The distance between a weighted shift operator and integral operators

In the sequel $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ are finite measure spaces. An operator $S\colon L_{\mu}^p \to L_{\nu}^p$, is called a weighted shift operator if it is representable as $\pi T_\varphi $, where $\pi$ is the operator of multiplication by a function $\pi$, $T_\varphi\colon x\to x\circ\varphi$ and $\varphi\colon Y\to X$ is a measurable mapping.
Theorem. {\it If $\pi T_\varphi\colon L_{\mu}^p\to L_{\nu}^p$ is a bounded weighted shift operator, $1\le p<\infty$, and$\pi\in L_{\nu}^p $, then $T_\varphi^\circ(|\pi|)\in L_{\mu}^\infty$ and}
$$ \|\pi T_\varphi\|=\bigl(\bigl\|T_\varphi^\circ(|\pi|^p)\bigr\|_{L^\infty}\bigr)^{1/p}. $$

Theorem. {\it If $T$, $S\colon L_{\mu}^p \to L_{\mu}^p$ are bounded operators, $1< p<\infty$, where $T$ is an integral operator and $S$ is a weighted shift operator then}
$$ \|S-T\|\ge\|S\|. $$