Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 327 (Distribution Theory and Applications)

For a compact convex set $K\subset {\mathbb{R}}^n$, letThe Paley–Wiener–Schwartz theorem says that if $u$ is a compactly supported distribution with support in $K$, its Fourier--Laplace transformis entire and, for some $C,N$,Conversely, every entire function satisfying such an estimate is the transform of a distribution supported in $K$.

For the forward direction, compact support lets $u$ act on the exponential after insertion of a cutoff equal to one near $K$. Differentiation in $z$ may be passed under the pairing, proving entire analyticity. The finite-order estimate for $u$ bounds derivatives of the exponential on $K$ by a polynomial in $|z|$ times $e^{H_K(\mathrm{Im}z)}$.

Conversely, restrict the entire function $F$ to ${\mathbb{R}}^n$. Its polynomial growth defines a tempered distribution $u$ by inverse Fourier transform. If a test function is supported outside $K$, separate its compact support from $K$ by a real vector $\eta$. Shifting the Fourier inversion contour from ${\mathbb{R}}^n$ to ${\mathbb{R}}^n+it\eta$ is allowed by entire analyticity. The exponential gained from the test function beats the bound $e^{t{H}_K(\eta )}$ as $t\to \infty$, so the pairing vanishes. Hence $\mathrm{supp}u\subset K$, completing the converse.

If $u\in {\mathcal{E}}^{\prime }(\mathbb{R})$ solves $P(D)u=v$, Fourier transformation givesThe Paley–Wiener–Schwartz theorem makes $\hat{u}$ entire, so $\hat{v}/P$ is entire.

Conversely, suppose $F(z)=\hat{v}(z)/P(z)$ is entire. Polynomial division estimates away from the finitely many zeros of $P$, together with the maximum principle on fixed disks around those zeros, show that $F$ retains a Paley--Wiener--Schwartz bound, with only the polynomial exponent changed. The converse theorem therefore gives $u\in {\mathcal{E}}^{\prime }(\mathbb{R})$ with $\hat{u}=F$. Then $P(D)u=v$. Thus

No. Non-entireness rules out a compactly supported solution, but the Malgrange–Ehrenpreis theorem gives a distributional fundamental solution of a linear differential operator $E$ with $P(D)E=\delta$. Since $v$ has compact support, the convolution $u=E\ast v$ is defined and satisfiesOne may choose a tempered fundamental solution for a constant-coefficient operator, so $u\in {\mathcal{S}}^{\prime }(\mathbb{R})$.

The Sobolev space $H^s({\mathbb{R}}^n)$ consists of $u\in {\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ for whichThe Local Sobolev space $H_{\mathrm{loc}}^s(X)$ consists of distributions $u$ such that $\chi u\in H^s({\mathbb{R}}^n)$ for every $\chi \in C_c^{\infty }(X)$.

If $P$ has degree $m$ and principal homogeneous part $P_m$, then $P(D)$ is an elliptic differential operator whenEquivalently, $|P(\xi )|\ge c|\xi {|}^m$ for all sufficiently large real $\xi$.

True. Multiplication by a Schwartz function is a bounded map $H^s({\mathbb{R}}^n)\to H^s({\mathbb{R}}^n)$ for every real $s$.

True. If $s>t$, then $⟨\xi {⟩}^{2t}\le ⟨\xi {⟩}^{2s}$, so $H^s({\mathbb{R}}^n)\subset H^t({\mathbb{R}}^n)$ continuously.

False. Ellipticity concerns the principal part and large frequencies; lower-order terms may create nonzero real roots. For example, $P(\xi )={\xi }^2-1$ is elliptic in one dimension but vanishes at $\xi =\pm 1$.

True. Repeated use of the Leibniz rule and the polynomial Taylor formula gives

True. The Sobolev embedding theorem gives $H^s({\mathbb{R}}^n)\hookrightarrow C^0({\mathbb{R}}^n)$ when $s>n/2$.

False. The constant distribution $u=1$ is tempered, but its Fourier transform is a multiple of the Dirac delta function, which is not an $L^2$ function after multiplication by any Sobolev weight. Thus $1\notin H^t({\mathbb{R}}^n)$ for every $t$.

True. The Fourier transform of a compactly supported distribution is a smooth function of at most polynomial growth. A sufficiently negative Sobolev weight makes its square integrable, so every $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$ belongs to $H^t({\mathbb{R}}^n)$ for some $t$.

True. Fourier transformation turns $D^{\alpha }$ into multiplication by ${\xi }^{\alpha }$, andHence $D^{\alpha }:H^s\to H^{s-|\alpha |}$ is bounded.

The derivative hypothesis implies that $1/Q(\xi )$ is a symbol of order $-\delta N$ at high frequency. Choose cutoffs $\chi \prec \psi$ in $X$ and a high-frequency cutoff in $\xi$. The corresponding Fourier multiplier is a parametrix for $Q(D)$, and the symbol calculus, together with the product formula from part (b), gives the localized estimatefor some sufficiently negative $t$. The commutator terms contain derivatives $Q^{(\alpha )}$; the assumed factor $|\xi {|}^{-\delta |\alpha |}$ lowers their order and lets them be absorbed inductively. Therefore

If $Q(D)u$ is smooth, it belongs locally to $H^s$ for every $s$. Starting from the fact that every compactly supported distribution has some negative Sobolev order and repeatedly applying the gain $\delta N>0$ places $u$ in every local Sobolev space. The Sobolev embedding theorem then gives $u\in C^{\infty }(X)$. Thus $Q(D)$ is a hypoelliptic differential operator.

The heat operatoris hypoelliptic: its symbol $i\tau +|\xi {|}^2$ satisfies the derivative estimates that yield local regularity by the argument in part (c). It is not elliptic as an operator of total order two, because its principal symbol is $|\xi {|}^2$, which vanishes at every nonzero covector $(\tau ,0)$ with $\tau \ne 0$. Hence it is a hypoelliptic differential operator that is not an elliptic differential operator.

A phase function is a real smooth function $\Phi$ on $X\times ({\mathbb{R}}^k\setminus \{0\})$, positively homogeneous of degree one in $\theta$, with $d_{x,\theta }\Phi \ne 0$. The symbol classconsists of smooth amplitudes satisfying, on each compact $K\subset X$,

To define the oscillatory integral, insert a cutoff $\chi (\epsilon \theta )$ equal to one near zero and setRepeated integration by parts with an operator $L$ satisfying $L{e}^{i\Phi }=e^{i\Phi }$ makes the integral absolutely convergent after enough iterations and shows that the limit defines a distribution.

Applying $D_x^{\alpha }$ changes only the constants in the defining estimates, while each $D_{\theta }$ lowers the power of $⟨\theta ⟩$ by one. Hence

The Leibniz rule writes every derivative of $a_1{a}_2$ as a finite sum of products of derivatives of the factors. Multiplying their symbol bounds adds the orders, so

Differentiating a function positively homogeneous of degree $M$ in $\theta$ makes $D_{\theta }^{\beta }b$ homogeneous of degree $M-|\beta |$, while $x$-derivatives preserve the degree. On the unit sphere these derivatives are bounded uniformly over compact subsets of $X$. Scaling then givesfor large $\theta$; smoothness controls the remaining compact frequency region. Thus

The singular support of $u\in {\mathcal{D}}^{\prime }(X)$ is the complement of the largest open subset of $X$ on which $u$ is represented by a smooth function.

Suppose $x_0$ has a neighborhood on which ${\nabla }_{\theta }\Phi (x,\theta )\ne 0$ for every $\theta \ne 0$. There one may integrate by parts repeatedly withEach adjoint application lowers the effective symbol order. After enough repetitions, the integral and all its $x$-derivatives converge absolutely and define a smooth function near $x_0$. Therefore

The claim is false because vanishing of the amplitude at one spatial point need not control its derivatives nearby. Take $X=\mathbb{R}$, $k=1$,Then $0\in Z(a)$, but distributionallyup to the Fourier-transform sign convention. Thus $0$ lies in the singular support of $I_{\Phi }(a)$ despite belonging to $Z(a)$.