Past exam of the distribution theory course of the University of Oxford

A distribution on $(a,b)$ is a continuous linear functional on the space of test functions $\mathcal{D}(a,b)=C_c^{\infty }(a,b)$. It has order of a distribution at most $m$ if, for every compact set $K\subset (a,b)$, there is $C_K$ such thatThe order of a distribution is the least such $m$. When saying “of order $m$” means merely that this estimate holds, the wording allows a smaller least order.

For each such $K$, the cutoff function theorem supplies a nonnegative $\chi \in \mathcal{D}(a,b)$ equal to one near $K$. If $M=\Vert \varphi {\Vert }_{\infty }$ and $\mathrm{supp}\varphi \subseteq K$, then $M\chi \pm \varphi \ge 0$ everywhere. Positivity of the linear functional $V$ therefore givesHere $V(\chi )\ge 0$ is finite and independent of $\varphi$. This is precisely the local bound for an order-zero distribution, so the given positive distribution is continuous on the space of test functions. The zero distribution also satisfies the estimate.

The distributional derivative of $u$ is defined byIt is a distribution because $\varphi \mapsto {\varphi }^{\prime }$ is a continuous map of the space of test functions into itself. If $f\in C^1(a,b)$ represents a regular distribution, integration by parts givesThe boundary term vanishes because $\varphi$ has compact support. Thus the distributional derivative agrees with the ordinary derivative.

For $\mathrm{supp}\varphi \subseteq K$, the assumed order of a distribution estimate givesConsequently $\boxed{\mathrm{ord}(u^{\prime })\le m+1}$.

On each interval $[k,k+1]$, $k\in \mathbb{Z}$, the function is $d(x)=x-k$ on $[k,k+\frac{1}{2}]$ and $d(x)=k+1-x$ on $[k+\frac{1}{2},k+1]$. Its ordinary derivative exists off $\frac{1}{2}\mathbb{Z}$ and is the locally integrable functionAssign arbitrary values at the endpoints. For a test function $\varphi$, split $-\int d{\varphi }^{\prime }$ at these corners and apply integration by parts on each piece. The adjacent boundary terms cancel because $d$ is continuous, and only finitely many pieces meet the compact support of $\varphi$. HenceThe distributional derivative is exactly the regular distribution of the ordinary derivative defined almost everywhere. There are no Dirac delta distributions in this first distributional derivative: corners change slope, while a jump of the function itself would create a point mass.

The bound $0\le d_n\le 2^{-n-1}$ proves uniform convergence of the series by the Weierstrass M-test. Each summand is continuous and satisfies $d_n(x+1)=d_n(x)$, so its sum $t$, the Takagi function, is continuous and has period one.

For $n\ge m$, both $2^n{u}_m$ and $2^n{v}_m$ are integers, giving $d_n(u_m)=d_n(v_m)=0$. ThereforeThe corners of $d_n$ occur on the grid $2^{-n-1}\mathbb{Z}$. When $n<m$, that grid is contained in $2^{-m}\mathbb{Z}$, so the interior of $[u_m,v_m]$ contains no corner of $d_n$. Its slope on this interval is consequently $+1$ or $-1$.

The interval $[u_{m+1},v_{m+1}]$ is a half of $[u_m,v_m]$. All the old slopes persist on this half, and the new summand contributes one more slope in $\{\pm 1\}$. ThusIf $t$ had a finite derivative $L$ at $x_0$, write $t(y)=t(x_0)+L(y-x_0)+r(y)$ with $r(y)=o(|y-x_0|)$. Since $u_m\le x_0<v_m$ and both endpoints tend to $x_0$,Indeed $|v_m-x_0|+|u_m-x_0|=v_m-u_m$; when $u_m=x_0$, its remainder is zero. This is the secants straddling a differentiability point argument, including dyadic $x_0$. But the displayed increments prevent $S_m$ from converging. Hence $\boxed{t\text{ is nowhere differentiable}}$, using periodicity to cover all real $x_0$.

For the distributional derivative, set $r_n(x)=s(2^{n}x)$, using $s$ from part (iii). The change of variable in the distributional derivative pairing gives $d_n^{\prime }=r_n$; these are the Rademacher functions. Uniform convergence implies weak convergence of distributions, and the continuity of distributional differentiation allows termwise differentiation in that topology. Explicitly,This series assertion is a distributional convergence statement. Its remainder satisfies the concrete estimateIt supplies a well-defined distributional derivative even though the ordinary derivative exists nowhere.

Choose $R>0$ with $\mathrm{supp}\varphi \subset [-R,R]$. Evenness of $f$ changes the symmetrically truncated integral intoThe mean value theorem gives $|\varphi (x)-\varphi (-x)|\le 2x\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. Since $f$ is continuous, it is bounded on $[0,R]$. Thus the Cauchy principal value exists andThis expression is linear in the test function and satisfiesFor each fixed compact support, choose one $R$ once and for all. The estimate proves that the even-weight principal-value distribution has order of a distribution at most one.

If $\varphi (0)=0$, the mean value theorem also gives $|\varphi (x)|\le |x|\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. Hence $f(x)\varphi (x)/x$ is absolutely integrable near zero and has compact support. The truncated integrals tend to its ordinary Lebesgue integral, so

No. Consider the continuous even functionThe formulas agree at $|x|=e^{-1}$, and $f(x)\to 0$ at zero. Part (i) defines its even-weight principal-value distribution. However,If this were a regular distribution represented by $g\in L_{\mathrm{loc}}^1(\mathbb{R})$, testing on either side of zero would force $g=f/x$ almost everywhere there, by uniqueness of the locally integrable function representing a regular distribution. Changing the value at zero cannot remove the displayed divergence, contradicting local integrability of $g$.

More generally, the exact condition isNecessity is the preceding argument; sufficiency follows because $f/x$ is then a locally integrable function and its truncated integrals converge absolutely. The value $f(0)=0$ alone provides no such rate of decay.

Convergence ${\varphi }_j\to \varphi$ in the space of test functions means that all supports lie in one compact set $K\subset \mathbb{R}$ and $\Vert {\varphi }_j^{(r)}-{\varphi }^{(r)}{\Vert }_{\infty }\to 0$ for every integer $r\ge 0$. An eventual common compact support is equivalent, since finitely many initial supports can be added to $K$.

Write the standard mollifier as ${\rho }_{\epsilon }(x)={\epsilon }^{-1}\rho (x/\epsilon )$, where $\rho \in \mathcal{D}(\mathbb{R})$ is nonnegative, supported in $[-1,1]$, and has integral one. For $0<\epsilon \le 1$, the convolution ${\rho }_{\epsilon }\ast \varphi$ has support in the fixed compact set $\mathrm{supp}\varphi +[-1,1]$. The given convolution derivative identity yieldsEach ${\varphi }^{(r)}$ is uniformly continuous; more explicitly, the mean value theorem bounds the absolute value by $\epsilon \Vert {\varphi }^{(r+1)}{\Vert }_{\infty }\int |y|\rho (y)dy$, uniformly in $x$. Thus

If ${\varphi }_j\to 0$ in the space of test functions and $K$ is a common compact support, then $\Vert {\varphi }_j{\Vert }_{L^1}\le |K|\Vert {\varphi }_j{\Vert }_{\infty }\to 0$, where $|K|$ is its Lebesgue measure.

The mollifiers themselves have no limit in the L1 space. If ${\rho }_{\epsilon }\to g$ in that norm, then for each $\delta >0$ the integral of $|g|$ on $\{|x|>\delta \}$ is zero, because ${\rho }_{\epsilon }$ vanishes there for $\epsilon <\delta$. Hence $g=0$ almost everywhere. This contradicts $\Vert {\rho }_{\epsilon }{\Vert }_{L^1}=1$. Therefore $\boxed{{\rho }_{\epsilon }\text{ does not converge in }{L}^1(\mathbb{R})}$.

The weak convergence of distributions $u_j\to u$ meansFor $\theta \in \mathcal{D}(\mathbb{R})$, its convolution with a distribution is the smooth functionThe subscript records the variable on which the distribution acts. With $\check{\theta }(x)=\theta (-x)$, its regular distribution satisfies the useful pairing identityIndeed integrating $\theta (x-y)\varphi (x)$ in $x$ gives $(\check{\theta }\ast \varphi )(y)$; continuity of the distribution justifies moving the pairing through that integral, by approximation with sums of test functions in one common compact support.

Part (i), applied to the reflected mollifier, gives ${\check{\rho }}_{\epsilon }\ast \varphi \to \varphi$ in the space of test functions. Continuity of $u$ therefore provesReflection is harmless even if the chosen standard mollifier is not even.

For the mollifiers themselves, a change of variable givesThus $\boxed{{\rho }_{\epsilon }\to {\delta }_0\text{ in }{\mathcal{D}}^{\prime }(\mathbb{R})}$, with limit the Dirac delta distribution, despite the failure of convergence in the L1 space.

The L1-normalized dilation of a distribution isFor a regular distribution this is the function $u_{\epsilon }(x)={\epsilon }^{-1}u(x/\epsilon )$. In terms of the wiki's dilation of a distribution convention, $u_{\epsilon }={\epsilon }^{-1}{\delta }_{1/\epsilon }u$; the normalization preserves the L1 space norm.

If $u\in L^1(\mathbb{R})$, the dominated convergence theorem, with dominating integrable function $|u(y)|\Vert \varphi {\Vert }_{\infty }$, givesConsequentlyAn arbitrary distribution need not have this behavior. Take the constant regular distribution $u=1$. Then $u_{\epsilon }={\epsilon }^{-1}$, so for a test function with integral one,Hence $\boxed{u=1\text{ gives no distributional limit}}$.

Put $A={\int }_0^{1}f$. ThenThe theorem on differentiation of an indefinite Lebesgue integral makes this an absolutely continuous function on $[0,1]$, withBoth $A-f$ and $Rf$ belong to the L1 space: for example, $|Rf(x)|\le |A|+\Vert f{\Vert }_{L^1}\le 2\Vert f{\Vert }_{L^1}$. The ordinary derivative of an absolutely continuous function is its weak derivative, as integration by parts against a test function shows. Therefore $\boxed{Rf\in W^{1,1}(0,1)}$, the indicated Sobolev space.

Using $(Rf{)}^{\prime }={\int }_0^{1}f-f$ and integration by parts, with $\varphi$ a test function, givesFor each $k\ge 1$, apply the same identity to $R^{k}f$ and the test function ${\varphi }^{(k)}$. Since ${\int }_0^1{\varphi }^{(k)}=0$, it givesStarting from $k=1$ and iterating provesThe intermediate means disappear because derivatives of a test function have zero integral; no assumption that $R^{k}f$ has zero mean is needed.

For the bound, writeThe triangle inequality and Fubini's theorem implyThus $\boxed{c=\frac{1}{2}\text{ is valid}}$, and every iterate $R^{k}f$ is in the L1 space.

To repeat integration by parts across whole periods, choose periodic primitives with zero mean. Set $h_0=f$, and recursively defineEvery $h_r$ has zero mean. Part (i) gives $h_{r+1}^{\prime }=-h_r$ almost everywhere, and $h_{r+1}(0)=h_{r+1}(1)$ because $R{h}_r$ vanishes at both endpoints. Its periodic extension is consequently locally an absolutely continuous function, with no jump at the integers. Part (ii) gives the uniform estimateThe weak derivative identity on the whole line is $D{\overline{h}}_{r+1}=-{\overline{h}}_r$. After scaling by the integer $j$ and repeating integration by parts $N$ times,The boundary terms vanish because $\varphi$ has compact support in $(0,1)$. The interval contains exactly $j$ periods, soTaking $N=m+1$ therefore yieldsThis rapid distributional decay of periodic oscillations uses only integrability of $f$ and smoothness of the test function.

Let $M={\sup }_j{j}^{-m}\Vert g_j{\Vert }_{L^1}<\infty$. For each $j$, apply the periodic primitives with zero mean construction of part (iii) to $g_j$, writing $h_{j,0}=g_j$ andThe estimates are independent of $j$: $\Vert h_{j,r}{\Vert }_{L^1}\le \Vert g_j{\Vert }_{L^1}$. With $N=m+1$, the same repeated integration by parts gives, for every test function $\varphi \in \mathcal{D}(0,1)$,This is precisely weak convergence of distributions, so $\boxed{u_j\to 0\text{ in }{\mathcal{D}}^{\prime }(0,1)}$. The extra integration by parts beyond the growth exponent $m$ is what ensures decay.

Let $\Omega \subset {\mathbb{R}}^n$ be an open set and $(U_{\alpha }{)}_{\alpha \in A}$ an open cover of $\Omega$. There exist nonnegative smooth functions $({\chi }_j{)}_{j\in J}$ and choices $\alpha (j)$ such thatand the family of supports is locally finite: every point has a neighborhood meeting only finitely many of them. This is a smooth partition of unity subordinate to the cover, allowing a refinement of its indexing. In Euclidean open sets, one can choose $J$ countable and the individual supports compact in their assigned cover members. Local finiteness ensures that the sum is smooth and that only finitely many terms meet any given compact set.

If $\varphi \in \mathcal{D}(\omega )$, let $\tilde{\varphi }$ be its extension by zero to $\Omega$. Its compact support is contained in $\omega$, so this extension remains a test function. The restriction of a distribution isExtension by zero preserves convergence in the space of test functions, so this defines a distribution on $\omega$.

Choose a partition of unity $({\chi }_j)$ subordinate to the neighborhoods ${\omega }_x$ from the hypothesis. For an arbitrary $\varphi \in \mathcal{D}(\Omega )$, local finiteness and compactness of $\mathrm{supp}\varphi$ give a finite decomposition $\varphi ={\sum }_j{\chi }_j\varphi$. Each term is a test function supported in one ${\omega }_x$. ThusSince this holds for every test function, $v=0$. This is the locality of distributions.

Define two open sets byThe support of a distribution and its singular support areBoth are relatively closed because the removed sets are open. The locality of distributions gives $u{|}_{Z(u)}=0$. On an overlap of two open smoothness regions, the two smooth functions represent the same distribution, so their difference is zero almost everywhere and hence everywhere by continuity. They therefore join to one smooth function on $G(u)$. The same partition of unity argument shows that it represents $u$ there.

For $u=1_{\omega }$, openness of $\omega$ impliesIndeed every neighborhood of a point in this relative closure meets $\omega$ in an open set containing a nonnegative test function with positive integral; outside the closure the indicator function is identically zero near the point.

The singular support of an indicator function needs a measure-sensitive boundary. For a Lebesgue measurable set $A\subseteq \Omega$, define its essential closure byThen the general answer isOutside this intersection, some ball has either no points of $\omega$ or a complement of Lebesgue measure zero. The regular distribution is respectively the constant zero or one on that ball and is smooth. Conversely, at a point of the intersection, every sufficiently small ball has positive Lebesgue measure in both sets. If a smooth function represented the indicator function there, its values would be in $\{0,1\}$ everywhere: any other value, by continuity, would give a ball of positive Lebesgue measure on which equality almost everywhere fails. On a connected ball such a continuous two-valued function is constant, contradicting the two positive measures.

In particular, the answer need not be the full topological boundary. For $\Omega ={\mathbb{R}}^n$ and $\omega ={\mathbb{R}}^n\setminus \{0\}$, the indicator function is the constant one almost everywhere, so its singular support is empty although the boundary is $\{0\}$. The full relative boundary is the answer when every neighborhood of each boundary point contains a positive Lebesgue measure portion of the complement, as for an open set with smooth boundary.

Let $u,v\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$, with at least one a compactly supported distribution, so their convolution is defined. The support and singular support rules for convolution areHere $A+B=\{a+b:a\in A,b\in B\}$ is the Minkowski sum. The compact-factor hypothesis also ensures that these sums of closed sets are closed. More generally, the same rules hold when addition is a proper map on $\mathrm{supp}u\times \mathrm{supp}v$, which is another sufficient condition for defining the convolution. The second inclusion explains why a compactly supported smooth function smooths every distribution under convolution.

Write $z=x+iy$, $r=|z|$, and $L=(4\pi {)}^{-1}\log (x^2+y^2)$. The Wirtinger derivatives satisfy $4{\partial }_{\bar{z}}{\partial }_z=\Delta$. Away from zero, ordinary differentiation givesThis equality also holds for distributional derivatives. To check that no Dirac delta distribution was lost at zero, integrate by parts outside the disk $r\le \epsilon$ against a test function. The new boundary term has absolute value at most $C\epsilon |\log \epsilon |\Vert \varphi {\Vert }_{\infty }\to 0$. Both $L$ and its displayed first derivatives, of size $O(r^{-1})$, are locally integrable functions in two dimensions, so the remaining integrals converge to their full-plane values.

Using the supplied fundamental solution of the Laplace equation identity $\Delta L={\delta }_0$, and commutation of distributional derivatives, now givesThus the Cauchy kernel fundamental solution has exactly the required normalization.

If $F$ is any other fundamental solution, then $H=F-E$ satisfies ${\partial }_{\bar{z}}H=0$ as a distribution. Applying $4{\partial }_z$ gives $\Delta H=0$. Elliptic regularity for the Laplacian implies that the initially arbitrary distribution $H$ is a smooth function. Its equation is then the ordinary Cauchy-Riemann equation, so $H$ is an entire function.

For a direct justification of the needed regularity, choose a radial mollifier ${\eta }_r$. Locally, $H\ast {\eta }_{\epsilon }$ is a smooth harmonic function. On a ball with sufficient room for both radii, the mean value property gives $(H\ast {\eta }_{\epsilon })\ast {\eta }_r=H\ast {\eta }_{\epsilon }$. Passing to the weak convergence of distributions limit $\epsilon \to 0$ yields $H\ast {\eta }_r=H$ on a smaller ball. The left side is smooth, proving the claim locally. This establishes the distributional Cauchy-Riemann lemma without assuming in advance that $H$ is a locally integrable function.

Conversely every entire function $h$ satisfies ${\partial }_{\bar{z}}h=0$ and can be added to $E$. Hence all fundamental solutions areNo growth condition at infinity was imposed, so $h$ need not be a polynomial.

Because $f$ is a compactly supported distribution, the convolution of distributions with a compactly supported factor defines $E\ast f$. The convolution derivative identity and part (i) giveThe difference between any two solutions has zero ${\partial }_{\bar{z}}$ distributional derivative. Part (ii) therefore proves that the general solution of the partial differential equation isThis formula is an identity of distributions; a pointwise integral is not required for arbitrary $f$.

If $u$ were a smooth function, its Wirtinger derivative ${\partial }_{\bar{z}}u=f$ would also be smooth, contradicting $\mathrm{singsupp}f\ne \varnothing$. In fact the singular support is exactly preserved here:One inclusion follows because a differential operator cannot create a singularity where $u$ is smooth. For the reverse inclusion, $E$ is smooth off zero, and its singular support is exactly $\{0\}$: if it were smooth near zero, ${\partial }_{\bar{z}}E={\delta }_0$ would be smooth there. The support and singular support rules for convolution give $\mathrm{singsupp}(E\ast f)\subseteq \{0\}+\mathrm{singsupp}f$, and adding the smooth entire function $h$ leaves this singular support unchanged. Thus there is no globally smooth solution when the given singular support is nonempty.

Interchange the two Wirtinger derivatives in part (i). The same real fundamental solution of the Laplace equation $L$ givesThe calculation is valid for distributions by the same vanishing boundary term estimate. Any other fundamental solution differs by a distribution $H$ with ${\partial }_{z}H=0$. Then $\Delta H=0$, so the regularity argument of part (ii) makes $H$ smooth, and its Cauchy-Riemann equations make it an antiholomorphic function. On the whole plane it can be written $H(z)=h(\bar{z})$ with $h$ an entire function. Conversely every such $H$ has ${\partial }_{z}H=0$. Therefore