(pp. 107-136)

The estimates proved in the previous chapter form a solid foundation for proving analogous results in higher dimensions for model operators of the form

$L_{b, m}\hspace{3pt}=\hspace{3pt}\sum_{j = 1}^{n}[x_{j} \partial_{x_j}^{2}\hspace{3pt}+\hspace{3pt}b_{j} \partial_{x_j}]\hspace{3pt}+\hspace{3pt}\sum_{k = 1}^{m} \partial_{y_k}^{2}, (7.1)$

here$b \in \mathbb{R}_{+}^{-n}$. In this context we exploit the fact that the solution operator for$L_{b, m}$is a product of solution operators for 1-dimensional problems.

In 2-dimensions we can write

$u\left\(x_{1}, x_{2}, t\right\)\hspace{3pt}-\hspace{3pt}u\left\(y_{1}, y_{2}, t\right\)\hspace{3pt}=\hspace{3pt}[u\left\(x_{1}, x_{2}, t\right\)\hspace{3pt}-\hspace{3pt}u\left\(x_{1}, y_{2}, t\right\)]\hspace{3pt}+\hspace{3pt}[u\left\(x_{1}, y_{2}, y\right\)\hspace{3pt}-\hspace{3pt}u\left\(y_{1}, y_{2}, t\right\)], \left\(7.2\right\)$

and in$n\hspace{3pt}\gt\hspace{3pt}2$dimensions we rewrite$u\left\(x, t\right\)\hspace{3pt}-\hspace{3pt}u\left\(y, t\right\)$as

$u(x, t)\hspace{3pt}-\hspace{3pt}u(y, t)\hspace{3pt}=\hspace{3pt}\sum_{j = 0}^{n-1}[u(x^{\prime}_{j}, x_{j+1}, y^{\prime\prime}_{j}, {t})\hspace{3pt}-\hspace{3pt}u(x^{\prime}_{j}, y_{j+1}, y^{\prime\prime}_{j}, t)], (7.3)$

where:

$\left. {x^{\prime}_{j}\hspace{3pt}=\hspace{3pt}\left\(x_{1},..., x_{j}\right\)\hspace{5pt}\text{if}\hspace{5pt} 1\hspace{3pt}\leq\hspace{3pt}j\hspace{5pt}\text{and}\hspace{5pt} \emptyset\hspace{5pt}\text{if}\hspace{5pt} j\hspace{3pt}\leq\hspace{3pt}0,\atop x^{\prime\prime}_{j}\hspace{3pt}=\hspace{3pt}\left\(x_{j+2},..., x_{n}\right\)\hspace{5pt}\text{if}\hspace{5pt} j\hspace{3pt}\lt\hspace{3pt}n\hspace{3pt}-\hspace{3pt}1\hspace{5pt} \text{and}\hspace{5pt} \emptyset\hspace{5pt} \text{if}\hspace{5pt} j\hspace{3pt}\geq\hspace{3pt}n\hspace{3pt}-\hspace{3pt}1.} (7.4)$

In this way we are reduced to estimating these differences 1-variable-at-a-time, which, in light of Lemma 6.1.1 suffices.

In the proofs of the 1-dimensional estimates the only facts about the data we use...