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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.

A distribution which vanishes on a neighborhood of each point vanishes globally. To prove this, choose a smooth partition of unity subordinate to the neighborhoods. Every test function is a finite sum of its products with the partition members, since its support is compact. The distribution vanishes on each summand. Applying the result to a difference proves that two distributions agreeing locally agree globally. This justifies defining the support of a distribution by removing the union of all open regions on which it is zero.

Target: Periodic primitive with zero mean. Parent: Periodic extension. Intended file: function.bigb.

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.

Use complex-linear distributions, so the pairing contains no complex conjugation. The space of smooth functions is $\mathcal{E}(X)=C^{\infty }(X)$, with seminormsThus $f_j\to f$ means uniform convergence of every derivative on every compact subset of $X$. A cofinal compact exhaustion with $K_j\subset \mathrm{int}{K}_{j+1}$ makes $p_j=p_{K_j,j}$ an increasing defining family. Such an exhaustion exists for every open set; for example, use bounded sets staying a positive distance from its complement and enlarge them slightly. This is a Fréchet space.

The compactly supported distribution space is the continuous dual space ${\mathcal{E}}^{\prime }(X)$. We use its weak-star topology: $u_j\to u$ precisely when $⟨u_j,f⟩\to ⟨u,f⟩$ for every $f\in \mathcal{E}(X)$. A stronger standard choice is the strong dual topology, which requires uniform convergence on bounded subsets of $\mathcal{E}(X)$; specifying the weak convention avoids conflating these definitions.

A continuous linear functional necessarily takes null sequences to zero. Conversely, suppose a linear functional $u$ fails to be continuous. For each $j$, it is unbounded on $\{f:p_j(f)\le 1\}$; otherwise scaling would give a continuity estimate. We can therefore choose $f_j$ withFor each fixed $l$, $p_l(f_j)\le p_j(f_j)\to 0$ once $j\ge l$. This contradicts the assumed null-sequence property. Hence sequential continuity characterizes the continuous dual:This is the sequential continuity criterion in a metrizable vector space applied to a linear functional.

Continuity also yields one compact $K⋐X$, an integer $m$ and a constant $C$ such that $|u(f)|\le C{p}_{K,m}(f)$. Thus $u$ has compact support in $K$ and finite order of a distribution. Conversely, a compactly supported distribution extends to $\mathcal{E}(X)$ by $u(f)=u(\eta f)$, where a cutoff function $\eta$ equals one near its support of a distribution. The local finite-order estimate makes this extension continuous and independent of $\eta$. This identifies the dual definition with compactly supported distributions.

Here is an explicit compact continuous-derivative representation of a distribution. Extend $u$ by zero to ${\mathbb{R}}^n$ using a cutoff function inside $X$, retaining finite order $m$. Put $r=m+2$ andThis locally $C^m$ function satisfies ${\partial }^{\gamma }E={\delta }_0$. Indeed, each one-dimensional factor has $r$th distributional derivative equal to the Dirac delta distribution. The convolution $F=u\ast E$ is continuous: the finite-order estimate extends $u$ to $C^m$ functions near its compact support, and translated $E$ varies continuously in their $C^m$ norms. Equivalently, mollify $E$ and use the estimate to obtain local uniform convergence of the convolved functions. Distributional differentiation gives ${\partial }^{\gamma }F=u$.

Choose $\chi \in C_c^{\infty }(X)$ equal to one near $\mathrm{supp}u$. Since $\chi u=u$, repeated Leibniz rule givesEvery coefficient function on the right is continuous and compactly supported in $X$. Thus the required representation is finite:The equality first holds on test functions and then on all smooth functions after inserting a common cutoff, so it holds in ${\mathcal{E}}^{\prime }(X)$.

A general distribution need not admit one finite such sum, even if compact support is not required of the continuous functions. On $X=\mathbb{R}$, considerThe sum is locally finite, so it defines a distribution. If it were a finite sum ${\sum }_{l=0}^M{\partial }^l{g}_l$ with all $g_l$ continuous, then on every fixed compact set its action would be bounded by test derivatives through order $M$. Near an integer $j>M$, however, it is exactly ${\delta }_j^{(j)}$, whose order is $j$. To see the contradiction directly, choose a test function $\psi$ with ${\psi }^{(j)}(0)\ne 0$ and put ${\psi }_{\epsilon }(x)={\epsilon }^j\psi ((x-j)/\epsilon )$. All derivative norms through order $M$ tend to zero, while $⟨{\delta }_j^{(j)},{\psi }_{\epsilon }⟩=(-1{)}^j{\psi }^{(j)}(0)$. This is a distribution of unbounded order. In dimensions $n>1$, the same example uses point masses at $(j,0,\dots ,0)$ and derivatives in the first coordinate.

The test-function space is $\mathcal{D}({\mathbb{R}}^n)=C_c^{\infty }({\mathbb{R}}^n)$. Its topology is the strict inductive limit topology of the spaces ${\mathcal{D}}_K$ of smooth functions supported in a fixed compact set $K$, each with the seminorms $p_{K,j}(\phi )={\max }_{|\alpha |\le j}{\sup }_K|{\partial }^{\alpha }\phi |$. In particular, a sequence converges in $\mathcal{D}$ precisely when its supports lie in one compact set and all its derivatives converge uniformly. The indices $\alpha$ use multi-index notation.

The distribution space ${\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ consists of continuous complex-linear forms on $\mathcal{D}$. Equivalently, for every compact $K$ there are $C_K$ and an integer $q_K\ge 0$ such thatWe use distributional convergence: $u_j\to u$ means $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every $\phi \in \mathcal{D}$. The pairings are bilinear, with no complex conjugation. A test function $f$ is identified with its regular distribution $⟨f,\phi ⟩=\int f\phi$.

For a distribution $u$ and a test function $\psi$, their smoothing convolution with a test function isIt is a smooth function, with ${\partial }_x^{\alpha }(u\ast \psi )(x)=⟨u_y,{\partial }^{\alpha }\psi (x-y)⟩$. Indeed, for $x$ in a compact neighborhood all translated tests have support in one compact set, and their difference quotients converge in the space of test functions. Notice that $u\ast \psi$ need not be compactly supported.

Choose a nonnegative mollifier $\rho \in C_c^{\infty }(B_1)$ with $\int \rho =1$, and write ${\rho }_{\epsilon }(x)={\epsilon }^{-n}\rho (x/\epsilon )$. If $\tilde{\rho }(x)=\rho (-x)$, thenThe right-hand test tends to $\phi$ in $\mathcal{D}$: its supports lie in $\mathrm{supp}\phi +{\overline{B}}_1$ for $\epsilon \le 1$, and every derivative converges uniformly by the approximate-identity argument. Thus the smooth regularizations converge to $u$ as distributions.

To obtain actual compactly supported approximants, choose a smooth cutoff function $\chi$ equal to one on $B_1$ and supported in $B_2$, and setFor each fixed test function $\phi$, the cutoff is identically one on its support once $j$ is large. Hence $⟨f_j,\phi ⟩=⟨u,{\tilde{\rho }}_{1/j}\ast \phi ⟩\to ⟨u,\phi ⟩$. This proves $\mathcal{D}$ is dense in ${\mathcal{D}}^{\prime }$, and in fact establishes the density of test functions in distributions and gives a convergent approximating sequence for each distribution. The expanding cutoff is essential when $u$ has noncompact support of a distribution.

For the radial limit, take a test function $\phi$ and introduce $t=r^2-1$ in polar coordinates. The Jacobian $rdr=dt/2$ givesWhen $\phi$ is supported away from the origin, $A$ vanishes near $t=-1$ and extends to a compactly supported smooth function on the whole line. The folded sine approximation to a Dirac delta is exposed by setting $B(s)=A(s)+A(-s)$: integration by parts givesby the Riemann-Lebesgue lemma. ThereforeThis surface delta distribution is arclength measure on the unit circle. By the level-set normalization of a surface delta, it is $2\delta (|x{|}^2-1)$: the factor two cancels the gradient magnitude $|\nabla (|x{|}^2-1)|=2$ on the circle. It is not twice arclength measure.

For a test function that can meet the origin, the endpoint cannot be discarded. Now $A(-1)=\pi \phi (0)$. Its right derivative is integrable, with $A^{\prime }(-1+s)=(\pi /4)\Delta \phi (0)+O(s)$ from differentiated Taylor expansion: the angular average cancels odd Taylor terms, so $A(t)=\pi \phi (0)+(\pi /4)(1+t)\Delta \phi (0)+O((1+t{)}^2)$ near $t=-1$. Applying integration by parts separately on the negative and positive intervals givesThe last two integrals tend to zero by the Riemann-Lebesgue lemma. Thus the radial quadratic oscillation defect in two dimensions gives the stronger oscillating point-mass defect formulaChoose $\phi$ supported in $B_{1/2}$ with $\phi (0)=1$. Its pairing is $-\pi \cos m+o(1)$, which does not converge. For completeness, if $\cos m$ had a limit $L$, the recurrence $\cos (m+1)+\cos (m-1)=2\cos 1\cos m$ would force $L=0$, whereas the even subsequence identity $\cos (2m)=2{\cos }^{2}m-1$ would then force $L=-1$. Consequently there is no limit in ${\mathcal{D}}^{\prime }({\mathbb{R}}^2)$.

Use the Fourier transform convention $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, with inverse factor $(2\pi {)}^{-n}$. The Paley–Wiener–Schwartz theorem can be stated with sharp convex support: for a nonempty compact convex set $K\subset {\mathbb{R}}^n$, let its support function be $H_K(\eta )={\sup }_{x\in K}x\cdot \eta$. Then $F$ is the Fourier transform of a unique distribution supported in $K$ if and only if it is entire andfor some $C$ and nonnegative integer $M$. For $K={\overline{B}}_R$, this is the familiar bound $C(1+|z|{)}^M{e}^{R|\mathrm{Im}z|}$. Allowing some $R$ characterizes all compactly supported distributions. Here the extension of the Fourier transform is $F(z)=⟨u,e^{-ix\cdot z}⟩$, interpreted with a cutoff equal to one near the support of a distribution.

First suppose $\mathrm{supp}u\subset K$. Fix one smooth cutoff function equal to one near $K$. Pairing the resulting compactly supported exponential with $u$ shows that $F$ is an entire function: differentiation with respect to $z_j$ inserts $-i{x}_j$, and the power series converges in the test-function seminorms uniformly on compact subsets of ${\mathbb{C}}^n$.

To keep the exponential type exactly $H_K$, rather than that of a fixed larger neighborhood, use a shrinking-cutoff exponential-type estimate. There are cutoffs ${\chi }_{\epsilon }$ equal to one on $K+\epsilon B_1$, supported in $K+3\epsilon {\overline{B}}_1$, and satisfying $|{\partial }^{\alpha }{\chi }_{\epsilon }|\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1$. One construction convolves the indicator of $K+2\epsilon B_1$ with a unit-mass mollifier supported in $\epsilon B_1$. Continuity of $u$ on a fixed compact neighborhood gives a finite order of a distribution $q$ there. By the product rule,Taking $\epsilon =(1+|z|{)}^{-1}$ proves the required bound with $M=2q$, since the extra exponential factor is at most $e^3$. The value of the pairing is independent of the chosen cutoff because all cutoffs agree near $\mathrm{supp}u$. Thus the forward direction has the exact asserted support function, without an unproved estimate on derivatives restricted only to $K$.

Conversely, suppose the entire $F$ has the stated bound. Its restriction to real frequency has polynomial growth, so define a tempered distribution $u$ byThe Schwartz space decay makes this integral absolutely convergent and continuous; by Fourier inversion, $\hat{u}=F$ on real frequency. It remains to prove the support assertion by mollifier regularization for contour recovery of support.

Choose a nonnegative unit-mass mollifier $\rho$ supported in $B_1$, and setOn real frequency, $F_{\epsilon }$ decays faster than any polynomial after enough applications of integration by parts to $\hat{\rho }$; hence $h_{\epsilon }$ is smooth, by differentiation under the integral sign. More generally, for every integer $L$ there is $C_{\epsilon ,L}$ such thatTo obtain this estimate, write the transform of ${\rho }_{\epsilon }$ as the real-frequency transform of $e^{x\cdot \eta }{\rho }_{\epsilon }(x)$ and integrate by parts; each derivative introduces at most one factor of $|\eta |$.

Fix a unit vector $v$ and take $L>M+n+1$. A contour-shift proof of the Paley–Wiener–Schwartz theorem moves the inverse-transform contour to ${\mathbb{R}}^n+itv$:Here is a justification of the shift. Rotate coordinates so $v$ is the first coordinate direction, apply the Cauchy integral theorem on a rectangle in the first complex variable, and integrate over the other real variables. For fixed $t$, the vertical sides at real part $\pm A$ have an integrated bound proportional to $A^{M-L+n-1}$, and hence vanish as $A\to \infty$. The same decay bounds give absolute convergence on the horizontal sides. No contour shift of an unregularized polynomially growing integral is needed.

Positive homogeneity of the support function now givesIf $x\notin K+\epsilon {\overline{B}}_1$, the Hahn-Banach separation theorem supplies a unit vector $v$ with $x\cdot v>H_K(v)+\epsilon$. Letting $t\to \infty$ proves $h_{\epsilon }(x)=0$. Therefore $h_{\epsilon }\in C_c^{\infty }$ and $\mathrm{supp}{h}_{\epsilon }\subset K+\epsilon {\overline{B}}_1$.

Since $\hat{\rho }(\epsilon \xi )\to 1$ and its modulus is at most one on real frequency, the dominated convergence theorem in the formula for $⟨u,\phi ⟩$ gives $h_{\epsilon }\to u$ as tempered distributions, and thus as distributions. A test function supported outside $K$ has positive distance from $K$, so its pairing with $h_{\epsilon }$ is zero for all sufficiently small $\epsilon$. This proves $\mathrm{supp}u\subset K$. The Fourier transform of a compactly supported distribution constructed in the forward direction equals $F$ on ${\mathbb{R}}^n$; applying the one-variable identity theorem successively in each coordinate extends the equality to ${\mathbb{C}}^n$. Injectivity of the Fourier transform of a tempered distribution proves uniqueness and completes both directions.

For the independence application, put $g_m(z)=e^{iz\cdot y_m}{f}_m(z)$ and $r_m=1/(m+1)$. The ball version of the Paley–Wiener–Schwartz theorem supplies a nonzero distribution $v_m$ supported in ${\overline{B}}_{r_m}(0)$ with ${\hat{v}}_m=g_m$. By the Translation property of the Fourier transform,where $⟨{\tau }_{y}v,\phi ⟩=⟨v,\phi (\cdot +y)⟩$. Distinct integer points are at least distance one apart. For distinct positive indices $j,k$, the largest possible radius sum is $1/2+1/3=5/6<1$. Thus these closed balls, and hence the distribution supports, are pairwise disjoint.

If ${\sum }_{m=1}^N{c}_m{f}_m=0$, injectivity of the Fourier transform gives ${\sum }_m{c}_m{w}_m=0$. For each $j$, choose a smooth cutoff function equal to one near its ball and zero near all other balls. Multiplying the distributional identity by this cutoff isolates $c_j{w}_j=0$. Since $f_j$ is not identically zero, $w_j\ne 0$, so $c_j=0$. Consequently $f_1,\dots ,f_N$ are linearly independent over $\mathbb{C}$. This Fourier independence from disjoint distribution supports uses the quantitative exponential types to establish support separation.

Use the Fourier transform convention $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$ and $D_j=-i{\partial }_j$, so $\widehat{P(D)u}=P(\xi )\hat{u}$. All distribution pairings use complex bilinearity, and the formal transpose is $P(-D)$, without complex conjugation of the coefficients. If $D_j$ instead denotes ${\partial }_j$, the Fourier multiplier is $P(i\xi )$; the argument below is unchanged after this substitution.

For a nonempty compact convex set $K\subset {\mathbb{R}}^n$, let $H_K(\eta )={\sup }_{x\in K}x\cdot \eta$ be its support function. The Paley–Wiener–Schwartz theorem says that the Fourier transform is a bijection between distributions supported in $K$ and entire functions $F$ on ${\mathbb{C}}^n$ for whichfor some $C>0$ and nonnegative integer $N$. In particular, support in ${\overline{B}}_R$ is equivalent to the bound with $e^{R|\mathrm{Im}\zeta |}$. Convexity is essential in the precise support formulation: the support function of a set equals that of its convex hull.

Suppose first that $u$ is a compactly supported distribution with support in $K$. Define $F(\zeta )=⟨u,e^{-ix\cdot \zeta }⟩$, inserting any cutoff function equal to one near $K$ when viewing $u$ on test functions. A fixed cutoff and the finite order of a distribution estimate permit differentiation of the pairing in each complex variable, giving ${\partial }_{\zeta }^{\alpha }F=⟨u,(-ix{)}^{\alpha }{e}^{-ix\cdot \zeta }⟩$. The exponential function's power series converges with every required derivative on a fixed compact set; hence $F$ is an entire function. On real frequencies it agrees with the Fourier transform of a compactly supported distribution.

The exact exponential indicator needs a shrinking cutoff function, rather than one fixed outside $K$. Choose ${\chi }_{\epsilon }=1$ near $K$, with support in $K+B_{2\epsilon }$ and $\Vert {\partial }^{\alpha }{\chi }_{\epsilon }{\Vert }_{\infty }\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1$. Such a cutoff is obtained by convolving the indicator of $K+B_{\epsilon }$ with a mollifier supported in $B_{\epsilon /2}$. On one fixed ball containing all these supports, the finite order of a distribution bound isThe Leibniz rule applied to $\theta ={\chi }_{\epsilon }{e}^{-ix\cdot \zeta }$ therefore givesTaking $\epsilon =(1+|\zeta |{)}^{-1}$ proves the required bound, with $N=2m$ and a fixed additional factor at most $e^2$.

Conversely, suppose $F$ is an entire function with the displayed bound. Its real restriction has polynomial growth, so its inverse Fourier transform defines a tempered distribution byRapid decay of the Schwartz function $\hat{\phi }$ makes this integral absolutely convergent and continuous in the Schwartz space topology.

To determine the support of a distribution, take a test function supported in a half-space $x\cdot \omega \ge H_K(\omega )+\delta$, where $|\omega |=1$ and $\delta >0$. The contour-shift proof of the Paley–Wiener–Schwartz theorem gives, for each $t\ge 0$,Here is the estimate that justifies both the shift and its limiting use. Since $\hat{\phi }(-\xi -it\omega )=\int e^{ix\cdot \xi }{e}^{-tx\cdot \omega }\phi (x)dx$, repeated integration by parts with $1-{\Delta }_x$ yieldsFor fixed $t$, the same estimate is uniform over the imaginary strip from $0$ to $t\omega$. Rotate coordinates so $\omega$ is the first coordinate direction, apply the Cauchy integral theorem on truncated rectangles, and let their real edges tend to infinity. If $2L>N+n$, the boundary integrals vanish and the horizontal integrals converge absolutely; thus the claimed shift is valid. Using $H_K(t\omega )=t{H}_K(\omega )$ now bounds the pairing bywhich tends to zero. A separating hyperplane exists at every point outside the compact convex set $K$. A finite partition of unity on the support of any exterior test function reduces it to such half-spaces, so $\mathrm{supp}u\subseteq K$. Fourier inversion gives uniqueness; the entire transform of this $u$ agrees with $F$ on ${\mathbb{R}}^n$ and hence everywhere, by applying the one-variable identity theorem successively in each coordinate. This completes both directions.

Now apply this theorem to compact support solvability for a constant-coefficient ordinary differential equation. If $P$ is a nonzero constant, its kernel is zero and $u=v/P$ is the unique solution. Otherwise writeThe solution space of $P(-D)\phi =0$ isIndeed these exponentials solve the equation, their initial derivative vectors have a nonzero Vandermonde determinant, and uniqueness for the order-$m$ ordinary differential equation makes them a basis. Pairing $P(D)u=v$ with any such solution proves necessity:

For sufficiency, choose $R$ with $\mathrm{supp}v\subseteq [-R,R]$ and let $V(\zeta )=⟨v,e^{-ix\zeta }⟩$. The assumed annihilation is precisely $V({\alpha }_j)=0$ for every $j$. Since all roots of a polynomial are simple, $U=V/P$ has removable singularities at every root and is an entire function. Outside one disk, $|P(\zeta )|\ge c(1+|\zeta |{)}^m$. Inside that disk, the extended $U$ is bounded. The polynomial division preservation of exponential type therefore givesThe Paley–Wiener–Schwartz theorem produces a compactly supported distribution $u$ with transform $U$, supported in the same interval. Then $\widehat{P(D)u}=PU=V$, so Fourier inversion gives $P(D)u=v$. ThusThe compactly supported solution is unique: $P\hat{u}=0$ forces the entire transform to vanish off the finitely many roots and hence everywhere. With repeated roots, the corresponding condition would also require derivatives $V^{(k)}(\alpha )=0$, or pairing with $x^k{e}^{-i\alpha x}$, up to one less than each multiplicity.

For $P(z)=a_N{z}^N+\cdots +a_0$ with $N\ge 1$ and $a_N\ne 0$, the unique retarded fundamental solution is $E=Hu$, where $P(D)u=0$, $u^{(j)}(0)=0$ for $j<N-1$, and $u^{(N-1)}(0)=1/a_N$. The distributional jump formula for a Heaviside product gives $P(D)E={\delta }_0$. Its explicit expression isThe residue theorem gives the initial derivatives from the coefficients at infinity. Equivalently, the inverse Fourier transform is integrated below all its poles, producing support of a distribution in $[0,\infty )$. Uniqueness follows because the difference of two retarded solutions is an analytic homogeneous solution vanishing on the negative half-line.

For $t>0$, the support of a distribution satisfiesThe duality formula $⟨{\delta }_{t}u,\phi ⟩=t^{-n}⟨u,\phi (\cdot /t)⟩$ proves one inclusion by transporting test functions; applying the inverse dilation proves the other. For a compact support, the Paley–Wiener–Schwartz theorem gives another proof: $\widehat{{\delta }_{t}u}(\zeta )=t^{-n}\hat{u}(\zeta /t)$ rescales the exponential growth indicator by $1/t$.