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Although $|u(x)|=e^{x^2}$ has superpolynomial growth, the rapidly varying phase makes $u$ an oscillatory tempered distribution. Split the integral against $\phi \in \mathcal{S}(\mathbb{R})$ into $|x|\le 1$ and the two tails. On a tail, with $\Phi (x)=e^{x^4}$,Integration by parts transfers the derivative toThis function and its derivative are integrable because $e^{x^2-x^4}$ dominates every polynomial, and the boundary term at infinity vanishes. The result is bounded by finitely many Schwartz space seminorms. The compact part has the same property. Thus the cutoff integrals converge and define a continuous linear functional:

Solved by gpt-5.6-sol high.

For any smooth amplitude $v(x,\lambda )$,because $D_x=-i{\partial }_x$. Applying each term of the differential operator under the oscillatory integral givesThe identity is justified distributionally by regularizing the frequency integral and integrating by parts.

Solved by gpt-5.6-sol high.

Iterate the correction: after constructing $b_{j-1}\in S^{-(j-1)}$, setThe same calculation improves the remainder by one order:Let $A_j$ and $B_N$ be the corresponding inverse oscillatory integrals. ThenWhen $N>n$, the frequency integral defining $B_N$ converges absolutely and depends continuously on $x$, so $B_N\in C(X)$. This completes the finite-order parametrix construction.

Solved by gpt-5.6-sol high.