Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 60 (Distribution Theory and Applications)

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$: its elements are smooth functions with compact support. For a fixed compact set $K$, put ${\mathcal{D}}_K=\{\phi \in C^{\infty }(\mathbb{R}):\mathrm{supp}\phi \subset K\}$ and use the seminorms $p_m(\phi )={\max }_{0\le j\le m}\Vert {\phi }^{(j)}{\Vert }_{\infty }$. The space of test functions carries the usual test-function inductive limit topology of these spaces. In particular, a sequence converges precisely when its supports eventually lie in one compact set and every derivative converges uniformly. Thus ${\phi }_j\to \phi$ means that a common $K$ contains their supports and $p_m({\phi }_j-\phi )\to 0$ for every $m$.

A distribution is a continuous linear functional on this space of test functions, and ${\mathcal{D}}^{\prime }(\mathbb{R})$ denotes their space. We use complex-linear, bilinear pairings $⟨u,\phi ⟩$. Equivalently, for every compact set $K$ there are $C_K$ and a nonnegative integer $m_K$ such thatThe usual weak convergence of distributions is $u_j\to u$ if $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function. This specifies the convergence used below; it does not require convergence in any norm.

The distributional derivative is defined byThe derivative map sends ${\mathcal{D}}_K$ continuously to itself, with $p_m({\phi }^{\prime })\le p_{m+1}(\phi )$. The preceding continuity estimate therefore proves that $u^{\prime }$ is again a distribution. This definition extends the ordinary derivative of a smooth function, by integration by parts.

Choose the translation of a distribution convention ${\tau }_{h}f(x)=f(x-h)$. Its action on a test function isFor fixed $h$, the translated test functions have translated compact support and unchanged derivative sup norms, so this defines a distribution. For the differentiability of distribution translations, apply the Taylor theorem in integral form:All supports lie in one slightly enlarged compact set for $|h|\le 1$, and the same identity for every derivative proves uniform convergence. The continuity of $u$ now givesThe minus sign in the translation parameter is necessary for this convention.

For (a), choose a test function $\eta$ with $\int \eta =1$. If $\int \psi =0$, then $F(x)={\int }_{-\infty }^x\psi (s)ds$ is a test function, since the zero integral makes it vanish beyond both ends of the compact support. If $u_1^{\prime }=0$, then $⟨u_1,\psi ⟩=⟨u_1,F^{\prime }⟩=0$. Decompose $\phi =(\int \phi )\eta +\psi$ to obtainConversely, these constant regular distributions have zero distributional derivative. This also proves the general fact that a distribution with zero derivative is constant.

For (b), choose a cutoff function $\rho$ equal to one near zero. Every test function decomposes asThe quotient extends as a smooth function at zero, and it has compact support. If $x{u}_2=0$, the definition of multiplication of a distribution by a smooth function gives $⟨u_2,\phi ⟩=⟨u_2,\rho ⟩\phi (0)$. Conversely, $x{\delta }_0=0$. Thus the kernel of multiplication by a coordinate is exactlyIn particular, derivatives of the Dirac delta distribution are not additional solutions: $x{\delta }_0^{\prime }=-{\delta }_0$.

Now write $D=d/dx$ and $P(z)=-z^n+{\sum }_{j=0}^{n-1}{a}_j{z}^j$, where $n\ge 1$. The characteristic roots of a constant-coefficient differential equation are the distinct roots of a polynomial ${\lambda }_1,\dots ,{\lambda }_r$ of $P$, with multiplicities $m_1,\dots ,m_r$. The distributional regularity of a constant-coefficient ordinary differential equation can be proved without assuming regularity in advance. Put $q_j(z)=(z-{\lambda }_j{)}^{m_j}$ and $Q={\prod }_j{q}_j=-P$. Since the $q_j$ are coprime polynomials, polynomial division and Bezout identity give polynomials $b_j$ such thatFor example, invert $Q/q_j$ modulo $q_j$, sum the resulting expressions, and absorb a remaining multiple of $Q$ into one coefficient. For $P(D)v=0$, the kernel decomposition for coprime polynomials therefore givesThe Leibniz rule for multiplication of a distribution by a smooth function implies $D^{m_j}(e^{-{\lambda }_{j}x}{v}_j)=0$. Repeatedly using a distribution with zero derivative is constant shows that a distribution with $m$th derivative zero is a polynomial of degree at most $m-1$: subtract the polynomial primitive of its constant $(m-1)$st derivative, and induct on $m$. Consequently the most general exponential polynomial solution of a constant-coefficient differential equation isEvery displayed term is annihilated by $P(D)$, so all coefficients are allowed. For the linear independence of these $n$ functions, on a relation, apply ${\prod }_{i\ne j}(D-{\lambda }_i{)}^{m_i}$; on $e^{{\lambda }_{j}x}$ times a polynomial of degree less than $m_j$, each remaining factor acts invertibly on that polynomial space. Hence the $j$th polynomial must vanish. Every distributional solution is an analytic function, and in particular a classical solution. For real coefficients and real-valued distributions, take complex conjugate coefficients at conjugate characteristic roots of a constant-coefficient differential equation, or equivalently use the corresponding real sine and cosine forms.

For the last equation, the operator is $xP(D)$: the coordinate multiplies the result of differentiation. The kernel of multiplication by a coordinate says precisely thatChoose the retarded fundamental solution of a constant-coefficient ordinary differential operator $E=Hw$, where $H$ is the Heaviside function and $w$ is the analytic solution of $P(D)w=0$ withSuch $w$ exists uniquely by the elementary initial value problem for a constant-coefficient ordinary differential equation; alternatively the residue construction in the next solution gives it explicitly. The distributional jump formula for a Heaviside product isIt follows by induction from $D(Hw)=H{w}^{\prime }+w(0){\delta }_0$, using integration by parts. With the chosen initial derivatives, every lower-order jump term vanishes and the leading coefficient $-1$ gives $P(D)E={\delta }_0$. Subtracting $CE$ reduces the last equation to the homogeneous one. Thus the coordinate-degenerate constant-coefficient differential equation has exactly the solutionsThere are $n+1$ independent constants. For $n\ge 2$, derivatives through order $n-2$ match at zero, while the derivative of order $n-1$ may jump; for $n=1$, the function itself may jump. An arbitrary distribution concentrated at zero cannot be added, because its image under the nonzero leading derivative would contain a nonvanishing highest derivative of the Dirac delta distribution.

The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear differential operator on Euclidean space has a distributional fundamental solution. With $D_j={\partial }_{x_j}$ and $P(D)={\sum }_{|\alpha |\le N}{a}_{\alpha }{D}^{\alpha }$, the conclusion is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^d)$ such that $P(D)E={\delta }_0$. If $N=0$, simply take $E=a_0^{-1}{\delta }_0$; hence assume $N\ge 1$.

We construct a Hörmander staircase in frequency space. The highest-degree homogeneous part $P_N$ is not identically zero on real vectors, since a polynomial vanishing on all real vectors has all coefficients zero. After an orthogonal transformation of coordinates, we can arrange that $P_N(e_1)\ne 0$. Thus, for each real ${\xi }^{\prime }\in {\mathbb{R}}^{d-1}$, the polynomialhas degree $N$ in $z$ and the same nonzero leading coefficient $b=i^N{P}_N(e_1)$, independent of ${\xi }^{\prime }$. This constant leading coefficient of a polynomial is what makes a uniform staircase possible.

For a fixed ${\xi }^{\prime }$, factor $p(z,{\xi }^{\prime })=b{\prod }_{\nu =1}^N(z-z_{\nu })$, counting repeated roots of a polynomial. Consider the $N+1$ heights $h_j=3j$, $j=0,\dots ,N$. A given root has imaginary part within distance less than one of at most one of these heights. By the pigeonhole principle, some height avoids every root, and thenThis is finite-height polynomial root avoidance. Root labels need not be chosen continuously or even measurably. Instead define the closed setsContinuity in $s$ extends the inequality from rational $q$ to every real $s$. These Borel sets partition ${\mathbb{R}}^{d-1}$. The Hörmander staircase assigns the horizontal contour $s+i{h}_j$ over $A_j$; its height is bounded by $3N$, and the denominator has the uniform lower bound $|b|$. For $d=1$, the transverse space is a single point and only one horizontal contour is needed.

Use the Fourier transform convention $\hat{\phi }(\zeta )=\int e^{-ix\cdot \zeta }\phi (x)dx$, with inverse factor $(2\pi {)}^{-d}$. Define the candidate fundamental solution of a linear differential operator byFor a test function supported in a fixed compact set $K$, Fourier decay in a bounded complex strip gives, for any integer $M$,Indeed, this is the real Fourier transform of $e^{-h_j{x}_1}\phi (x)$ with reversed frequency, and repeated integration by parts with $(1-\Delta {)}^M$ proves the estimate. Taking $2M>d$ and using the denominator bound proves absolute convergence and a continuity estimate on ${\mathcal{D}}_K$. Thus $E$ is a distribution; no unsupported interpretation of a divergent inverse Fourier transform is being used.

For its distributional derivatives, the formal transpose of a differential operator is $P(-D)$. Sinceapplying $P(D)$ cancels the denominator. For each fixed ${\xi }^{\prime }$, the numerator is an entire function of $z$. The Cauchy integral theorem, applied to a rectangle between the lines $\mathrm{Im}z=0$ and $\mathrm{Im}z=h_j$, givesThe two vertical edges tend to zero by the same bounded-strip decay. The resulting contour deformation is performed separately for each transverse frequency, so discontinuities of the staircase height introduce no additional boundary terms. Absolute convergence permits integration over the partition $A_j$. The Fourier inversion theorem then yieldsThis proves the Malgrange–Ehrenpreis theorem. Undoing the orthogonal change of coordinates gives the fundamental solution of a linear differential operator for the original operator; the Dirac delta distribution is unchanged by that change of coordinates.

For a one-dimensional operator $L=P(D)$ of degree $N\ge 1$, with leading coefficient $a_N$, we may choose the single staircase contour below every pole. Take $\gamma >\max \{\mathrm{Re}\lambda :P(\lambda )=0\}$ and integrate on $\mathrm{Im}\zeta =-\gamma$. The corresponding formula is the Bromwich contour version of the construction above, after putting $\lambda =i\zeta$. Its poles all lie above the frequency contour. For $x<0$, close that contour downwards; the exponential function $e^{ix\zeta }$ decays and there are no enclosed poles. For $x>0$, close upwards, where the exponential function again decays, and apply the residue theorem. The bound $1/P(i\zeta )=O(|\zeta {|}^{-N})$ justifies the large arcs, including $N=1$ by the Jordan lemma away from their endpoints. Thus away from $x=0$ the retarded fundamental solution is $H(x)u(x)$, whereThe sum is over distinct roots of a polynomial, with the residue including the whole multiplicity. For simple characteristic roots of a constant-coefficient differential equation, it reduces to $u(x)={\sum }_{\lambda }{e}^{\lambda x}/P^{\prime }(\lambda )$. Repeated roots give exponential polynomial solutions of a constant-coefficient differential equation through differentiation of $e^{zx}$.

To establish the equality also at the origin, rather than leave a possible point-supported term undecided, verify the distributional jump formula for a Heaviside product. The residue expression is an entire function of $x$, andIts initial derivatives areThese identities follow by integrating over a large circle: for $j\le N-2$ the integrand is $O(|z{|}^{-2})$, whereas for $j=N-1$ its coefficient of $z^{-1}$ is $a_N^{-1}$. The distributional jump formula for a Heaviside product therefore gives $L(Hu)={\delta }_0$. Moreover, since $e^{-\gamma x}Hu$ is a tempered distribution, its Fourier transform satisfiesThere are no real zeros of this polynomial, so its reciprocal is the transform. This is exactly the shifted-contour construction, proving equality there as a distribution as well. HenceThe sign of $a_N$ matters: for the operator in the preceding solution, $a_N=-1$ and the last initial derivative is $-1$.

An oscillatory integral defines a distribution by cancellation, even when its oscillatory integral amplitude is not integrable in frequency. We use symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$, where $X\subset {\mathbb{R}}^d$ is open and $N\in \mathbb{R}$. An oscillatory integral amplitude $a$ belongs to this class if it is smooth and, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,In symbol calculus, spatial derivatives preserve the order and frequency derivatives lower it. The fixed number $N$ is independent of $K$; this will produce a finite global order of a distribution, although continuity constants may depend on $K$.

A phase function $\Phi$ is real and smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, is positively homogeneous of degree one in $\theta$, and has nonzero total differential there:This positive homogeneity is required at nonzero frequency; arbitrary smooth low-frequency modifications give the equivalent version homogeneous only for large $|\theta |$. In particular, smoothness at zero is not an additional requirement on a general homogeneous phase function. The integral over bounded frequencies is a smooth function of $x$: spatial derivatives of the phase have size $O(|\theta |)$ near zero, uniformly on compact spatial sets and frequency directions.

Choose a cutoff function $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero. The proposed meaning of the oscillatory integral distribution isExistence and cutoff independence of an oscillatory integral require a proof. Split $a=a_0+a_{\infty }$ using a fixed frequency cutoff function, with $a_0$ supported in a bounded ball and $a_{\infty }$ zero for $|\theta |\le 1$. The low-frequency part is already absolutely integrable. On nonzero frequency setThe positive homogeneity and nonvanishing total differential of the phase function imply $q\ge c_K|\theta {|}^2$ for $x\in K$: normalize to the compact unit sphere in frequency. Direct differentiation gives $L{e}^{i\Phi }=e^{i\Phi }$.

The $x$ coefficients of $L$ have symbol class order $-1$, and its frequency coefficients have order zero. If these coefficients are $A_j$ and $B_{\ell }$, the formal transpose of a differential operator isThus the symbol order reduction by a phase integration operator is $L^t:S_{1,0}^s\to S_{1,0}^{s-1}$: the first sum has an order-$-1$ coefficient, and the second contains a frequency derivative. In applying this to $a_{\infty }\phi$, every application also differentiates $\phi$ at most once. Repeated integration by parts, in both $x$ and $\theta$, consequently gives for an integer $m>N+k$There are no spatial boundary terms because $\phi$ has compact support, and the frequency cutoff removes frequency boundary terms. Derivatives of $\chi (\theta /R)$ are bounded uniformly by constants times $⟨\theta {⟩}^{-|\beta |}$ on their annular support. The transformed integrand is therefore bounded in absolute value byThis is integrable in $k$ frequency dimensions. Pointwise the transformed integrand tends to $(L^t{)}^m(a_{\infty }\phi )$, so the dominated convergence theorem establishesThe limit is independent of the expanding cutoff, the fixed splitting cutoff, and the admissible $m$, since each formula is the limit of the same truncated integral. Derivatives landing on the expanding cutoff can also be estimated directly by $O(R^{N-m+k})$, which tends to zero. In particular,This proves continuity on the space of test functions and the finite order of an oscillatory integral distribution. The same $m$ works for all compact sets. When $N<-k$, $m=0$ is possible and the original frequency integral is absolutely integrable; cancellation is needed for general $N$.

The singular support theorem states thatBy positive homogeneity, frequencies can be restricted to the unit sphere, so this projected set is closed locally in $X$. More precisely, the stationary-direction bound for singular support restricts the frequency directions to the closed conic support of an oscillatory amplitude; using a closed directional support avoids losing limits occurring at arbitrarily high frequency. The simpler displayed bound is sufficient here.

To prove the bound, take $x_0$ outside the displayed stationary set. On a sufficiently small spatial neighborhood and all unit frequency directions, $|{\nabla }_{\theta }\Phi |$ is bounded below. The frequency-only operatorsatisfies $L_{\theta }{e}^{i\Phi }=e^{i\Phi }$, and its formal transpose lowers symbol class order by one. A spatial derivative of order $|\alpha |$ of $e^{i\Phi }a$ has amplitude order at most $N+|\alpha |$. Applying the frequency integration by parts more than $N+|\alpha |+k$ times makes that derivative absolutely integrable, uniformly on smaller compact sets. Every spatial derivative therefore exists and is continuous there. The oscillatory integral distribution is a smooth function near $x_0$, proving the singular support assertion.

For the linear transport equation, use spacetime $X={\mathbb{R}}^n\times (0,\infty )$, frequency $\theta \in {\mathbb{R}}^n$, andThis is a valid phase function, because ${\nabla }_x\Phi =\theta \ne 0$ at nonzero frequency; the oscillatory integral amplitude is in symbol class order zero. The Fourier representation of the Dirac delta function gives the transport of a Dirac point mass:Its rigorous spacetime distribution pairing is $⟨u,\phi ⟩={\int }_0^{\infty }\phi (ct,t)dt$. Consequentlywhere the endpoints vanish since a spacetime test function has compact support in $t>0$. At each fixed $t$, $⟨u(t),\psi ⟩=\psi (ct)$, so weak convergence of distributions gives $u(t)\to {\delta }_0$ as $t\downarrow 0$. This is the required initial trace.

Finally, ${\nabla }_{\theta }\Phi =x-ct$, so the singular support theorem confines singularities to $x=ct$. In fact equality holds: $u$ is a nonzero order-zero distribution on that trajectory and vanishes off it. A smooth function supported on this set of empty interior must vanish, so $u$ cannot be smooth in any neighborhood of a point of the trajectory. ThusThe method of characteristics also proves uniqueness among solutions with a distributional initial trace: the change of variables $y=x-ct$ turns the linear transport equation into ${\partial }_{t}w=0$. Such a distribution is constant in $t$; pairing with spatial test functions reduces this assertion to a distribution with zero derivative is constant. Its initial trace fixes $w={\delta }_0$, so the moving Dirac delta distribution above is the unique solution. There is transport of the singularity along the characteristic and no smoothing.