Estimate a singular integral using a dyadic decomposition

ca.classical-analysis-and-odesfourier analysisharmonic-analysis

Let $0<\alpha_{j}<1$, $j=1,\dots,d+1$. I am trying to estimate the following singular integral:
$$I(y_{1},\dots,y_{d},z):=\int_{\substack{ x\in[0,1]^{d}\\
\frac{d x_{1} \dots d x_{d}}{|x_{1}-y_{1}|^{\alpha_{1}}\dots

where $0<y_{j}<1/2$, $0<z<1$.

If this is a standard integral that has a known estimate, please refer me to some reference.

I can show precisely that in the one-dimensional case $d=1$, one has
$$I\lesssim \frac{1}{(y_{1}+z)^{\alpha_{2}}
|z-y_{1}|^{\alpha_{1}+\alpha_{2}-1}}\quad (1).$$

To achieve that, I simply consider all the nine possibilities that come from combining one of the cases
$x>>y_{1}$, $y_{1}>>x$, and $x\sim y_{1}$
with one of the cases
$x>>z$, $z>>x$, and $x\sim z$.
In the case ($x\sim y_{1}$ and $x\sim z$), I use a dyadic decomposition. I put one of the singular factors in a dyadic interval and integrate the other singular factor explicitly.

I don't know how to generalize that to higher dimensions.

Let $\lambda_{1},\dots,\lambda_{d},\lambda\leq 1$ be dyadic numbers. If we set $|x_{j}-y_{j}|\sim \lambda_{j}$ and $||x|-z|\sim \lambda$, then the problem reduces to estimating the measure of $\Omega=R\cap S$, the intersection of the rectangle $$R:=\{[y_{1}+\lambda_{1}/2,y_{1}+\lambda_{1}]\times\dots \times
with the spherical shell
For $\Omega$ to be nonempty, we necessarily have
$$z+\lambda \sim y_{1}+\dots+y_{d}+\lambda_{1}+\dots+\lambda_{d}.$$
How proceed from here ?

Best Answer

It is a bit of a long story, but if you are still interested, I can teach you how to make a fairly accurate estimate for the same integral $\mathcal I_0$ without the condition $|x|\ge\frac 12$ (it is possible to reintroduce that condition, but that will just make the computation a bit cumbersome and in all honesty, I don't think you care much whether it is $1/2$ or $1/5$ anyway). At this moment I'll just post the bound $$ \mathcal I_0(y,z)\lesssim \sum_I\sum_{\Gamma_I\le\gamma\le 1}\gamma^{\frac{|J|}2-\beta}\prod_{i\in I}\min(y_i,\frac \gamma{y_i})^{1-\alpha_i}\prod_{j\in_J}\max(\sqrt\gamma,y_j)^{-\alpha_j} $$ where I assume that $y=(y_1,\dots,y_n)\in [0,1]^n$, $z\in[0,1]$ and my $\alpha_j$ are the same as yours except I call the last one $\beta$: it obviously plays a special role in this game. $I$ here runs over all subsets of $\{1,\dots,n\}$ (including the empty set and the full set), $J$ is the complement of $I$, $\Gamma_I=\min(|\,|y_I|^2-z^2\,|,1)$ where $y_I$ is the projection of $y$ to the coordinate subspace associated with $I$ (so $y_\varnothing=0$ and $y_{\{1,\dots,n\}}=y$) and $\gamma$ runs over the powers of $2$ (dyadic scales) as usual.

Note that if $\alpha_i,\beta$ are generic, this sum consists of finitely many geometric progressions and, thus, reduces to the finite sum over the scales at which the behavior changes and the endpoints. In general, however, there may be flat pieces, which will produce logarithmic factors.

Note also that each term by itself is a trivial estimate from below of the integral over some single dyadic box (trivial in the sense that it is just the volume of the box divided by the product of the maxima of the denominator factors in that box). So, in the generic case, this gives you a finite sum that is equivalent to the integral. I haven't tried to see if the chains of overlapping boxes corresponding to the flat parts in the summation create an overcount somewhere or are sharp too).

So, if this type of bound is interesting or useful for you, let me know and I'll post the details :-).