In Bounded linear operators that commute with translation
It has: Let $m(\xi)=(T^{\;*}\psi)^\wedge(\xi)/\psi(\xi)$, then we have $$ (Tf)^\wedge(\xi)=m(\xi)\hat{f}(\xi) $$ and therefore, $\|m\|_{L^\infty}=\|T\|_{L^2}$.
Why $\|m\|_{L^\infty}=\|T\|_{L^2}$. I can't see this.
Hint: $\|T\|_{L^{2}} \leq \|m\|_{\infty}$ is obvious from the definition of the norm. Now let $N$ be a positive integer and an $\epsilon >0$ and consider $I_A$ where $A=\{\xi: |m(\xi)| > \|m\|_{\infty} -\epsilon, \|x\|\leq N\}$. (We can choose $N$ such that $A$ has positive measure). This is an $L^{2}$ function and hence it can be written as $\hat {f}$ for some $L^{2}$ function $f$. Now $\|T\|_{L^{2}} \geq \frac {\|m(\xi)\hat {f}(\xi)\|_2} {\|f(\xi)\|_2} $ by definition of the norm of $T$. Now just simplify the last expression and let $\epsilon \to 0$.