# Support distances and admissible convolution domains *Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026. Public domain (CC0).* A convolution equation has a solution domain and an equation domain. Their compatibility says that every translated kernel support fits. A stronger support condition says that tests cannot escape toward the equation boundary while their transpose convolutions stay in a fixed compact set. We prove that this condition is exactly preservation of the distance from the two domain complements. Basic references are [Grubb's Fourier-distribution notes](https://web.math.ku.dk/~grubb/dist5.pdf), [Melrose's tempered-distribution notes](https://math.mit.edu/~rbm/iml/Chapter1.pdf), and Hörmander's *The Analysis of Linear Partial Differential Operators*, volumes I and II. The ordinary compact convolution-support identity is Theorem 3.1 of Convex supports and convolution cancellation. Read Smooth forcing, invertible kernels and support confinement for the necessary support condition supplied by smooth solvability. The complete argument below proves the geometric equivalence and its domain consequences. ## 1. The distance criterion Fix a nonzero compact distribution \(\mu\), let \(S=\operatorname{supp}\mu\), and assume \[ X_2-S\subset X_1 \tag{L1} \] for nonempty open \(X_1,X_2\subset\mathbb R^n\), \(n\ge1\). Here \(A-B=\{a-b:a\in A,\ b\in B\}\). Reflection is \(\check\mu(\phi)=\mu(\phi(-\,\cdot))\). For compact \(v\) supported in \(X_2\), write \[ T=\operatorname{supp}v,\qquad W=\operatorname{supp}(\check\mu*v). \tag{L2} \] We use distance to the full complement, and set an infimum over an empty set equal to \(+\infty\). The pair is called \(\mu\)-convex for supports when, for every compact \(K_1\subset X_1\), all compact smooth tests whose image support lies in \(K_1\) have their original supports in one compact \(K_2\subset X_2\). The exact equivalent criterion is \[ d(T,\mathbb R^n\setminus X_2) =d(W,\mathbb R^n\setminus X_1) \quad(v\in\mathcal E'(X_2)). \tag{L3} \] Here \(\mathcal E'(X_2)\) means compact distributions supported inside \(X_2\). Smooth tests suffice to check (L3); the proof passes to all compact distributions by mollification. It also extends the compact confinement to distributional tests. Compatibility alone always gives the inequality from right to left: \[ d(W,\mathbb R^n\setminus X_1) \ge d(T,\mathbb R^n\setminus X_2). \tag{L4} \] To prove equality from confinement, suppose the image has a larger margin. Translate the original test toward a nearest equation-boundary point. Before it reaches that point, every test stays in \(X_2\), while every image stays in one compact subset of \(X_1\). Uniform confinement would put the limiting boundary point inside a compact subset of \(X_2\), which is impossible. For the reverse implication, the ordinary support theorem gives \[ \operatorname{conv}W =\operatorname{conv}T-\operatorname{conv}S . \tag{L5} \] If \(W\subset K_1\), then \(T\subset\operatorname{conv}K_1+s_0\) for any \(s_0\in S\). This gives a fixed bounded set. The equal-distance condition gives a fixed positive margin inside \(X_2\). Together they give the required compact \(K_2\). ## 2. The maximal equation domain The largest domain where a smooth function on \(X_1\) can be convolved with this kernel is \[ Y=\{x:x-S\subset X_1\}. \tag{L6} \] It is open because the translated compact support has a positive margin inside \(X_1\). Every compatible equation domain is contained in \(Y\). If \(X_1\) is convex, then \(Y\) is convex. When \(Y\) is nonempty, \((X_1,Y)\) is \(\mu\)-convex for supports. For arbitrary \(X_1\), every \(\mu\)-convex equation domain must be a union of entire connected components of \(Y\). A sequence of points in the equation domain cannot tend to an excluded point inside \(Y\): their translated kernel supports still have a uniform margin in \(X_1\), and (L3), applied to point masses, transfers that margin back to the equation domain. Consequently, when \(X_1\) is convex and the equation domain is nonempty, the only \(\mu\)-convex choice is \(X_2=Y\). These are support conditions. They do not by themselves assert invertibility of the kernel or full solvability on an arbitrary pair. ## 3. Four worked examples ### Example 1: an asymmetric two-point kernel on an interval Take \[ \mu=\delta_{-1/2}-2\delta_{5/4},\qquad X_1=(-3,4). \tag{L7} \] The maximal equation domain is the intersection of two translates: \[ Y=(X_1-1/2)\cap(X_1+5/4)=(-7/4,7/2). \tag{L8} \] For a nonnegative smooth \(v\) supported exactly in \(T=[-1/2,3/4]\), its two transpose copies are disjoint. Hence \[ W=(T+1/2)\cup(T-5/4) =[0,5/4]\cup[-7/4,-1/2]. \tag{L9} \] The coefficient \(-2\) changes the values on the second interval, but does not remove any part of its support. The boundary distances are \[ d(T,\mathbb R\setminus Y)=\min(5/4,11/4)=5/4, \quad d(W,\mathbb R\setminus X_1)=\min(5/4,11/4)=5/4. \tag{L10} \] For image supports inside \(K_1=[-2,3/2]\), the convex-domain proof gives the compact confinement set \[ K_2=\{x:x-[-1/2,5/4]\subset[-2,3/2]\} =[-3/4,1]\subset Y. \tag{L11} \] Indeed the leftmost translated point \(x-5/4\) must be at least \(-2\), and the rightmost point \(x+1/2\) must be at most \(3/2\). This applies to every compact distribution whose transpose image stays inside \(K_1\), including distributions with cancellations. ![Exact equation and solution intervals, two reflected support copies, equal boundary margins and a compact confinement interval](figures/asymmetric-kernel-support-margins.png) The rows show the sets in (L7)–(L11), using one real coordinate \(x\). Open circles mark domain endpoints; filled circles mark compact-support endpoints. The marked left margins are both \(5/4\). Vertical row positions are labels, not a second spatial coordinate. ### Example 2: cancellation and regularization Let \(\mu=\delta_0-\delta_1\), \(X_1=(-2,3)\), and \(Y=(-1,3)\). Take the compact distribution \(v=\delta_0+\delta_1\). Then \[ \check\mu*v=(\delta_0-\delta_{-1})*(\delta_0+\delta_1) =\delta_1-\delta_{-1}. \tag{L12} \] The coefficient at zero cancels. Thus \(T=\{0,1\}\), \(W=\{-1,1\}\), while \(T-S=\{-1,0,1\}\). Convolution support need not equal the full difference set. Its convex hull is exactly the difference of the convex hulls, as in (L5). Both distances equal one. Choose a positive smooth mollifier \(\rho\) supported exactly in \([-1,1]\), with integral one, and \(0<\varepsilon<1/2\). The regularized supports are \[ \begin{aligned} T_\varepsilon&=[-\varepsilon,\varepsilon] \cup[1-\varepsilon,1+\varepsilon],\\ W_\varepsilon&=[-1-\varepsilon,-1+\varepsilon] \cup[1-\varepsilon,1+\varepsilon]. \end{aligned} \tag{L13} \] The image remains the difference of two disjoint mollifier copies, so the middle cancellation persists. The two domain distances are both \(1-\varepsilon\), converging to one. This is an example where the supports can be computed exactly. The general mollification lemma only uses outer support inclusion and convergence on tests; it does not assume equality with an outer parallel set. ![A cancelled middle atom, the actual convolution support and the two disjoint regularized support sets](figures/cancellation-and-mollified-support-distances.png) The left panel records the atoms and the cancelled zero coefficient. The right panel uses \(\varepsilon=1/8\), with closed support intervals in (L13). Their domain margins are \(7/8\). Dashed lines join convex hulls; they do not assert that the gaps belong to the support. ### Example 3: selecting an entire component Take \(\mu=\delta_{2/3}\) and the disconnected solution domain \[ X_1=(-4,-2)\cup(0,3),\qquad Y=X_1+2/3=(-10/3,-4/3)\cup(2/3,11/3). \tag{L14} \] Select the whole right component \(X_2=(2/3,11/3)\). This pair is \(\mu\)-convex for supports. The transpose shifts every support by \(-2/3\). For each compact \(K_1\subset X_1\), take \[ K_2=(K_1\cap[0,3])+2/3. \tag{L15} \] The intersection is compact and lies inside \((0,3)\), since \(K_1\) contains no domain endpoint. Thus \(K_2\) is compact in \(X_2\), and contains every original support whose translated image lies in \(K_1\). Cutting this component down to \(X'_2=(1,3)\) breaks the condition. With \(v=\delta_2\), the image support is \(\{4/3\}\), and \[ d(\{2\},\mathbb R\setminus X'_2)=1,\qquad d(\{4/3\},\mathbb R\setminus X_1)=4/3. \tag{L16} \] Compatibility still holds, but the unequal margins disprove support convexity. The general component condition is necessary; this point-mass example proves sufficiency for its particular selected component. ### Example 4: translating a test to an excluded boundary Let \(\mu=\delta_0\), \(X_1=(-2,2)\), \(X_2=(-1,1)\), and choose a positive smooth test supported exactly in \(T=[-1/4,1/2]\). Its image support is the same set. The two distances are \(t=1/2\) and \(w=3/2\). For \(0\le s<1/2\), its translated support is \[ T+s=[-1/4+s,1/2+s]\subset X_2. \tag{L17} \] Every image support lies in the single compact \(K_1=[-1/4,1]\subset X_1\). If one compact \(K_2\subset X_2\) contained all the original supports, closedness would put the limit of their right endpoints, \(1\), in \(K_2\). But \(1\notin X_2\). This is the translation contradiction from the full proof. The identity kernel is invertible, and every compact smooth forcing on \(X_2\) extends by zero to \(X_1\). The failed support condition concerns arbitrary smooth forcing, as shown by the explicit boundary obstruction in Smooth forcing, invertible kernels and support confinement. ## 4. Exercises and full solutions The ten exercises total 100 points. The solutions include every required argument. **Exercise 1 (10 points; introductory).** Let \(X_1=(-5/2,9/2)\) and let a nonzero kernel have support \(S=\{-3/4,3/2\}\). Determine \(Y\). For \(T=[0,1]\), compute the distance of \(T\) from the complement of \(Y\), and the distance of \(T-S\) from the complement of \(X_1\). *Solution.* Intersect the two translates: \[ Y=(X_1-3/4)\cap(X_1+3/2) =(-13/4,15/4)\cap(-1,6)=(-1,15/4). \tag{S1} \] The distance of \(T\) is \(\min(1,15/4-1)=1\). Its two reflected translates are \([3/4,7/4]\) and \([-3/2,-1/2]\), with full difference-set extrema \(-3/2,7/4\). Their distance from the complement of \(X_1\) is \(\min(-3/2+5/2,9/2-7/4)=1\). This computes the outer difference set; if an actual convolution has overlapping copies and cancellations, its support must be analyzed or controlled by the ordinary-support theorem. Here the chosen intervals are disjoint, so nonzero point coefficients and a test with full interval support give exactly these two support intervals. **Exercise 2 (8 points; intermediate).** Take \(\mu=\delta_0+i\delta_2\), \(v=\delta_0+i\delta_2\), \(X_1=(-3,5)\). Compute the transpose convolution, the maximal domain and both support distances. *Solution.* Reflection does not conjugate the coefficient: \(\check\mu=\delta_0+i\delta_{-2}\). Therefore \[ \check\mu*v=(1+i^2)\delta_0+i\delta_{-2}+i\delta_2 =i\delta_{-2}+i\delta_2. \tag{S2} \] The maximal domain is \(Y=X_1\cap(X_1+2)=(-1,5)\). The original support is \(\{0,2\}\), at distance \(\min(1,3)=1\) from its complement. The image support is \(\{-2,2\}\), at distance \(\min(1,3)=1\) from the complement of \(X_1\). Conjugating the reflected coefficient would produce a different coefficient at zero and would violate the bilinear convention used throughout. **Exercise 3 (10 points; intermediate).** Let \(X_1=(-4,5)\times(-3,6)\) and \(\mu=\delta_{(0,0)}-3\delta_{(1,0)}+i\delta_{(0,2)}\). Determine \(Y\), and verify the distance equality for \(v=\delta_{(0,1)}\). *Solution.* The conditions \(x\), \(x-(1,0)\), and \(x-(0,2)\) in \(X_1\) give \[ Y=(-3,5)\times(-1,6). \tag{S3} \] The original point \((0,1)\) has distance \(\min(3,5,2,5)=2\) from the rectangle complement. Its transpose image is \(\delta_{(0,1)}-3\delta_{(-1,1)}+i\delta_{(0,-1)}\); the three locations are distinct, so all remain in the support. The coordinate extrema are \(x=-1,0\), \(y=-1,1\), giving distance \(\min(3,5,2,5)=2\) from the complement of \(X_1\). For a set in an open rectangle, its distance to the complement is the least of its coordinate gaps to the four sides: the corresponding perpendicular segment attains the least gap, and every point outside has at least one coordinate beyond a side. This proves the distance formula used here. **Exercise 4 (10 points; advanced).** Reconstruct the translation argument when \(00\), while the lower limit is at least \(d\). They agree. This argument retains possible cancellations and holes in the actual support. **Exercise 7 (10 points; advanced).** Prove that a \(\mu\)-convex equation domain is relatively closed in \(Y\), including the case \(X_1=\mathbb R^n\), and conclude that it is a union of components. *Solution.* Let \(x_j\in X_2\) tend to \(x\in Y\). The compact set \(x-S\) lies inside \(X_1\) with a positive margin \(\delta\); if the complement of \(X_1\) is empty, take \(\delta=1\). For sufficiently large \(j\), the translated support \(x_j-S\) has margin at least \(\delta/2\). Apply the distance identity to \(\delta_{x_j}\), whose image is the translate of \(\check\mu\) with exactly that support. It gives \(d(\{x_j\},\mathbb R^n\setminus X_2)\ge\delta/2\). If \(x\) were excluded from \(X_2\), this distance would be at most \(|x_j-x|\), tending to zero. Thus \(x\in X_2\). The equation domain is already relatively open in \(Y\), because it is open in the real space. Its intersection with each connected component of \(Y\) is therefore both open and closed in that component. Connectedness makes this intersection empty or the entire component. **Exercise 8 (8 points; intermediate).** Explain why \(Y\) is open even though it is written as an intersection of possibly infinitely many translates. If \(X_1\) is convex and \(Y\ne\varnothing\), prove uniqueness of a nonempty \(\mu\)-convex equation domain. *Solution.* For \(x\in Y\), the entire compact set \(x-S\) lies in the open \(X_1\). It has one positive distance from the complement. Any sufficiently small perturbation of \(x\) translates all those support points by less than that distance, so the perturbation remains in \(Y\). This proves openness; an arbitrary infinite intersection of open sets would not suffice. Each translate \(X_1+s\) is convex, so their intersection \(Y\) is convex and, when nonempty, connected by its line segments. The full convex-domain proof gives \((X_1,Y)\) as a \(\mu\)-convex pair. Every other nonempty such equation domain must be a union of components of \(Y\); since there is only one component, it must equal \(Y\). **Exercise 9 (8 points; introductory).** For \(\mu=\delta_0-\delta_3\) and \(X_1=(-1,1)\), determine \(Y\). What happens to the distance equality for the zero test, and why does this not produce a nonempty admissible equation domain? *Solution.* One needs both \(x\in(-1,1)\) and \(x-3\in(-1,1)\), so \[ Y=(-1,1)\cap(2,4)=\varnothing. \tag{S5} \] Compatibility requires \(X_2\subset Y\); hence there is no nonempty compatible equation domain. A test on the empty domain can only be zero. Its original and image supports are empty, giving two infinite distances by the infimum convention. This vacuous identity does not meet the definition's requirement of two nonempty open domains. **Exercise 10 (18 points; advanced).** Classify the zero-kernel case for both compact confinement and the distance identity. Then, for the cancellation example with a symmetric probability mollifier, compute the zeroth, first and third moments of its image after mollification. *Solution.* If \(\mu=0\), every transpose image is zero. For \(K_1=\varnothing\), confinement would require one compact \(K_2\subset X_2\) containing every compact smooth test support. A bump at each point forces \(X_2\subset K_2\), hence \(X_2=K_2\) compact. In positive dimension a nonempty open subset of the real space cannot be compact: compactness makes it closed, and connectedness would make a nonempty clopen set the whole, unbounded space. Thus confinement always fails. Every image distance is infinite. If \(X_2=\mathbb R^n\), every original distance is infinite too, so the identity holds. If the domain is proper and nonempty, a point mass inside it has a finite distance from its nonempty closed complement, and the identity fails. This explains the nonzero-kernel hypothesis in the standalone equivalence. The zero kernel also cannot solve nonzero forcing. For Example 2, the image density is \(\rho_\varepsilon(x-1)-\rho_\varepsilon(x+1)\). Its total mass is \(1-1=0\). Symmetry gives the first centered mollifier moment zero, so the first image moment is \(1-(-1)=2\). Let \(m_2=\int t^2\rho(t)\,dt\). The third moments of the two translated mollifiers are respectively \(1+3\varepsilon^2m_2\) and \(-1-3\varepsilon^2m_2\), by expanding \((\pm1+\varepsilon t)^3\) and using the vanishing odd moments. Their difference is \[ \int x^3[\rho_\varepsilon(x-1)-\rho_\varepsilon(x+1)]\,dx =2+6\varepsilon^2m_2. \tag{S6} \] These moments agree with the limiting atoms as \(\varepsilon\to0\). They do not replace the support-distance proof. ## References - Gerd Grubb, *Distributions and Operators*, Chapter 5, “Fourier transformation of distributions,” freely readable [lecture notes](https://web.math.ku.dk/~grubb/dist5.pdf). - Richard B. Melrose, *Introduction to Microlocal Analysis*, MIT, 2007, Chapter 1, “Tempered distributions and the Fourier transform,” freely readable [notes](https://math.mit.edu/~rbm/iml/Chapter1.pdf). - Lars Hörmander, *The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis*, second edition, Springer, 1990. - Lars Hörmander, *The Analysis of Linear Partial Differential Operators II: Differential Operators with Constant Coefficients*, Springer, 1983. - The complete proof below supplies the distance equivalence, actual regularized-support convergence, maximal convex-domain pair, component necessity and degenerate zero-kernel classification. The ordinary-support identity is reused through the exact internal proof linked in the introduction.