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?
