Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 327 (Distribution Theory and Applications)

The Schwartz space iswhereA sequence ${\phi }_j$ converges to $\phi$ in $\mathcal{S}$ exactly when every one of these seminorms of ${\phi }_j-\phi$ tends to zero.

The space of tempered distributions ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is the continuous dual of $\mathcal{S}({\mathbb{R}}^n)$. Its standard weak convergence is

Continuity immediately implies sequential continuity. Conversely, suppose a linear form $u$ is sequentially continuous but not continuous at zero. Enumerate an increasing family of seminorms that generates the Schwartz space topology, and letSince $u$ is unbounded on every neighborhood of zero, choose ${\phi }_k\in U_k$ with $|u({\phi }_k)|\ge 1$. For each fixed $j$, $p_j({\phi }_k)<1/k$ once $k\ge j$, so ${\phi }_k\to 0$ in $\mathcal{S}$. Sequential continuity would imply $u({\phi }_k)\to 0$, a contradiction. Hence $u$ is continuous and belongs to ${\mathcal{S}}^{\prime }$.

Each continuous $f_{\alpha }$ of polynomial growth defines a regular tempered distribution byChoosing an integer $L>M_{\alpha }+n$ giveswhich is bounded by finitely many Schwartz space seminorms. Its distributional derivative satisfiesand is therefore tempered. A finite sum of continuous linear forms is continuous, so

Let $u$ be a compactly supported distribution. It has some finite order $m$. Choose $k$ so large that the Bessel potential kernel $G_k={\mathcal{F}}^{-1}(1+|\xi {|}^2{)}^{-k}$ has enough continuous derivatives forto be bounded and continuous. Compact support of $u$ makes boundedness uniform under translation. Since $(1-\Delta {)}^k{G}_k={\delta }_0$ distributionally,Expanding each power of $\Delta$ expresses $u$ as a finite sum of derivatives of the bounded continuous function $f$. This proves the structure theorem for compactly supported distributions.

Although $|u(x)|=e^{x^2}$ has superpolynomial growth, the rapidly varying phase makes $u$ an oscillatory tempered distribution. Split the integral against $\phi \in \mathcal{S}(\mathbb{R})$ into $|x|\le 1$ and the two tails. On a tail, with $\Phi (x)=e^{x^4}$,Integration by parts transfers the derivative toThis function and its derivative are integrable because $e^{x^2-x^4}$ dominates every polynomial, and the boundary term at infinity vanishes. The result is bounded by finitely many Schwartz space seminorms. The compact part has the same property. Thus the cutoff integrals converge and define a continuous linear functional:

Use $D=-i\partial$, so $P(D)e^{i\lambda \cdot x}=P(\lambda )e^{i\lambda \cdot x}$. A concrete distributional division construction isThe locally integrable family, initially defined for sufficiently large $\mathrm{Re}s$, has a meromorphic continuation; $\mathrm{FP}$ denotes its finite part at zero. Multiplication before continuation givesThe right side is holomorphic at $s=0$ with value $1$, so comparison of constant Laurent coefficients yields $PT=1$. Thereforesatisfies $P(D)E={\delta }_0$. This finite-part formula explicitly realizes the Malgrange–Ehrenpreis theorem.

For each one-dimensional factor $P_j(D_j)$, choose its retarded fundamental solution $E_j=H(x_j)g_j(x_j)$. Partial-fraction decomposition of the reciprocal polynomial giveswhere each ${\alpha }_{rj}$ is a polynomial whose degree is one less than the multiplicity of the associated root. Constants, including powers of $i$ from $D=-i\partial$, can be absorbed into the polynomials and exponents.

Take the tensor productIt vanishes unless every $x_j>0$, has the required polynomial-exponential form there, and satisfies

Factor the operator asSet $f(x)=H(x)\sin x$ and $g(x)=H(x)\sinh x$. Since both vanish at zero and have right derivative one, their distributional second derivatives areBecause $D^2=-{\partial }^2$,Consequently $E(x_1,x_2)=f(x_1)g(x_2)$ obeysIt equals $\sin x_1\sinh x_2$ in the positive quadrant and zero otherwise.

Write $P=P_N+P_{N-1}+\cdots +P_0$ as a sum of homogeneous parts. The operator is elliptic whenContinuity on the unit sphere gives $c={\min }_{|\omega |=1}|P_N(\omega )|>0$. Uniformly in $\omega$,so for sufficiently large $t$, $|P(t\omega )|\ge (c/2)t^N$. Since $t^N\asymp ⟨t\omega {⟩}^N$ at large $t$,for sufficiently large $|\lambda |$.

A parametrix for $P(D)$ is a distribution $E$ for whichwith $r\in C^{\infty }$. Choose a smooth cutoff $\chi$ that vanishes on a large ball containing every real zero of $P$ and equals one outside a slightly larger ball. Ellipticity makesa symbol of order $-N$. For $E={\mathcal{F}}^{-1}q$,Since $1-\chi$ is smooth and compactly supported, its inverse Fourier transform is smooth. Thus $E$ is a parametrix.

The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^n;N)=S_{1,0}^N$ consists of $c\in C^{\infty }(X\times {\mathbb{R}}^n)$ such that, for every compact $K\subset X$ and all multi-indices $\alpha ,\beta$,The defining estimate givesThe Leibniz rule givesand the triangle inequality givesThese are the basic rules of symbol calculus.

The assumed lower bound gives $|a|=|Q{|}^{-1}{\lesssim }_K⟨\lambda {⟩}^{-M}$ at large frequency. Differentiating $1/Q$ repeatedly expresses every derivative as a finite sum of products of derivatives of $Q$ divided by powers of $Q$. Since $D_{\lambda }^{\beta }Q$ has polynomial order at most $M-|\beta |$, induction and symbol calculus giveThe interpolation region $R\le |\lambda |\le R+1$ is compact in frequency and causes no problem. Hence

For any smooth amplitude $v(x,\lambda )$,because $D_x=-i{\partial }_x$. Applying each term of the differential operator under the oscillatory integral givesThe identity is justified distributionally by regularizing the frequency integral and integrating by parts.

Since $Q(x,\lambda )$ is a polynomial of degree at most $M$ in $\lambda$, its Taylor formula is exact:The $\alpha =0$ term is $Q{a}_1=1$ outside the cutoff region; its difference from one is a symbol of order $-\infty$. For $|\alpha |\ge 1$,so their product lies in $S^{-|\alpha |}\subset S^{-1}$. Defining $b_1$ as the negative of the cutoff remainder and these lower-order terms yields

TakeBy symbol calculus, $a_2\in S^{-M-1}=S^{N_Q-1}$. Its zeroth-order contribution is $Q{a}_2=b_1$ outside the compact transition region and therefore cancels the order $-1$ remainder. Every term incontains at least one $\lambda$ derivative of $Q$ and has order at most $(M-1)+(-M-1)=-2$. Absorbing these terms and the smoothing cutoff contribution into $b_2$ gives

Iterate the correction: after constructing $b_{j-1}\in S^{-(j-1)}$, setThe same calculation improves the remainder by one order:Let $A_j$ and $B_N$ be the corresponding inverse oscillatory integrals. ThenWhen $N>n$, the frequency integral defining $B_N$ converges absolutely and depends continuously on $x$, so $B_N\in C(X)$. This completes the finite-order parametrix construction.