I'm currently studying Classical and Multilinear Harmonic Analysis. Vol. 1 by Camil Muscalu, Wilhelm Schlag. I need to verify following calculus inequality (Eq. 9.27, at page 255) $$\int_{-\infty}^\infty \omega_I(x)\omega_J(x)dx\le C\min(|I|,|J|)\left(\frac{|I|+|J|}{|I|+|J|+\operatorname{dist}(I,J)}\right)^{1+\delta}$$ where $$\omega_I(x):=C\left(\frac{|I|}{|I|+\operatorname{dist}(x,I)}\right)^{1+\delta}$$ Here $I$ and $J$ are dyadic intervals in $[0,1]$. Precisely from Definition 8.12 of the same book.
We set $$A_n:=\{[(k-1)2^{-n},k2^{-n}|1\le k\le2^n\}$$ for each $n\ge0$ and $D_n:=\bigcup^n_{n=0}A_m$ for each $0\le n\le\infty$. Each $I$ and $J$ are an element of $A_n$.
There is a hint for this it says
To prove this reduce to $I=[0,1]$ and $|J|\le 1$ by scaling and then consider different cases depending on the size of $\operatorname{dist}(0,I)$.
I don't know how to approach the "different cases".
I think you may be misreading the text. The asserted calculus inequality is for all intervals $I,J\subset\mathbb{R}$, not just dyadic subintervals of $[0,1]$. Also, note that the constant $C=C_{\delta}$ is just some absolute constant, which may depend on $\delta>0$. It's not the constant used in the authors' earlier definition of the functions $\omega_{I}$; otherwise, we should have a $C^{2}$ in the RHS of the inequality. In fact, $C$ must depend on $\delta$. To see this, observe that if $J=[-1,1]$ and $I=[-r,r]$, where $r\geq 1$, then \begin{align*} \int_{\mathbb{R}}\omega_{I}\omega_{J}\geq\int_{1}^{r}\dfrac{1}{(1+\left|x-1\right|)^{1+\delta}}+\int_{-r}^{-1}\dfrac{1}{(1+\left|x+1\right|)^{1+\delta}}=\int_{-(r-1)}^{r-1}\dfrac{1}{(1+\left|x\right|)^{1+\delta}} \end{align*} Letting $r\uparrow\infty$ and $\delta\downarrow 0$ together with the divergence of the integral $\int1/(1+\left|x\right|)$ shows that we cannot hope a constan which holds for all $\delta>0$.
Suppose we have reduced to the case where $I=[0,1]$ and $J\subset\mathbb{R}$ is an interval with $\left|J\right|\leq 1$. You didn't indicate any issue with this part, so I will omitit. We need to show that there exists a constant $C>0$, which may depend on $\delta$, such that $$\int_{\mathbb{R}}\omega_{I}(x)\omega_{J}(x)\mathrm{d}x\leq C_{\delta}\left|J\right|\left(\dfrac{1+\left|J\right|}{1+\left|J\right|+\text{dist}(I,J)}\right)^{1+\delta},\qquad\forall\left|J\right|\leq 1\tag{1}$$
First, consider all $J=[a,b]$ such that $\text{dist}(I,J)\leq\left|J\right|$. Then $$\dfrac{1}{3}\leq\dfrac{1+\left|J\right|}{1+2\left|J\right|}\leq\dfrac{1+\left|J\right|}{1+\left|J\right|+\text{dist}(I,J)}\leq 1$$ So, \begin{align*} \int_{\mathbb{R}}\omega_{I}(x)\omega_{J}(x)&\leq\int_{J}\omega_{I}(x)\omega_{J}(x)\mathrm{d}x+\int_{-\infty}^{a}(1+\left|x-a\right|/\left|J\right|)^{-1-\delta}\mathrm{d}x+\int_{b}^{\infty}(1+\left|x-b\right|/\left|J\right|)^{-1-\delta}\mathrm{d}x\\ &\leq\left|J\right|+2\left|J\right|\int_{\mathbb{R}}(1+\left|x\right|)^{-1-\delta}\mathrm{d}x\\ &\leq\underbrace{\left(3^{1+\delta}\left(1+2\int_{\mathbb{R}}(1+\left|x\right|)^{-1-\delta}\mathrm{d}x\right)\right)}_{C_{1}}\left|J\right|\left(\dfrac{1+\left|J\right|}{1+\left|J\right|+\text{dist}(I,J)}\right)^{1+\delta} \tag{2} \end{align*} where we use the translation and dilation invariance of the integral in the second inequality and our previous analysis in the third inequality.
Now, consider all $J$ such that $\left|J\right|\leq\text{dist}(I,J)$. Without loss of generality, suppose $1\leq a< b<\infty$. Let $c=(a+1)/2$. We split the integral over the intervals $(-\infty,0]$, $I=[0,1]$, $[1,c]$, $[c,a]$, $J=[a,b]$, and $[b,\infty)$.
We deal with the terms over $[0,1]$ first. If $\text{dist}(I,J)\leq 1$, then $$\dfrac{1}{3}\leq\dfrac{1+\left|J\right|}{1+\left|J\right|+\text{dist}(I,J)}\leq 1$$ Since \begin{align*} \int_{0}^{1}\omega_{I}\omega_{J}\leq\int_{0}^{1}\left(1+\left|x-a\right|/\left|J\right|\right)^{-1-\delta}\mathrm{d}x\leq\left|J\right|\int_{\mathbb{R}}(1+\left|x\right|)^{-1-\delta}\mathrm{d}x \end{align*} The same constant $C_{1}$ as above works to bound this term. If $\text{dist}(I,J)>1$, then $$\int_{0}^{1}\omega_{I}\omega_{J}\leq\dfrac{\left|J\right|^{1+\delta}}{(\left|J\right|+\text{dist}(I,J))^{1+\delta}}$$ Since $$\dfrac{1+\left|J\right|+\text{dist}(I,J)}{\left|J\right|+\text{dist}(I,J)}=1+\dfrac{1}{\left|J\right|+\text{dist}(I,J)}\leq 1+\dfrac{1}{1+\left|J\right|}\leq 2,$$ we conclude that $$\int_{0}^{1}\omega_{I}\omega_{J}\leq 2\dfrac{\left|J\right|^{1+\delta}}{(1+\left|J\right|+\text{dist}(I,J))^{1+\delta}}\leq 2\left|J\right|\dfrac{(1+\left|J\right|)^{1+\delta}}{(1+\left|J\right|+\text{dist}(I,J))^{1+\delta}} \tag{3}$$ where we use $\left|J\right|\leq 1$.
Now consider the integral over the two subintervals $[1,c]$ and $[c,a]$. Observe that \begin{align*} \int_{1}^{c}\omega_{I}\omega_{J}&=\int_{1}^{c}\dfrac{1}{(1+\left|x-1\right|)^{1+\delta}}\dfrac{\left|J\right|^{1+\delta}}{(\left|J\right|+\left|a-x\right|)^{1+\delta}}\\ &\leq\dfrac{\left|J\right|^{1+\delta}}{(\left|J\right|+\left|a-c\right|)^{1+\delta}}\int_{\mathbb{R}}(1+\left|x\right|)^{-1-\delta}\\ &=\dfrac{\left|J\right|^{1+\delta}}{(\left|J\right|+\text{dist}(I,J)/2)^{1+\delta}}\int_{\mathbb{R}}(1+\left|x\right|)^{-1-\delta}\\ \end{align*} The same case analysis for $\text{dist}(I,J)\leq 1$ and $\text{dist}(I,J)>1$ as above, shows that $$\int_{1}^{c}\omega_{I}\omega_{J}\leq\max\left\{C_{1},2^{2+\delta}\int_{\mathbb{R}}(1+\left|x\right|)^{-1-\delta}\right\}\left|J\right|\dfrac{(1+\left|J\right|)^{1+\delta}}{(1+\left|J\right|+\text{dist}(I,J))^{1+\delta}}\tag{4}$$
Observe that by dilation and translation invariance, we have the estimate \begin{align*} \int_{c}^{a}\omega_{I}\omega_{J}&\leq\dfrac{1}{(1+\left|c-1\right|)^{1+\delta}}\int_{c}^{a}\dfrac{1}{(1+\left|a-x\right|/\left|J\right|)^{1+\delta}}\\ &\leq\dfrac{\left|J\right|}{(1+\text{dist}(I,J)/2)^{1+\delta}}\int_{\mathbb{R}}(1+\left|x\right|)^{-1-\delta}\\ &\leq \underbrace{\left(2^{1+\delta}\int_{\mathbb{R}}(1+\left|x\right|)^{-1-\delta}\right)}_{C_{2}}\left|J\right|\dfrac{(1+\left|J\right|)^{1+\delta}}{(1+\left|J\right|+\text{dist}(I,J))^{1+\delta}}\tag{5}, \end{align*} since $\left|J\right|\leq 1$.
For the integral over $J$, we have estimate $$\int_{J}\omega_{I}\omega_{J}\leq\left|J\right|\dfrac{1}{(1+\text{dist}(I,J))^{1+\delta}}$$ Since $$\dfrac{1+\left|J\right|+\text{dist}(I,J)}{1+\text{dist}(I,J)}\leq1+\dfrac{\left|J\right|}{1+\text{dist}(I,J)}\leq 1+\left|J\right|,$$ we conclude that $$\int_{J}\omega_{I}\omega_{J}\leq\left|J\right|\dfrac{(1+\left|J\right|)^{1+\delta}}{(1+\left|J\right|+\text{dist}(I,J))^{1+\delta}}\tag{6}$$
Lastly, \begin{align*} \int_{b}^{\infty}\omega_{I}\omega_{J}&\leq\dfrac{1}{(1+\left|b-1\right|)^{1+\delta}}\int_{b}^{\infty}\dfrac{1}{(1+\left|x-b\right|/\left|J\right|)^{1+\delta}}\\ &\leq\dfrac{\left|J\right|}{(1+\left|J\right|+\text{dist}(I,J))^{1+\delta}}\int_{\mathbb{R}}(1+\left|x\right|)^{-1-\delta}\tag{7}\\ \end{align*} And \begin{align*} \int_{-\infty}^{0}\omega_{I}\omega_{J}&\leq\dfrac{\left|J\right|^{1+\delta}}{(\left|J\right|+a)^{1+\delta}}\int_{\mathbb{R}}\dfrac{1}{(1+\left|x-1\right|)^{1+\delta}}\\ &\leq\dfrac{\left|J\right|^{1+\delta}}{(1+\left|J\right|+\text{dist}(I,J))^{1+\delta}}\int_{\mathbb{R}}(1+\left|x\right|)^{-1-\delta}\tag{8}, \end{align*} we we note $\left|J\right|^{1+\delta}\leq\left|J\right|$, since $\left|J\right|\leq 1$.
Finally, taking the desired constant $C$ to be $$C:=\max\left\{C_{1},C_{2},2^{2+\delta}\int_{\mathbb{R}}(1+\left|x\right|)^{-1-\delta},2\right\}\tag{9}$$ yields the desired inequality.