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$\|f\|_{B^{0}_{p,p}}=(\sum_{j\geq -1} \|\Delta_j f\|_p^p)^{1/p}$ is the Besov norm of $f$.

Here the Fourier transform of $\Delta_jf~(j\geq 0)$ is $\psi(2^{-j}\xi)\hat{f}(\xi)$ and $\psi$ is a smooth function with support in an annulus and $\hat{\Delta_{-1}f}(\xi)=\chi(\xi)\hat{f}(\xi)$. $$\chi(\xi)+\sum_{j\geq 0}\psi(2^{-j}\xi)=1.$$

Suppose there is a family of operator $\{K_{\sigma}\}_{\sigma\in \mathbb{M}^{d\times d}, ~~\|\sigma\|\leq 1}$ and we have $\|K_\sigma f\|_{B^0_{p,\infty}}\leq C \|f\|_{B^0_{p,\infty}}$ (C is independent with $\sigma$). Now suppose $\sigma(x): \mathbb{R}^d\rightarrow \mathbb{M}^{d\times d}$ is continous, satisfying $$|\sigma(x)-\sigma(0)|\leq \epsilon_0,~~~~ \sigma(x)=\sigma(0) ~~~\forall |x|>\delta$$ where $\epsilon_0, \delta$ are small. Can we get $$\|K_{\sigma(x)}f(x)-K_{\sigma(0)}f(x)\|_{B^0_{p,\infty}}\leq C(\epsilon_0)\|f\|_{B^0_{p,\infty}}~~~?$$

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  • $\begingroup$ Is $K_\sigma$ supposed to be continuous in $\sigma$? Otherwise $K_{\sigma(x)}$ can translate $f(x)$ to $f(x+h(x))$, where $h$ is any function of $x$. In particular it can translate any garbage to a place arbitrarily close to $x$. $\endgroup$
    – Fan Zheng
    Mar 4, 2016 at 19:07

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