A Calderon Zygmund kernel $K$ is a function $K :\mathbb{R}^d-\{0\}\longrightarrow \mathbb{C}$ satisfying, for some constant $B$,
1)$|K(x)|\leq B|x|^{-d}$ for all $x\in\mathbb{R}^d$
2)$\displaystyle\int_{|x|>2|y|}|K(x)-K(x-y)|\,dx\leq B$ for all $y\leq 0$
3)$\displaystyle\int_{r<|x|<s}K(x)\,dx=0$ for all $0<r<s<\infty$
Now for positive real number $r,s(r<s)$ and a Calderon Zygmund kernel $K :\mathbb{R}^d-\{0\}\longrightarrow \mathbb{C}$ we define
$T_{r,s}f(x)=\displaystyle\int K(y)\chi_{r<|y|<s}f(x-y)\,dy$
and $m_{r,s}(\xi)=\displaystyle\int K(x)\chi_{r<|x|<s}e^{-2\pi ix\cdot\xi}\, dx$(A Fourier coefficient of $K\chi$).
The following is the book of Camil Muscalu, "Classical and Multilinear harmonic analysis" page 168.
In the middle of the page it says,
"$\lVert T_{r,s} \rVert_{2\rightarrow 2}=\lVert m_{r,s}\rVert_\infty$"
And I don't know how the equality holds there. It says "by Plancherel's theorem" but doesn't plancherel theorem only say about L^2 norm?
This is how I tried. $T_{r,s}f$ is a convolution of $K\chi$ and $f$, so I wrote down young's inequality to get the estimate $\lVert T_{r,s} \rVert_{2\rightarrow 2}\leq \lVert K\chi\rVert_1$. Since $m_{r,s}$ is a Fourier coefficient of $K\chi$, I have $\lVert m_{r,s}\rVert_\infty \leq\lVert K\chi\rVert_1$ but this doesn't seem to help me. I need your help!
Since $T_{r,s} f = (K\chi)* f$, taking the Fourier transform we have $$ \widehat{T_{r,s}f} = \widehat{K\chi} \cdot \widehat{f} = m_{r,s} \hat{f} $$
By Plancherel $$ \|T_{r,s} f\|^2_2 = \|\widehat{T_{r,s} f}\|^2_2 $$ we estimate the RHS pointwise $$ \int |m_{r,s} \hat{f}|^2 \mathrm{d}\xi \leq \|m_{r,s}\|_\infty^2 \int |\hat{f}|^2 \mathrm{d}\xi = \|m_{r,s}\|_\infty^2 \|\hat{f}\|_2^2 $$ Applying Plancherel again you get the desired estimate.