Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 327 / 1

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

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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 ${\mathcal{S}}^{\prime }$.

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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

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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.

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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:

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