Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 327 (Distribution Theory and Applications)

The space of test functions isA sequence ${\phi }_m$ converges to $\phi$ in $\mathcal{D}(\mathbb{R})$ when all supports lie in one compact set $K$ andfor every nonnegative integer $j$. The distribution space ${\mathcal{D}}^{\prime }(\mathbb{R})$ is the continuous dual space of $\mathcal{D}(\mathbb{R})$, and $u_m\to u$ in ${\mathcal{D}}^{\prime }$ meansfor every test function $\phi$.

If a linear form $u$ is continuous, it clearly maps every null sequence to a scalar sequence tending to zero. Conversely, suppose it has this sequential property. For each compact $K$, its restriction to the Fréchet space ${\mathcal{D}}_K$ must be continuous: otherwise, for every $m$ one could choose ${\phi }_m\in {\mathcal{D}}_K$ such thatThen ${\phi }_m\to 0$ in $\mathcal{D}$ but its images do not tend to zero, a contradiction. Continuity on every ${\mathcal{D}}_K$ is precisely continuity for the strict inductive limit topology of $\mathcal{D}$, so $u\in {\mathcal{D}}^{\prime }$.

Use the convention ${\tau }_h\phi (x)=\phi (x-h)$. Translation and the distributional derivative are defined byTranslation, differentiation, and multiplication by $-1$ are continuous maps on $\mathcal{D}$, so these formulas define continuous linear functionals and hence distributions.

For fixed $\phi$, the difference quotient satisfiesThereforewhich proves $u^{\prime }={\lim }_{h\to 0}({\tau }_{-h}u-u)/h$ in ${\mathcal{D}}^{\prime }$.

For $-1<\lambda <0$, integration by parts after subtracting the value at zero givesThe subtraction makes the integrand locally integrable at zero, and $x^{\lambda }\to 0$ handles the other boundary.

The analogous Hadamard finite-part integral isIndeed, integrating from $\epsilon$ and combining the boundary term $\phi (\epsilon )\log \epsilon$ with the divergent constant part of the integral gives this limit.

Because $x_{+}^{\lambda }$ is a locally integrable function, its first distributional derivative has order at most one. It is not of order zero. Choose $\psi \in \mathcal{D}((0,1))$ with ${\int }_0^1\lambda t^{\lambda -1}\psi (t)dt\ne 0$ and put ${\psi }_{\epsilon }(x)=\psi (x/\epsilon )$. The sup norms stay bounded whileis unbounded as $\epsilon \downarrow 0$. This contradicts the local sup-norm estimate required of an order-zero distribution. Hence $(x_{+}^{\lambda }{)}^{\prime }$ has order exactly $\boxed{1}$.

A phase function is a real smooth functionthat is positively homogeneous of degree one in $\theta$ and has no critical point in all variables:The symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)$ consists of all $a\in C^{\infty }(X\times {\mathbb{R}}^k)$ such that, for every compact $K⋐X$ and multi-indices $\alpha ,\beta$,

For a cutoff $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero, define the oscillatory integral byRepeated integration by parts makes the limit meaningful and independent of the cutoff.

The singular support is the complement of the largest open set on which the distribution is represented by a smooth function. Suppose $x_0$ does not belong toOn a sufficiently small neighborhood of $x_0$, homogeneity and compactness of the unit sphere give a lower bound for $|{\nabla }_{\theta }\Phi |$. The differential operatorsatisfies $L{e}^{i\Phi }=e^{i\Phi }$. Repeatedly transferring $L$ to the amplitude lowers its symbol order until the integral and all its $x$-derivatives converge absolutely. Hence $I_{\Phi }(a)$ is smooth near $x_0$, proving

For the stated distribution on ${\mathbb{R}}^3$, rotational symmetry lets us align the polar axis with $x$ and write $\rho =|x|$. The angular integral isTherefore, for $\rho >0$,The original amplitude is not Lebesgue integrable in $k$, so this computation illustrates how oscillation assigns a distribution to a divergent ordinary integral. The result is smooth away from the origin and has singular support $\{0\}$. It is a constant multiple of the three-dimensional Yukawa potential, satisfying $(-\Delta +1)u=4{\pi }^2{\delta }_0$ with the normalization used in the question.

Differentiation in $x$ does not change the allowed growth order in $\theta$, while every $\theta$-derivative lowers it by one. More precisely, for further multi-indices $\gamma ,\delta$,Thus

The multivariable Leibniz rule writes each derivative of $a_1{a}_2$ as a finite sum of productsEach term is bounded by a constant timesHence

Using multi-index notation, the Schwartz space isIts topology is generated by the displayed seminorms. The tempered distribution ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is its continuous dual, with weak convergence defined by convergence of every pairing with a Schwartz function.

For the conventiondifferentiation under the integral and integration by parts giveThese identities bound every Schwartz seminorm of $\hat{\phi }$ by finitely many seminorms of $\phi$, proving continuity. The Fourier inversion theorem gives the continuous inverseso the Fourier transform is a continuous isomorphism of $\mathcal{S}$. By duality,defines a continuous isomorphism of ${\mathcal{S}}^{\prime }$.

For $\omega \ne 0$, insert the Gaussian damping factor $e^{-\epsilon x^2/2}$ and use the Gaussian integral:Taking $\epsilon \downarrow 0$ in ${\mathcal{S}}^{\prime }$ with the continuous square-root branch gives the Fresnel integralFor $\omega =0$, the function is constant and its Fourier transform is $\boxed{2\pi {\delta }_0}$.

For a locally integrable $u$, the change of variables formula $y=Ax$ givesThis motivatesBecause pullback by an invertible linear map acts continuously on the Schwartz space, the right side is a continuous linear functional of $\phi$. It therefore defines a tempered distribution and agrees with ordinary pullback when $u$ is a function.

For every $\phi \in \mathcal{S}$, the change of variables formula gives the identityUsing this, the distributional Fourier transform, and the pullback formula from part b,Therefore

By the spectral theorem for real symmetric matrices, write $A=O^{t}DO$ with $O$ orthogonal and $D=\mathrm{diag}({\xi }_1,\dots ,{\xi }_n)$. Part a and the tensor-product property of the Fourier transform giveThusApplying the pullback rule from part c to the orthogonal change of variables, for which $|\det O|=1$, replaces $D$ by $A$ and $D^{-1}$ by $A^{-1}$. Since determinant and signature are invariant under orthogonal conjugation,