Convergence in $\mathcal{D}'(\mathbb{R})$ via stationary phase method

Question

Im so confuse about how i can use the stationary phase method to find the limit of $$u_\lambda(x)=\lambda^Ne^{-i\lambda x}$$ when $\lambda\to\infty$, where N is a positive integer. The idea is take a test function $\phi$ and evaluate $$\langle u_\lambda,\phi\rangle=\lambda^N\int e^{-i\lambda x}\phi(x)dx,$$ I try to do the same but with the distribution $v_\lambda(x)=\lambda^{1/2}e^{i\lambda x^2/2}$ and I bellieve the final result is $(2\pi)^{1/2}e^{-i\pi/4}\phi(0)$, but I'm really very unsure of my answer. Thanks for your attention.

Answer 1

The phase has no stationary points, so you actually want to use the method of non-stationary phase which guarantees arbitrarily rapid polynomial decay. Since the proof is simple enough, I'll summarize the salient points here.

You can use integration by parts to find that $$\langle u_\lambda,\phi\rangle=\lambda^N\int_{-\infty}^{+\infty}e^{-i\lambda x}\phi(x)dx=\lambda^N\int_{-\infty}^{+\infty} (-i\lambda)^{-1}\frac{d}{dx}e^{-i\lambda x}\phi(x)dx=\lambda^N\int_ {-\infty}^{+\infty}(i\lambda)^{-1}e^{-i\lambda x}\phi'(x)dx.$$ Do this $N$ more times, and you'll get a factor of $\lambda^{-1}$ in front. In particular, $$\langle u_\lambda,\phi\rangle =\lambda^{-1}i^{N+1}\widehat{\phi^{(N+1)}}(\lambda),$$ where $\hat{\cdot}$ denotes the Fourier transform. Since $\phi$ is a test function, so is $\phi^{(N+1)}$. By the Riemann-Lebesgue lemma, its Fourier transform decays to $0$ as $\lambda\rightarrow\infty$ (or just take a supremum over the support of $\phi^{(N+1)}$), and hence $$\langle u_\lambda,\phi\rangle\rightarrow 0\qquad\text{ as } \lambda\rightarrow\infty.$$