case by analyticity. The resulting exponent isThus the Mehler formula for Hermite functions isAs a normalization check, its $t\downarrow 0$ limit is $H_0(x)H_0(y)$. Setting $t=e^{-2s}$ and multiplying by $e^{-s}$ yields the harmonic oscillator transition kernel with eigenvalues $2n+1$.

For a closed manifold $M$ with an oriented Riemannian metric, define $\delta$ to be the formal $L^2$ adjoint operator of the exterior derivative, equivalently $\delta =(-1{)}^{n(q+1)+1}\ast d\ast$ on $q$-forms. The Hodge Laplacian is ${\Delta }_q=d\delta +\delta d$. The Hodge decomposition theorem states that its space ${\mathcal{H}}^q=\ker {\Delta }_q$ of harmonic differential forms is finite-dimensional and smooth, and thatThe last isomorphism sends a harmonic differential form to its de Rham cohomology class: every class has exactly one harmonic representative. Out-of-range degree spaces are zero.

We first justify the analytic facts needed in the proof. In local coordinates and a smooth frame for the form bundle, the principal symbol of ${\Delta }_q$ is $g^{ij}{\xi }_i{\xi }_{j}I$. This follows from the exterior/contraction identities for $d$ and $\delta$. Freezing the leading coefficients, the Fourier transform estimate for the constant positive quadratic form gives, for each nonnegative integer $s$,Here is how to pass to variable coefficients. On a sufficiently small coordinate ball the difference of leading coefficients has arbitrarily small supremum norm. After differentiating $s$ times, its top-order contribution is bounded by $\epsilon \Vert u{\Vert }_{H^{s+2}}$; derivatives of the coefficients and first-order terms contribute $C_s\Vert u{\Vert }_{H^{s+1}}$. A partition of unity gives the same types of terms from commutators with the cutoffs. Absorb the epsilon term and use Sobolev interpolation to absorb $\Vert u{\Vert }_{H^{s+1}}$ into a small multiple of $\Vert u{\Vert }_{H^{s+2}}$ plus $C\Vert u{\Vert }_2$. This proves the estimate.

The first-order energy form $Q(u,u)=\Vert du{\Vert }_2^2+\Vert \delta u{\Vert }_2^2$ satisfies $Q(u,u)+c\Vert u{\Vert }_2^2\ge c^{\prime }\Vert u{\Vert }_{H^1}^2$ for a sufficiently large $c>0$. This follows from the same frozen-symbol argument at first order, or by integrating the principal part and absorbing the lower-order terms. The Lax-Milgram theorem thus constructs a bounded inverse $R=({\Delta }_q+c{)}^{-1}:L^2\to H^1$ in the weak sense. To obtain $H^2$ regularity of this weak solution, use a local difference quotient ${\delta }_{h}u$ and test the weak equation with $-{\delta }_{-h}({\chi }^2{\delta }_{h}u)$. Positivity of $g^{ij}$ bounds $\Vert \chi {\delta }_h\nabla u{\Vert }_2^2$; derivatives of $\chi$ and difference quotients of the smooth coefficients give terms bounded by $C\Vert u{\Vert }_{H^1}\Vert \chi {\delta }_h\nabla u{\Vert }_2$ and $C\Vert u{\Vert }_{H^1}^2$. The forcing contributes at most $C\Vert f{\Vert }_2(\Vert \chi {\delta }_h\nabla u{\Vert }_2+\Vert u{\Vert }_{H^1})$. Absorption gives a bound independent of $h$, so the difference-quotient criterion yields $u\in H^2$. Differentiating the equation and repeating gives the higher estimates and smoothness for smooth forcing. Thus the use of elliptic regularity here has been justified rather than assumed.

The inverse $R$ is compact as an $L^2$ operator by the Rellich-Kondrachov compactness theorem, and self-adjoint by symmetry of the weak energy form. It is positive and injective with dense range. The spectral theorem for compact self-adjoint operators now gives an orthonormal eigenbasis of ${\Delta }_q$, with nonnegative eigenvalues tending to infinity. Each eigenfunction is smooth by the estimates above. In particular ${\mathcal{H}}^q$ is finite-dimensional and there is a positive gap above its zero eigenvalue.

Let $H_q$ be the orthogonal projection onto ${\mathcal{H}}^q$. Define the Green operator of the Hodge Laplacian by multiplying a positive-eigenvalue coefficient by $1/\lambda$ and sending the zero coefficients to zero. The gap makes this bounded on $L^2$, and the elliptic estimate shows it maps $H^s$ to $H^{s+2}$; hence it sends smooth forms to smooth forms. It obeysFor a smooth $q$-form $\alpha$, expand the left side to obtainThese three summands are mutually orthogonal. The energy identityshows that a harmonic differential form has both $dh=0$ and $\delta h=0$, making it orthogonal to the exact and coexact summands by integration by parts. Also $⟨d\beta ,\delta \gamma ⟩=⟨d^2\beta ,\gamma ⟩=0$. This proves the claimed direct sum and its uniqueness.

Finally let $d\alpha =0$ and write $u=G_q\alpha$. We have $d{\Delta }_{q}u={\Delta }_{q+1}du$ because $d^2=0$. Since $d(\alpha -H_q\alpha )=0$, $du$ is harmonic. It is also exact, so its orthogonality to harmonic forms gives $du=0$. The decomposition therefore reduces to $\alpha =H_q\alpha +d\delta u$: every closed differential form is cohomologous to a harmonic one. If a harmonic $h$ is exact, $h=d\beta$, then $\Vert h{\Vert }_2^2=⟨\delta h,\beta ⟩=0$. Hence its class is zero only when $h=0$, establishing the isomorphism with de Rham cohomology.