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.

  • $\begingroup$ What is the meaning of $y$ here? $\endgroup$ Aug 9 '14 at 22:19
  • 4
    $\begingroup$ 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. $\endgroup$
    – Terry Tao
    Aug 10 '14 at 2:00
  • $\begingroup$ @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. $\endgroup$
    – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.