Let $f : \mathbb{R} \rightarrow \mathbb{C}$ be a Schwartz function. For $\varepsilon > 0$ consider the integral $$F(\varepsilon) = \int_{\mathbb{R}}{\hat{f}(x) e^{i \varepsilon x^2}dx}$$

By Fourier inversion, we have for every integer $N \geq 0$, the identity

$$ F(\varepsilon) = \sum_{n = 0}^N{c_n f^{(2n)}(0) \frac{1}{n!} \varepsilon^n} + R_N(\varepsilon) \, , \quad R_N(\varepsilon) := \int_{\mathbb{R}}{\hat{f}(x)\left(e^{i \varepsilon x^2}- \sum_{n = 0}^N{\frac{(i \varepsilon x^2)^n}{n!}} \right)dx} $$ for some constants $c_n \in \mathbb{C}$.

I'd like to know wheter this describes an asymptotic expansion of $F(\varepsilon)$, as $\varepsilon \rightarrow 0$. Given $N \geq 0$, can we find $C = C(f,N) > 0$ and $\varepsilon_0 > 0$ such that $$ |R_N(\varepsilon)| \leq C\left(\frac{1}{(N+1)!} \varepsilon^{N+1} \right)\,, $$ for all $0 < \varepsilon < \varepsilon_0$?

This is easy to prove if $\hat{f}$ has compact support, using for instance the bound $$ \left|\exp(z) - \sum_{n = 0}^{N}{\frac{z^n}{n!}}\right| \leq \frac{2|z|^{N+1}}{(N+1)!} $$ valid for all $z \in \mathbb{C}$ with $|z| \leq 1 + N/2$. I don't see how to generalize this to all Schwartz functions $f$ using an appropriate truncation of the integral $R_N(\varepsilon)$.


We have $$ e^{i\varepsilon x^2} - \sum_{n=0}^N \frac{(i\varepsilon x^2)^n}{n!} = \frac{(i\varepsilon x^2)^{N+1}}{N!} \int_0^1 (1-u)^N e^{i\varepsilon x^2 u} \, du, $$ by fiddling about with the integral form of the remainder. In particular, the remaining integral is bounded above by $1/(N+1)$ for any $x$, so we have $$ \lvert R_N(\varepsilon) \rvert \leqslant \int_{\mathbb{R}} \lvert \hat{f}(x) \rvert \left\lvert e^{i\varepsilon x^2} - \sum_{n=0}^N \frac{(i\varepsilon x^2)^n}{n!} \right\rvert dx \leq \frac{\varepsilon^{N+1}}{(N+1)!} \int_{\mathbb{R}} \lvert \hat{f}(x) \rvert \lvert x \rvert^{2N+2} dx . $$ The remaining integral exists and is finite since $f$, and hence $\hat{f}$, is Schwartz, and hence we have $R_N(\varepsilon) = O(\varepsilon^{N+1}) = o(\varepsilon^N)$ as $\varepsilon \to 0$ as required.

  • $\begingroup$ Very nice, thanks! $\endgroup$ – m.s Apr 8 '18 at 8:09

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