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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.