Let $F(\mathbf{x}) \in \mathbb{Z}[x_1, ..., x_n]$ be a degree $d$ homogeneous form. Let $$ I(\alpha) = \int_{[0,1]^n} e^{2 \pi i F(\mathbf{x}) \alpha} dx_1...dx_n. $$ Then the singular integral is defined as $$ \sigma_{\infty} = \int_{\mathbb{R}} I(\alpha) d\alpha. $$

It says in an article I am reading that $\sigma_{\infty} > 0$ if the equation $F(\mathbf{x}) =0$ has a non-singular real solution in $(0,1)^n$. I would greatly appreciate if someone could give me a hint or some explanation on how I can prove this statement. Thank you very much.


Here's a sketch. First of all, proceeding formally, if we start with the $\alpha$ integral, then we obtain $$ \sigma_{\infty} = \int \delta(F(x))\, dx , $$ and near an $a$ with $F(a)=0$, $\nabla F(a)\not= 0$, this looks like $\delta(v\cdot t)$, so we obtain a positive contribution.

To make a proof out of this, fix a $\varphi\ge 0$ with $\widehat{\varphi}\ge 0$, $\varphi(0)=1$ (Fejer kernel would work), and consider $$ \int I(\alpha)\varphi(\alpha/n)\, d\alpha = n \int \widehat{\varphi}(nF(x))\, dx . $$ Now near an $a=(b,c)$ as above, with $b\in\mathbb R$, $c\in\mathbb R^{d-1}$, we find solutions to $F=0$ in a whole neighborhood $U\subseteq \mathbb R^{d-1}$ of $c$ (if $\partial F/\partial x_1(a)\not=0$), by the implicit function theorem. As before, integration over $(a-\delta,a+\delta)\times U$ gives a contribution that stays positive in the limit $n\to\infty$.

  • $\begingroup$ I was revisiting this solution again to go through in more detail. Does this argument then imply that $\int I(\alpha) d \alpha$ goes to infinity at the end? (Because of $n$ in front of the integral on the right hand side of the last equation) $\endgroup$ – Johnny T. Sep 26 '16 at 21:33
  • $\begingroup$ @JohnnyT.: No, that's not clear, because the $n$ inside the argument has the opposite effect. $\endgroup$ – Christian Remling Sep 26 '16 at 22:03
  • $\begingroup$ I guess this is what I thought you were doing: $n \int \widehat{\phi}(n F(x)) dx \geq n \int_{B} \widehat{\phi}(n F(x)) dx = n \int_{B} 1 dx$, where $B = (b - \delta, b +\delta) \times U$ which goes to infinity... Could you possibly explain how the very end of the argument works? $\endgroup$ – Johnny T. Sep 26 '16 at 23:07
  • $\begingroup$ No, your estimate doesn't work because the set of $x$ where the integrand is close to $1$ shrinks as $n\to\infty$. The quick summary is that $\widehat{\varphi}(nF(x))\ge 1/2$, say, on set of measure $\gtrsim 1/n$. This follows because $F=0$ on a hypersurface $(b(c),c)$ near $a$, so $\widehat{\varphi}(\ldots)=1$ there, and then just use derivatives (or Taylor expansions) to control how fast the integrand can decrease. $\endgroup$ – Christian Remling Sep 27 '16 at 2:31
  • $\begingroup$ Ok, I think I am missing something pretty big here. I would like to clarify two things: 1. On the set $B = (b - \delta, b + \delta) \times U$ the function $F(x) = 0$? 2. The construction of $B$ depended only on $F$ so in particular this set was chosen independent of $n$? $\endgroup$ – Johnny T. Sep 27 '16 at 16:50

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