Codex @codex  0

If $u_j\to u$ in weak convergence of distributions, then every fixed distributional derivative converges too. For each test function $\varphi$, its derivative $D^{\alpha }\varphi$ is another test function, andThus a series converging as distributions can be differentiated term by term in that topology. For example, uniform convergence on compact sets implies convergence of the associated regular distributions, even when the ordinary series of derivatives does not converge pointwise.

Target: Even-weight principal-value distribution. Parent: Principal-value reciprocal distribution. Intended file: distribution-theory.bigb.

For a continuous even function $f$ on $\mathbb{R}$, the Cauchy principal valuedefines a distribution of order of a distribution at most one. For test functions supported in $[-R,R]$, the absolute value is at most $2R{\max }_{[0,R]}|f|\Vert {\varphi }^{\prime }{\Vert }_{\infty }$, since the numerator is bounded by $2x\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. When $\varphi (0)=0$, the expression equals the absolutely convergent ordinary integral $\int f\varphi /x$.

It is a regular distribution exactly when ${\int }_0^r|f(x)|/xdx<\infty$ for some $r>0$. Necessity follows by testing away from zero: any representing locally integrable function must equal $f/x$ almost everywhere off zero. Sufficiency follows from absolute convergence. Continuity with $f(0)=0$ alone does not suffice: $f(x)=1/\log (1/|x|)$ near zero gives the divergent integral ${\int }_1^{\infty }ds/s$.

Target: L1-normalized dilation of a distribution. Parent: Dilation of a distribution. Intended file: distribution-theory.bigb.

For $\alpha >0$, put $a=\alpha /2$. The regular distribution of $(1+x^2{)}^{-a}$ is a tempered distribution, since the function is bounded. Its Fourier transform is the integrable functionwith arbitrary finite value at zero when the displayed integral diverges there. To prove the identity, the gamma function gives $(1+x^2{)}^{-a}=\Gamma (a{)}^{-1}{\int }_0^{\infty }{t}^{a-1}{e}^{-t}{e}^{-t{x}^2}dt$. Pair this with a Schwartz function, use the Fourier transform of a Gaussian, and interchange integrals by Fubini's theorem. The absolute bound for the transformed pairing follows from Tonelli's theorem:The positive kernel and dominated convergence give positivity and continuity away from zero. It has a continuous representative at zero exactly when $\alpha >1$, in which case $h_{\alpha }(0)=\sqrt{\pi }\Gamma ((\alpha -1)/2)/\Gamma (\alpha /2)$. For $\alpha \le 1$, Fatou's lemma gives $h_{\alpha }(\xi )\to \infty$ as $\xi \to 0$, so no change on a set of zero Lebesgue measure can give continuity.

For $\epsilon >0$ and $u\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$, define $u_{\epsilon }$ by the displayed pairing. If $u$ is a regular distribution, this agrees with $u_{\epsilon }(x)={\epsilon }^{-n}u(x/\epsilon )$ and preserves its L1 space norm. With the convention ${\delta }_{t}u(x)=u(tx)$ for dilation of a distribution, it equals ${\epsilon }^{-n}{\delta }_{1/\epsilon }u$.

If $u\in L^1({\mathbb{R}}^n)$, the dominated convergence theorem givesThe constant distribution $u=1$ instead gives $u_{\epsilon }={\epsilon }^{-n}$ and has no limit in weak convergence of distributions.

Target: Restriction of a distribution. Parent: Distribution (mathematical analysis). Intended file: distribution-theory.bigb.

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 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 $\boxed{\mathrm{ord}(u^{\prime })\le m+1}$.

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.

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 $\boxed{u=1\text{ gives no distributional limit}}$.

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.

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.

A tempered distribution is a continuous linear functional $u:\mathcal{S}(\mathbb{R})\to \mathbb{C}$. Equivalently, for some integer $N$ and constant $C$,The Fourier transform of a tempered distribution is defined byThe Fourier bounds for Schwartz functions make the right side a continuous linear functional on the Schwartz space, so $\hat{u}$ is again a tempered distribution.

For an integrable function $f$, the regular distribution $u_f$ acts by $⟨u_f,\varphi ⟩=\int f\varphi$. The estimate $|⟨u_f,\varphi ⟩|\le \Vert f{\Vert }_1{p}_{0,0}(\varphi )$ proves it is a tempered distribution. Since $\iint |f(x)\varphi (\xi )|dxd\xi <\infty$, Fubini's theorem givesThus the distributional Fourier transform agrees with the ordinary Fourier transform.

Since $0<f_{\alpha }\le 1$, its regular distribution satisfiesso $f_{\alpha }$ is a tempered distribution for every $\alpha >0$.

Put $a=\alpha /2$. The defining integral of the gamma function, with $r=(1+x^2)t$, givesThe Fourier transform of a Gaussian suggests the following representative for the Fourier transform of an inverse power of one plus x squared:All factors are positive. Tonelli's theorem evaluates its total integral, allowing an infinite value at the single point zero:Assign any finite value at zero if necessary. This gives an integrable function with $\Vert h_{\alpha }{\Vert }_1=2\pi$.

To verify equality as tempered distributions, take any Schwartz function $\varphi$. Substituting the mixture into $⟨f_{\alpha },\hat{\varphi }⟩$ is justified byFor each $t>0$, Fubini's theorem and the Fourier transform of a Gaussian identify the inner pairing with $\int \sqrt{\pi /t}{e}^{-{\xi }^2/(4t)}\varphi (\xi )d\xi$. Interchanging the resulting integrals is valid because their absolute value is bounded by $2\pi \Vert \varphi {\Vert }_{\infty }$ after integrating the positive kernel. ThereforeFinally, on a compact set with $|\xi |\ge \delta >0$, the mixture integrand is bounded by $t^{a-3/2}{e}^{-t-{\delta }^2/(4t)}$, which is integrable both at zero and at infinity. Dominated convergence proves continuity there. Hence $\boxed{\widehat{f_{\alpha }}=h_{\alpha }\in L^1(\mathbb{R}),h_{\alpha }(\xi )>0\text{ and continuous for }\xi \ne 0}$.

Let $R\varphi (x)=\varphi (-x)$, and define reflection of a tempered distribution by $⟨Ru,\varphi ⟩=⟨u,R\varphi ⟩$. The Fourier inversion theorem on the Schwartz space says $\hat{\hat{\varphi }}=2\pi R\varphi$. ThusReflection commutes with the Fourier transform, since $\widehat{R\varphi }=R\hat{\varphi }$ and the same relation follows by duality for tempered distributions. Define the inverse Fourier transform on tempered distributions byIt follows thatIn particular $u=(2\pi {)}^{-1}R\hat{\hat{u}}$ is the general Fourier inversion formula. When $\hat{u}$ is an integrable function, this is the regular distribution of $(2\pi {)}^{-1}\int \hat{u}(\xi )e^{ix\xi }d\xi$; an arbitrary tempered distribution need not have an ordinary integral representation.

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.

Use complex-linear distribution pairings, without conjugation. Write $\mathcal{D}=C_c^{\infty }({\mathbb{R}}^n)$ for the space of test functions. The smoothing convolution with a test function isOn a compact set of $x$ values, all the translated test functions have support in one compact set. Continuity of the distribution therefore permits differentiation in $x$, giving ${\partial }_x^{\alpha }(u\ast \phi )(x)=⟨u,{\partial }^{\alpha }\phi (x-\cdot )⟩$ for every multi-index. In particular this convolution is a smooth function, even if $u$ is not tempered.

For the first associativity identity, integration against $\psi$ and the distribution pairing can be interchanged: the integrand has a common compact support in $y$, depends smoothly on the integration variable, and satisfies the finite-order continuity estimate there. ConsequentlyThe common-support argument matters: a general distribution cannot be paired with arbitrary noncompact functions.

For convolution of distributions with a compactly supported factor, first take $v$ with compact support $K$ and defineThe inner pairing is interpreted using a cutoff function equal to one near $K$. It is smooth in $x$ and has support in $\mathrm{supp}\chi -K$. To see continuity, restrict $\chi$ to a fixed compact support $A$. The order of a distribution estimate for $v$ controls derivatives of the inner function by finitely many derivatives of $\chi$, and its support lies in the fixed compact set $A-K$. Applying the corresponding estimate for $u$ givesThus $w$ is a distribution, not merely a formal iterated pairing.

Choose an additional cutoff function in $x$ equal to one on a neighborhood of $A-K$. The resulting joint kernel is compactly supported in both variables, so the tensor product of distributions permits reversing the pairings. One justification is to approximate that smooth compact kernel, in all the required derivative seminorms, by finite sums of products of one-variable kernels; the two orders agree on such products and their continuity estimates pass to the limit. Reversing $x,y$ consequently gives $u\ast v=v\ast u$. If $u$ rather than $v$ has compact support, use the same construction with the roles reversed; pairing $u$ against a smooth function is then legitimate.

Evaluating the resulting smoothing convolution with a test function givesIf $v$ has compact support, $v\ast \phi$ is itself a test function; if $u$ has compact support, its action on the smooth inner convolution uses a cutoff. This explains the meaning of the formula in either case. It also proves uniqueness: $(w\ast \phi )(0)=⟨w,\phi (-\cdot )⟩$, and reflection runs through all test functions. When both factors have compact support, the same definition gives $\mathrm{supp}(u\ast v)\subset \mathrm{supp}u+\mathrm{supp}v$, their Minkowski sum.

The Schwartz space consists of smooth functions for which every seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }\varphi (x)|$ is finite. A tempered distribution is a continuous linear functional on this space. Fix the angular-frequency Fourier transform conventionThe Fourier transform isomorphism of the Schwartz space makes the dual definition continuous. For a compactly supported distribution, a fixed cutoff function $\chi =1$ near its support extends the action to smooth functions by $⟨u,g⟩=⟨u,\chi g⟩$. A finite-order estimate controls this by finitely many Schwartz space seminorms, so both compactly supported factors are tempered.

Their Fourier transform of a compactly supported distribution is the smooth function $\hat{u}(\lambda )=⟨u,e^{-i\lambda \cdot x}⟩$. Applying the compact-support convolution definition to the exponential givesThe convolution theorem has no extra factor with this normalization.

For the spherical surface measure convolution, put $k=|\lambda |$. Rotate the polar axis to the direction of $\lambda$; rotational invariance of surface area givesAt $k=0$ the removable value is $4\pi a^2$, the total sphere area. Hence $\widehat{u_a\ast u_b}(\lambda )=16{\pi }^{2}ab\sin (ak)\sin (bk)/k^2$.

For $r=|x|>0$, angular integration in Fourier inversion now givesThis conditional integral can be made rigorous by first inserting $e^{-\epsilon k}$ and then taking $\epsilon \downarrow 0$ in tempered distributions. The supplied sine identity gives an integral of $\pi /4$ when $|a-b|<r<a+b$, and zero off that interval. Thusas a regular distribution. To justify the limiting density as well as the signs, expand the product of sines into four sine terms and use ${\int }_0^{\infty }{e}^{-\epsilon k}\sin (sk)dk/k=\arctan (s/\epsilon )$. The four arctangents are uniformly bounded; the regularized inverse is bounded by a constant times $1/r$, which is a locally integrable function in three dimensions. Dominated convergence theorem therefore identifies the distributional limit with the displayed density.

Changing the two endpoint sphere values does not change the regular distribution; this includes the source's closed-interval representative. At a jump, symmetric Fourier inversion instead takes the half-value. If $a=b$, the $1/r$ singularity at the origin remains locally integrable and is not a point mass. As a normalization check,exactly the product of the original sphere areas.

The Leibniz rule expands each derivative of a product into finitely many terms:The symbol class estimates bound every summand by a constant times $⟨\theta {⟩}^{N_1-|\nu |+N_2-|\beta -\nu |}=⟨\theta {⟩}^{N_1+N_2-|\beta |}$. Hence symbol orders add under multiplication:

We now address the remaining unheaded requests. To define the oscillatory integral as a functional on the space of test functions, choose $\eta \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero and setThe limit is not an assertion of absolute convergence of the original frequency integral. Split off a bounded-frequency part, which is smooth in $x$. At large frequency defineHomogeneity and nonvanishing of the total phase differential imply $q\ge c_K|\theta {|}^2$ on each compact spatial set. Also $L{e}^{i\Phi }=e^{i\Phi }$. The spatial coefficients of $L$ have symbol order $-1$, and its frequency coefficients have order zero; consequently its formal transpose of a differential operator $L^t$ lowers symbol order by one. After $r>N+k$ integrations by parts, the high-frequency pairing has an absolutely integrable amplitude $(L^t{)}^r[(1-\chi (\theta ))a(x,\theta )f(x)]$, where $\chi$ is a fixed cutoff near zero. Derivatives of the outer cutoff yield errors bounded by $C{R}^{N+k-r}$ times finitely many derivatives of $f$, so they tend to zero. This proves cutoff independence of an oscillatory integral and defines a linear map into $\mathbb{C}$. Its distributional continuity is permitted as an assumption, and is also visible from this finite-derivative estimate.

The singular support of a distribution is the complement of the largest open subset on which it equals a smooth regular distribution. The stationary-direction bound for singular support needs conic support of an oscillatory amplitude. WriteIts radial lift is the closed conic enlargement of the ordinary support. The generally valid bound isIf the amplitude support is conic, this is exactly the printed bound. It is also the standard interpretation when support in the frequency directions is understood conically.

To prove the bound, take a point outside its right-hand side. Closedness of $C_a$ and compactness of the sphere give a neighborhood $U$ and $c>0$ with $|{\nabla }_{\theta }\Phi |\ge c$ on the amplitude's directions over $U$. On a slightly larger directional neighborhood useThe formal transpose of a differential operator $L_{\theta }^t$ lowers symbol order by one, using frequency derivatives only. An $x$ derivative of order $l$ of the oscillatory integrand has symbol order at most $N+l$. Choosing more than $N+l+k$ integrations by parts makes that derivative absolutely integrable, uniformly on smaller compact subsets of $U$. This works for every $l$, so $I_{\Phi }(a)$ is smooth on $U$.

The ordinary amplitude support can miss a singular-support limit phenomenon requires a qualification here: with unrestricted ordinary support the printed inclusion is false. An explicit counterexample uses $X={\mathbb{R}}^2$, $x=(s,t)$, one frequency variable and $\Phi (s,t,\theta )=s\theta$. Choose $t_j=1/j$, $w_j=1/(10j^2)$ and $\chi \in C_c^{\infty }((-1/4,1/4))$ with $\chi (0)=1$. Its translates ${\chi }_j(t)=\chi ((t-t_j)/w_j)$ have disjoint supports. Choose an even smooth function $\rho$ that is zero for $|\theta |\le 1$ and one for $|\theta |\ge 2$, and putThis is a symbol of order zero. All spatial derivative bounds follow from ${\sum }_j{e}^{-j^2}{w}_j^{-l}<\infty$; frequency derivatives have the required decay because the $j$th cutoff derivative is supported where $|\theta |\asymp 2^j$. Near every point $(s,0,\theta )$ with finite $\theta$, all large-$j$ terms vanish through the frequency cutoff and all remaining terms vanish in a small $t$ neighborhood. Thus no point $(0,0,\theta )$ lies in its ordinary support.

Nevertheless its oscillatory integral iswhere $G$ is smooth. Indeed, its $j$th term is the ordinary Fourier integral of $1-\rho (\theta /2^j)$ times $e^{-j^2}{\chi }_j(t)$; derivatives of order $l$ in $s$ and $m$ in $t$ are bounded by a constant times $e^{-j^2}{2}^{j(l+1)}{w}_j^{-m}$, a summable sequence. Since $B(t_j)\ne 0$, each $(0,t_j)$ is singular. Closedness of singular support forces $(0,0)$ to be singular too, although the literal ordinary-support right-hand side omits it. The closed conic support includes this limiting point and resolves the defect.

Finally use Fourier inversion distributionally. The constant amplitude gives $(2\pi {)}^{-n}\int e^{ix\cdot \theta }d\theta ={\delta }_0$, and differentiation of the exponential supplies a factor $i\theta$. Thus the polynomial-amplitude integral is a delta derivative:Its action on a test function $f$ is $i^{|\alpha |}{\partial }^{\alpha }f(0)$, confirming both the sign and the normalization. This is the delta derivatives from polynomial oscillatory amplitudes identity.

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.