Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 327 (Distribution Theory and Applications)

The Schwartz space on the real line isThese seminorms define its Fréchet space topology. The space of tempered distributions is its continuous dual space,

With the angular-frequency convention, the Fourier transform of a Schwartz function isIt maps $\mathcal{S}$ continuously to itself. The transform of $u\in {\mathcal{S}}^{\prime }$ is defined through the dual pairing:up to the fixed reflection and $2\pi$ factor if the inverse-transform convention is used for the test function.

For $f\in L^1(\mathbb{R})$, its regular tempered distribution acts by integration, and the distributional transform agrees with the functionThe triangle inequality gives the uniform boundIf ${\lambda }_j\to \lambda$, then $f(x)e^{-i{\lambda }_{j}x}\to f(x)e^{-i\lambda x}$ pointwise and every integrand is bounded in absolute value by the Lebesgue integrable function $|f|$. The Dominated convergence theorem therefore proves that $\hat{f}$ is continuous. In fact the Riemann-Lebesgue lemma also gives $\hat{f}(\lambda )\to 0$ as $|\lambda |\to \infty$.

LetThe integral is absolutely convergent. For a Schwartz function $\phi$, Fubini's theorem and integration by parts in $\lambda$ giveEquivalently, $v^{\prime }=-\hat{u}$ as a distributional derivative, and hence

Because $x^{-2}$ is integrable on $[1,\infty )$, the Dominated convergence theorem applied to the defining integral proves that $v$ is continuous on $\mathbb{R}$.

For $\lambda >0$, the change of variables formula $t=\lambda x$ givesand for $\lambda <0$ the analogous formula is obtained with $e^{it}$. On either open half-line the lower endpoint stays away from zero locally, so repeated differentiation under the integral sign proves smoothness. Thus

For $0<\lambda <1$, the preceding change of variables and one integration by parts giveSplit the last improper integral at one and add and subtract one on $(\lambda ,1)$. Since ${\int }_{\lambda }^{1}dx/x=-\log \lambda$,

The definition also gives the complex conjugate relationConsequently, for $-1<\lambda <0$,

On either open half-line, part (i) says $\hat{u}=-v^{\prime }$. Differentiating the formula in part (iii) makes all elementary terms cancel except the logarithm and the bracket. ThereforewhereandwhereBoth functions are smooth on their respective half-lines, proving the required assertion.

For $\lambda >0$, subtracting the two formulas from part (iv) yieldsUsing the stated Dirichlet integral and taking the one-sided limit gives

Remove the two reciprocal tails by settingThe assumed $O(x^{-2})$ remainder at each end and local integrability on bounded intervals imply $r\in L^1(\mathbb{R})$. Its Fourier transform $\hat{r}$ is therefore continuous at zero. Since $\widehat{u(-\cdot )}(\lambda )=\hat{u}(-\lambda )$,Parts (iv) and (v) show that both requested one-sided limits exist. If $L_{+}$ denotes the limit from positive frequencies and $L_{-}$ the limit from negative frequencies, then the common value $\hat{r}(0)$ cancels andThis is the universal jump caused by reciprocal tails.

The space of test functions isA sequence ${\phi }_j$ converges to $\phi$ in $\mathcal{D}$ when all supports lie eventually in one compact set $K$ andfor every multi-index $\alpha$. A distribution is a linear functional $u:\mathcal{D}\to \mathbb{C}$ such that, for every compact $K$, there are $C_K>0$ and $m_K\in {\mathbb{N}}_0$ withwhenever $\mathrm{supp}\phi \subseteq K$. Convergence in ${\mathcal{D}}^{\prime }$ is pointwise convergence on test functions.

Continuity plainly implies sequential continuity. Conversely, suppose the linear form is sequentially continuous but the displayed estimate fails for some compact $K$. For every $j$, choose ${\phi }_j$ supported in $K$ such thatThen ${\phi }_j\to 0$ in $\mathcal{D}$, while $u({\phi }_j)\nrightarrow 0$, a contradiction. Hence the seminorm estimate holds on every compact set and $u\in {\mathcal{D}}^{\prime }$.

For a translation vector $h$ and a multi-index $\alpha$, defineThese definitions extend ordinary translation and differentiation to distributions.

If ${\tau }_{t{e}_i}u=u$ for every $t$, differentiating its pairing at $t=0$ gives ${\partial }_{i}u=0$. Conversely, if ${\partial }_{i}u=0$, then for every test functionThe pairing is constant in $t$, hence ${\tau }_{t{e}_i}u=u$. This proves both directions of translation invariance and vanishing distributional derivative.

Finally, the distributional differentiation obeys the linear chain rule under the linear coordinates $s=x-y$ and $t=x+y$. Thusand likewiseas distributions. Adding the two identities givesthe one-dimensional wave equation; this is the low-regularity form of the travelling waves in the D'Alembert formula.

A phase function is a real smooth function $\Phi$ on $X\times ({\mathbb{R}}^k\setminus \{0\})$ that is positively homogeneous of degree one in $\theta$ and has nonzero total differential $d_{x,\theta }\Phi$. The symbol classconsists of smooth amplitudes for which, for every compact $K\subset X$ and all multi-indices $\alpha ,\beta$,

Choose a smooth cutoff function $\chi$ equal to one near zero. The associated oscillatory integral is defined on a test function $f$ byOn the compact $x$-support of $f$, use an integration-by-parts operator $L$ satisfying $L{e}^{i\Phi }=e^{i\Phi }$. Repeated application of its formal adjoint lowers the effective symbol order until the integral is absolutely convergent. The resulting bounds involve only finitely many derivatives of $f$, prove that the limit is independent of $\chi$, and give the seminorm estimate required for

Applying $D_x^{\alpha }$ changes only the constants in the defining symbol estimates, while every $D_{\theta }$ 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 two factors. Multiplying the corresponding symbol estimates adds their orders, so

If $b$ is positively homogeneous of degree $M$ for large $|\theta |$, then $D_{\theta }^{\beta }b$ is positively homogeneous of degree $M-|\beta |$, while $x$-derivatives preserve the degree. These derivatives are uniformly bounded on the unit sphere when $x$ ranges over a compact subset of $X$. Rescaling $\theta$ givesat large frequency, and smoothness controls the remaining compact region. Therefore

For a test function $f\in \mathcal{D}(\mathbb{R})$, define the proposed integral by reversing the order of integration:The Fourier transform of a test function is a Schwartz function, while $\sqrt{{\theta }^2+1}\asymp 1+|\theta |$. The final integral is consequently absolutely convergent. Repeated integration by parts in $x$ bounds it by finitely many seminorms of the test function, so it defines a continuous linear functional on $\mathcal{D}(\mathbb{R})$.

Equivalently, put $\lambda =\sqrt{{\theta }^2+1}$ on the two half-lines. ThenThe density has only an integrable inverse-square-root singularity at one and is bounded at infinity, so it is a regular tempered distribution. Its inverse Fourier transform is exactly the proposed oscillatory integral. Thus