# Is there an asymptotic bound for this oscillatory integral?

I have an oscillatory integral:

$$\int u(x,y) e^{i\lambda f(x,y)} dx$$

with $f(x,y)\in \mathbb{C}^{\infty}$ a complex-valued function in a neighborhood of $(0,0)$ satisfying:

$$\text{Im} f \geq 0 \quad \text{Im} f(0,0) =0 \quad f'_x(0,0) = 0 \quad \text{det}f'_{xx}(0,0) \neq 0$$

I also have $u(0,0)=0$. However, $u$ is not differentiable at $(0,0)$. This does not allow me to apply stationary phase theorems from Hormander (The analysis of linear partial differential operators, v.1). Is there an asymptotic ($\lambda \rightarrow +\infty$) bound on this integral?

Edit I also checked that $u$ has the form of inner product on Hilbert space:

$$u(x,y) = \langle g(y), P_{(0,\infty)}(\text{sgn}(x) A) h(y) \rangle$$

with $P_{(0,\infty)}$ spectral projector onto positive eigenvalues, $\text{sgn}(x)$ sign function, $A$ linear self-adjoint operator. The spectrum of $\text{sgn}(x) A$ is discrete.

• What is the meaning of $y$ here? – Christian Remling Aug 9 '14 at 22:19
• The question is not well-posed without specifying the full set of hypotheses on $u$. For instance, if $u$ is not absolutely integrable then the integral does not make sense, let alone have any asymptotics. If all one has is continuity on $u$, then in analogy with the Riemann-Lebesgue lemma, the best one can hope for is a qualitative decay as $\lambda \to \infty$ without any decay rate. If one has stronger control on $u$, I recommend approximating $u$ by a smooth function plus a small error, using stationary phase for the former, and crude estimates for the latter. – Terry Tao Aug 10 '14 at 2:00
• @teagut It works with $u$ bounded compactly supported such that $u'$ is in $L^1$, by a van der Corput method. I simplified my first answer. – Bazin Jul 3 '18 at 21:53

Let me assume that $u$ is depending only on the $y$ variable, that $f$ is smooth and depends also only on $y$ and is such that $$\Im f\ge 0,\quad f(0)=0, df(0)=0,\quad \det f''(0)\not=0.$$ Then there exists a neighborhood $V$ of $0$ such that for $u\in L^\infty_{\text{comp}}(V)\cap C^1 (V\backslash\{0\})$, $u'\in L^1$, $$I(\lambda, u)=\int_{\mathbb R} e^{i\lambda f(x)} u(x) dx=O(\lambda^{-1/2}).$$ To prove this, we can use a van der Corput method. Let $\chi$ be a smooth compactly supported function equal to $1$ on $[-1/2, 1/2]$, supported in $[-1,1]$. We have $I(\lambda, \chi(\cdot \lambda^{1/2})u)=O(\lambda^{-1/2})$ since the integration takes place on $x\in [-\lambda^{-1/2}, \lambda^{-1/2}]$ and $u e^{if}$ is bounded. Moreover, checking $$I(\lambda, (1-\chi(\cdot \lambda^{1/2})) u)= \int_{\mathbb R} e^{i\lambda f(x)} (1-\chi(x\lambda^{1/2}))u(x) dx,$$ we use the identity (note that $f'(x)\not=0$ on $V\backslash\{0\}$) $$e^{i\lambda f(x)}=\frac{1}{i\lambda f'(x)}\frac{d}{dx}\bigl(e^{i\lambda f(x)}\bigr),$$ and we integrate by parts. We need to check $$\lambda^{-1}\frac{d}{dx}\left(\frac{u(x) \bigl(1-\chi(\lambda^{1/2}x)\bigr)}{f'(x)}\right).$$ Since $\vert (1-\chi(\lambda^{1/2}x))u'/f'\vert$ is bounded above by $\vert u'(x)\vert \frac{1}{\vert x\vert}$ and supported on $\vert x\vert\lambda^{1/2}\ge 1/2$, $\vert x\vert\lesssim 1$, we need to check $$\lambda^{-1}\int_{\lambda^{-1/2}\le \vert x\vert\lesssim 1}\frac{\vert u'(x) \vert}{\vert x\vert} dx\le \lambda^{-1/2}\int\vert u'(x) \vert dx,$$ which is $O(\lambda^{-1/2})$ thanks to the assumption $u'\in L^1$. Another term comes from $\lambda^{-1}(1-\chi)uf''/(f')^2$, and amounts to bound from above, since $u$ is bounded $$\lambda^{-1}\int_{\vert x\vert\ge \frac12\lambda^{-1/2}} \frac{dx}{\vert x\vert^2} \lesssim \lambda^{-1+\frac12},$$ which is $O(\lambda^{-1/2})$. We are left with $$\left\vert\lambda^{-1}\int \frac{u(x)}{f'(x)}\chi'(\lambda^{1/2}x)\lambda^{1/2} dx\right\vert \le C\lambda^{-1/2} \int_{\frac12 \lambda^{-1/2}\le \vert x\vert \le \lambda^{-1/2}}\frac{\vert u(x)\vert}{\vert x\vert} dx\lesssim\int_{\frac12 \lambda^{-1/2}\le \vert x\vert \le \lambda^{-1/2}} \vert{u(x)}\vert dx=O(\lambda^{-1/2}),$$ since $u$ is bounded.