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Recently, I am reading "Inequalities for strongly singular convolution operator" written by Fefferman. I have some question on the detail of proof of Theorem 2'.

Let $\theta \in (0,1)$.

Let $f \in L^{1}(\mathbf{R}^n)$ and $\alpha > 0 $ be given.
Applying the Calderón–Zygmund decomposition on $f$, we have a collection of cube $ \{I_j\}$. Let $d_j=diam(I_j)$

Let $\phi \in C^{\infty}(\mathbf{R}^n)$ be a positive function with $\int_{\mathbf{R}^n} \phi =1$ and $\phi$ vanish outside unit ball. Let $\phi_j(x)= d_j^{\frac{-n}{1-\theta}}\phi(d_j^{\frac{-1}{1-\theta}}x).$

Let $J^{\frac{n\theta}{2}}$ be the Bessel potential of order $\frac{n\theta}{2}$

In the thesis, the notation "$x \sim I_j$" means $x\in I_l$ for some $I_l$ from the collection which touches or coincides with $I_j$.

Somewhere in the proof:
For $d_j<1$ , $x\nsim I_j$ ,
$$\sup_{y \in I_j}|J^{\frac{n\theta}{2}}*\phi_j(x-y)||I_j| \lesssim \int_{I_j}(J^{\frac{n\theta}{2}}*\phi_j)(x-y) dy$$ since $(J^{\frac{n\theta}{2}}*\phi_j)(x-y)$ is rounghly constant over $I_j$.
(The implicit constant should be independent of $j$ and $x$)

My question is how can I prove $(J^{\frac{n\theta}{2}}*\phi_j)(x-y)$ is rounghly constant over $I_j$ for $x\nsim I_j$?

It is clear that $(J^{\frac{n\theta}{2}}*\phi_j)(x-y)$ is bounded above. For bounded below, my idea is to find the lower bound of $J^{\frac{n\theta}{2}}(x-y-d_j^{\frac{1}{1-\theta}}z)$ for $x\nsim I_j, y\in I_j, |z|<1$ since $$(J^{\frac{n\theta}{2}}*\phi_j)(x-y) = \int_{|z|<1}J^{\frac{n\theta}{2}}(x-y-d_j^{\frac{1}{1-\theta}}z)\phi(z)dz$$.

Since $$J^{\frac{n\theta}{2}}(x)=c_\alpha\int_0^{\infty} e^{\frac{-\pi |x|^2}{\delta}}e^\frac{-\delta}{4\pi}\delta^{\frac{{-n-\frac{n\theta}{2}}}{2}}\frac{d\delta}{\delta}$$ ,

we need to prove $e^{\frac{-\pi |x-y-d_j^{\frac{1}{1-\theta}}z|^2}{\delta}}$ is bounded below.

But I have no idea. Do I misunderstand anything?

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  • $\begingroup$ Are you able to show $\sup_{y\in I_j, |z|<1} J^{n\theta/2} (x-y-d_j^{1/(1-\theta)}z)<C\inf_{y\in I_j, |z|<1} J^{n\theta/2} (x-y-d_j^{1/(1-\theta)}z)$? $\endgroup$
    – Fan Zheng
    Oct 17, 2016 at 18:11
  • $\begingroup$ It is equivalent to prove $J^{\frac{n\theta}{2}}(x-y-d_j^{\frac{1}{1-\theta}}z)$ is bounded which I have no idea. $\endgroup$
    – user134927
    Oct 18, 2016 at 4:51
  • $\begingroup$ Not really. It is equivalent to prove $J^{n\theta/2}(x-y')$ is comparable when $y'=y+d_j^{1/(1-\theta)}$ ranges over the double of $I_j$. Morally this follows from the behavior of the Bessel potential near 0 and infinity. Near 0 it diverges by a power law, so if its argument varies within a constant factor then its values are comparable. Near infinity it decays by an exponential law, so if its argument varies within a constant distance then its values are comparable. All these are satisfied in your situation, but I have to look up the asymptotics more carefully before i can write up an answer. $\endgroup$
    – Fan Zheng
    Oct 18, 2016 at 15:21
  • $\begingroup$ Is $x-y'$ really varies within a constant factor? $x \nsim I_j$ means x either outside the collection of cube or belongs to one of the cube which is not $I_j$ and not touch $I_j$ . $x$ can be far away from $y'$ or near $y'$. $\endgroup$
    – user134927
    Oct 19, 2016 at 3:39

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