 Solution

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.