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For a weak solution of a uniformly elliptic homogeneous divergence-form elliptic operator, testing with ${\eta }^2(u-c)$ and choosing the gradient of the cutoff function in $B_{2R}\setminus B_R$ gives ${\int }_{B_R}|\nabla u{|}^2\le C{R}^{-2}{\int }_{B_{2R}\setminus B_R}|u-c{|}^2$. The arbitrary constant $c$ can be chosen as an annular average, which makes the estimate suitable for a hole-filling argument.

For a bounded Lipschitz domain, functions can be extended with the displayed bound and common compact support. Reflect across local Lipschitz boundary graphs, combine the reflected functions with a partition of unity, and multiply by a fixed cutoff function. Lipschitz coordinate changes control derivative measures; the product rule bounds cutoff function terms by the $L^1$ norm. Zero extension is also possible with an estimate for BV traces on a hypersurface, but its boundary jump must be included in the variation.

For a smooth function $g$ compactly supported in a disc $D\subset \mathbb{C}$, the Cauchy-Green operator isHere $w=x+iy$ and $dw\wedge d\bar{w}=-2idx\wedge dy$, fixing the sign between the two forms of the integral. The kernel is locally integrable, andIndeed, the distribution $1/(\pi z)$ is a fundamental solution of ${\partial }_{\bar{z}}$, so differentiating its convolution with $g$ gives ${\delta }_0\ast g=g$. Smooth compactly supported input makes the convolution smooth and the identity classical.^[ref]

For local data without compact support, multiply by a smooth cutoff function equal to one on a smaller disc. The resulting operator solves the original equation there. Applying this construction to coefficients in one coordinate at a time gives the local primitives in the Dolbeault-Poincaré lemma. The solution need not itself have compact support.

For a uniformly elliptic operator $Lu=a^{ij}(x)D_{ij}u$ with continuous coefficients and $W⋐U$,Uniform continuity lets one freeze $a^{ij}$ on sufficiently small open balls and use the perturbation of a constant-coefficient elliptic second-derivative estimate. A partition of unity and cutoff functions reduce the global interior estimate to finitely many such balls; derivatives of the cutoffs produce only the displayed lower-order norm.

Every compactly supported distribution on an open set $X$ is a finite sum $u={\sum }_{\alpha }{\partial }^{\alpha }{f}_{\alpha }$ with continuous $f_{\alpha }$ compactly supported inside $X$. One can first write $u={\partial }^{\gamma }F$ for a continuous global convolution primitive, then multiply by a cutoff function equal to one near the support. The identity $\chi {\partial }^{\gamma }F={\sum }_{\beta \le \gamma }(-1{)}^{|\beta |}(\genfrac{}{}{0ex}{}{\gamma }{\beta }){\partial }^{\gamma -\beta }[({\partial }^{\beta }\chi )F]$ localizes every coefficient. Finite global representation can fail for a distribution of unbounded order.

Let a continuous function satisfy both viscosity inequalities for the Laplace operator. On any relatively compact ball, take its harmonic replacement $h$ with matching boundary values. If $u-h$ has a positive maximum, the function $h+\epsilon (R^2-|x-a{|}^2)$ still lies below $u$ somewhere for sufficiently small positive $\epsilon$, while matching it on the boundary. At an interior maximum of their difference it is a valid upper test, but its Laplacian is $-2n\epsilon <0$, a contradiction. The opposite perturbation excludes a negative minimum. Thus $u=h$ throughout the ball and is a smooth harmonic function. A locally defined test is extended near its touching point using a smooth cutoff function when global tests are required.

Every distribution on an open set $U$ is a limit of test functions in weak convergence of distributions. On the whole space ${\mathbb{R}}^n$, choose a mollifier $\rho$ and compactly supported smooth cutoffs ${\chi }_j$ equal to one on expanding balls. Thenare test functions and $u_j\to u$: for each fixed test function $\varphi$, the cutoff is eventually one near its support, and the reflected mollifier approximates $\varphi$ with every derivative in one common compact set.

For a general open set $U$, ordinary fixed-radius convolution need not be defined near its boundary. Choose an exhaustion $K_j$ of $U$ and cutoff functions ${\chi }_j$ compactly supported inside $U$, equal to one near $K_j$. Extend the compactly supported distribution ${\chi }_{j}u$ by zero to the whole space, then convolve with a mollifier of radius ${ϵ}_j\to 0$ small enough that the resulting support stays inside $U$. For every fixed test, the cutoff eventually has no effect, so these smooth compactly supported approximants converge to $u$.

Suppose a degree-$N$ polynomial obeys $|{\partial }^{\alpha }Q(\xi )|\le C_{\alpha }|\xi {|}^{-\delta |\alpha |}|Q(\xi )|$ at large real frequency, with $\delta >0$. A nonzero constant derivative of total order $N$ gives $|Q|\gtrsim ⟨\xi {⟩}^{\delta N}$. Reciprocal differentiation then bounds ${\partial }^{\alpha }(1/Q)$ by $⟨\xi {⟩}^{-\delta N-\delta |\alpha |}$. The high-frequency reciprocal parametrix kernel therefore gains $\delta N$ Sobolev derivatives. Localize the distribution by a cutoff function equal to one near the target; the commutator lies outside that target and the reciprocal kernel makes its contribution smooth. This proves the displayed gain without demanding global regularity of the original distribution.

For $\Delta v=\mathrm{div}F+G$ and $U\subset \subset V$, a cutoff function energy estimate gives $\Vert v{\Vert }_{H^1(U)}\le C(\Vert v{\Vert }_{L^2(V)}+\Vert F{\Vert }_{L^2(V)}+\Vert G{\Vert }_{L^2(V)})$. One derivative of forcing is permitted because integration by parts places it on the test function. This estimate allows a mollified distributional solution initially in $L^2$ to gain its first weak derivative.

For a Lebesgue measurable set $A$ in an open set $\Omega \subseteq {\mathbb{R}}^n$, its essential closure consists of the points whose every sufficiently small ball has positive Lebesgue measure intersection with $A$. Equivalently,where the right side is the support of a distribution. To see the equivalence, a ball on which $A$ is null gives zero pairings with all test functions. If $A$ has positive measure in a ball, cover that ball by smaller compactly contained balls and choose a nonnegative cutoff function positive on one meeting $A$ in positive measure; its pairing is positive. Thus essential closure is relatively closed and unchanged by modifying $A$ on a Lebesgue null set. For an open set $A$ it equals its relative closure.

Target: Singular support of an indicator function. Parent: Singular support. Intended file: distribution-theory.bigb.

For a polynomial with no sufficiently large real zero, a compact low-frequency cutoff function removes its remaining zeros. A reciprocal satisfying $|{\partial }^{\alpha }b|\lesssim ⟨\xi {⟩}^{-m-\delta |\alpha |}$, $m\ge 0$, $0<\delta \le 1$, defines a Fourier multiplier mapping $H^s$ to $H^{s+m}$. Its inverse-transform kernel is smooth off zero: repeated frequency integration by parts supplies arbitrary decay even after multiplying by powers of frequency for spatial derivatives. Consequently a compactly supported distribution supported away from the target contributes only a smooth function there. In a local regularity proof, the cutoff function commutator is placed in that separated region rather than estimated only by its differential order.

In one dimension, $xw=0$ if and only if $w=c{\delta }_0$. Choose a cutoff function $\eta =1$ near zero and write every test function as $\phi =\phi (0)\eta +x\psi$. The pairing with $x\psi$ vanishes, leaving $⟨w,\phi ⟩=\phi (0)⟨w,\eta ⟩$.

For zero initial displacement and unit delta initial velocity, the spatial Fourier transform of the wave equation gives the displayed multiplier. A compact cutoff function around zero frequency produces an ordinary smooth function after inversion. The remaining sine splits into two oscillatory integrals with phases $x\cdot \xi \pm ct|\xi |$ and symbols proportional to $(1-\chi )/|\xi |$ of order $-1$. Their stationary directions lie on $|x|=c|t|$, proving a light-cone bound for singular support. This does not claim vanishing of smooth tails inside the cone.

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.

A distribution $u$ is a $2\pi$ periodic distribution whenIteration gives invariance under all integer multiples of $2\pi$. For a tempered distribution the identity also holds for every Schwartz function, since test functions are dense in the Schwartz space: multiplying by cutoff functions $\chi (x/R)$ equal to one near zero gives such an approximation in every Schwartz seminorm.

For $\varphi \in \mathcal{S}(\mathbb{R})$ define $a_k(x)=\Psi (x+2\pi k)\varphi (x)$. The smooth periodic partition of unity gives $\varphi ={\sum }_k{a}_k$ with convergence in the Schwartz space. Indeed, each $a_k$ is supported where $|x+2\pi k|\le 2\pi$, and the generalized Leibniz rule and arbitrarily rapid decay of the derivatives of $\varphi$ give, for every $m,n,M$,Consequently continuity and periodicity of the tempered distribution giveFor the last interchange, all summands have the fixed compact support of $\Psi$ and their derivatives decrease faster than any power of $|k|$. Hence the series converges in the space of test functions, as well as in the Schwartz space. Notice that $P\varphi$ itself need not be a Schwartz function; the factor $\Psi$ makes the pairing legitimate.

Original solutions to all three questions of the functional analytic methods for partial differential equations course of the University of Oxford Part C examination, held on 2 June 2023. Source: released parent PDF, bundle pages 339–342; paper pages 1–4. The native PDF confirms three substantive printing issues: the cutoff function bound in 2(b)(ii), the missing boundary restriction in 3(a)(ii), and the Sobolev space used for the asserted convergence in 3(b)(iv). Each is addressed explicitly at the relevant leaf.

First repeat the Caccioppoli inequality calculation on two concentric balls. If $h$ is a harmonic function on $B_R$, choose a cutoff function equal to one on $B_r$, supported in $B_R$, and with gradient bounded by $C/(R-r)$. The argument of (ii), with $A=I$ and $k=0$, givesHere $C_{\ast }$ depends at most on the dimension. When $h$ is only known to be smooth inside $B_R$ and in $L^2(B_R)$, the weak formulation and all integrals involving the cutoff function are still legitimate, since its support is strictly inside $B_R$.

Every classical partial derivative of the given smooth harmonic function $u$ is again a harmonic function, since constant-coefficient partial derivatives commute with the Laplacian. Choose indices $i_1,\dots ,i_m$ representing the multi-index $\alpha$, and defineFor each $j\ge 1$, the preceding Caccioppoli inequality yieldsStarting with $u\in L^2(B_1)$, these estimates also establish the integrability needed at each subsequent step. Iterating $m$ times givesThis is the desired interior derivative estimate for a harmonic function, with a constant depending only on $n$ and $m$, uniformly over every multi-index of size $m$.

There is a small incompatibility in the printed cutoff function assumptions. No $\phi \in C_c^{\infty }(B_1)$ can equal one on $B_{1/2}$ and satisfy $|\nabla \phi |\le 2$: for a unit vector $e$, choose $t<1$ beyond its support. Continuity gives $\phi (e/2)=1$, while the fundamental theorem of calculus would give $1\le 2(t-1/2)<1$. Replace the number $2$ by a fixed sufficiently large absolute constant $C_0$; such a cutoff function exists, and the requested Caccioppoli inequality is unchanged apart from its constant. We keep $M=\Vert \nabla \phi {\Vert }_{\infty }$ in the calculation, which also yields the formal bound under the printed assumptions.

The weak formulation is ${\int }_{B_1}A\nabla u\cdot \nabla \zeta =0$ for test functions $\zeta$. Since $A$ is bounded, this extends by density of smooth functions in a Sobolev space to every $\zeta \in H_0^1(B_1)$. Part (a)(ii), followed by mollification inside the compact support, makes ${\phi }^2(u-k)$ an admissible choice. The product rule givesPut $X=\int {\phi }^2|\nabla u{|}^2$ and $Y=\int (u-k{)}^2|\nabla \phi {|}^2$. The lower and upper uniform ellipticity bounds, together with Cauchy-Schwarz inequality, implyIf $X=0$ the conclusion is immediate. Otherwise divide by $X^{1/2}$ and square to obtain $X\le 4{\Lambda }^2{\lambda }^{-2}Y$. Since $\phi =1$ on $B_{1/2}$,For a fixed valid cutoff function, $M\le C_0$, this has the required constant independent of $k$. Formally inserting the printed value $M=2$ gives $16{\Lambda }^2/{\lambda }^2$. No derivative of $A$ was taken, so differentiability of the coefficients is unnecessary.

Set $w=(u-k{)}^{+}$, the positive part of $u-k$. The allowed Sobolev chain rule givesUse the admissible zero-boundary Sobolev space function ${\phi }^{2}w$ in the weak formulation, with a valid cutoff function as explained in (ii). Its gradient is ${\phi }^2\nabla w+2\phi w\nabla \phi$, soIndeed the integrands vanish off $\{u>k\}$, and on that set $\nabla w=\nabla u$. The uniform ellipticity and Cauchy-Schwarz inequality estimates from (ii) now apply to $w$ and giveFixing the cutoff function gives the required constant independent of $k$. This is a superlevel Caccioppoli inequality; it uses the positive part as a test and does not require the truncated function itself to solve the original equation.

A hyperbolic Riemann surface in potential theory is a noncompact Riemann surface admitting a positive Green function on a Riemann surface. With pole $p$, this is a positive function harmonic off $p$, having local form $-\log |\zeta |$ plus a harmonic function in a coordinate $\zeta (p)=0$, and obtained as the minimal such positive function by exhaustion. This is a noncircular definition for the present proof. For a simply connected surface it is equivalent to disk conformal type; for arbitrary surfaces potential-theoretic hyperbolicity and disk universal-cover type should be distinguished.

The unit disk is an example. At pole zero its function is $G(z,0)=-\log |z|$: it is positive in the punctured disk, harmonic there, has the correct logarithmic singularity, and tends to zero at the boundary. If another positive function has the same singularity and is harmonic elsewhere, subtract $G$ and apply the maximum principle for harmonic functions on disks of radius $r<1$; the boundary lower bound is $\log r$. Letting $r\uparrow 1$ shows that the other function dominates $G$. Thus $G$ is indeed the positive Green function on a Riemann surface. At a general pole it is

Here is a Green-function exhaustion proof of disk uniformization. The main analytic idea is to build conformal maps on bordered disks from their Green functions and pass to an extremal limit. The positive global Green function prevents that limit from becoming constant.

First exhaust the noncompact simply connected surface $R$ by relatively compact domains $R_n$ containing $p$, each with a smooth Jordan boundary. This topological step does not use conformal uniformization: take regular neighborhoods of finitely many coordinate disks covering successive compact sets and fill bounded complementary components. Simple connectedness and the Jordan curve theorem make the filled neighborhoods topological disks; enlarge them to be nested.

Each $R_n$ has a zero-boundary Green function on a Riemann surface $G_n(\cdot ,p)$. One analytic construction, independent of conformal uniformization, is as follows. Equip the compact bordered domain with a smooth conformal metric. Multiply $-\log |\zeta |$ by a smooth cutoff function equal to one near $p$ and zero near the boundary; call the result $s$. Distributionallywhere $k$ is smooth and supported away from $p$. Solve $\Delta v=k$ with zero boundary data. This auxiliary Dirichlet problem can be obtained by minimizing $\frac{1}{2}\int |\nabla v{|}^2+\int kv$ over functions with zero boundary trace: the Poincaré inequality gives coercivity, minimization in a Hilbert space gives a weak solution, and interior and smooth-boundary elliptic regularity give a smooth solution. Then $G_n=s+v$ is harmonic away from $p$, has the logarithmic pole, and vanishes at the boundary. It is positive by the maximum principle for harmonic functions, applied after removing a sufficiently small circle around its positive pole. This construction uses ordinary linear elliptic existence, not the desired uniformization theorem. Equivalently, the same bordered-domain Green functions can be constructed by the Perron method for the Dirichlet problem using local logarithmic barriers.

On $R_n\setminus \{p\}$ choose local harmonic conjugates $G_n^{\ast }$. Their period around the pole is $-2\pi$, and every closed loop is homologous to an integer multiple of that loop. Thereforeis a single-valued holomorphic function. It extends across $p$ with one simple zero and has no others. Since $G_n>0$, $|f_n|<1$; since $G_n=0$ on the boundary, $|f_n|\to 1$ there. Thus $f_n:R_n\to \mathbb{D}$ is proper. For any $a\in \mathbb{D}$, choose a smooth interior contour sufficiently close to the boundary that $|f_n|>|a|$ everywhere on it. The contour surrounds $p$ and all possible preimages of $a$, because $|f_n|>|a|$ throughout a thin boundary collar. By Rouché's theorem, $f_n-a$ has the same number of zeros inside as $f_n$, namely one. Hence $f_n$ is onto and one-to-one, with nonzero derivative: it is a biholomorphism from $R_n$ to the disk.

Write near $p$Multiply $f_n$ by a unit constant so that, in the fixed coordinate, $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity gives $G_n\le G_{n+1}$: their difference is harmonic even at $p$ and nonnegative on $\partial R_n$. The global positive Green function on a Riemann surface $G_R$ similarly majorizes every $G_n$. Comparing the regular parts shows that $c_n$ increases and is bounded above by the finite regular part of $G_R$ at $p$. Consequently

On each fixed compact subdomain, the $f_n$ are eventually bounded holomorphic functions. Montel theorem and a diagonal subsequence therefore give a locally uniform convergence limit $F:R\to \mathbb{D}$ with $F(p)=0$ and $F^{\prime }(p)=d>0$. The maximum modulus principle keeps its values strictly inside the disk. It is injective: for fixed $q$, the functions $f_n-f_n(q)$ have no zeros away from $q$; Hurwitz's theorem implies that their nonconstant limit has no other zeros. This is the locally uniform limit of univalent functions argument.

There is also an extremality property. For any holomorphic function $g:R\to \mathbb{D}$ with $g(p)=0$, the function $g\circ f_n^{-1}$ is a disk self-map fixing zero. The Schwarz lemma gives $|g^{\prime }(p)|\le f_n^{\prime }(p)$. Passing to the limit yields $|g^{\prime }(p)|\le d=F^{\prime }(p)$.

Finally suppose $F$ omits $a\in \mathbb{D}$. Necessarily $a\ne 0$. Apply the disk Möbius transformation ${\varphi }_a(w)=(w-a)/(1-\overline{a}w)$. The nowhere-zero function ${\varphi }_a\circ F$ has a global holomorphic logarithm because $R$ is simply connected, hence a holomorphic square root $h$. It maps into the disk and $|h(p)|=\sqrt{|a|}$. Normalize it by setting $g={\varphi }_{h(p)}\circ h$, so $g(p)=0$. Direct differentiation givesThe strict inequality follows from $(1-\sqrt{|a|}{)}^2>0$. This contradicts extremality. Therefore $F$ omits no point, and we conclude