$L^2$ boundness of the operator $T$ associated to the Calderon-Zygmund kernel Given a Calderon-Zygmund kernel $K$ we can define the convolution $$Tf = \lim_{\epsilon \to 0} \int_{|x-y|>\epsilon} K(x-y) f(y)dy$$
For $0<r<s<\infty$, Define the operator $T_{r,s}$ as :
$$T_{r,s}f(x) = \int_{\Bbb{R}^d} K(y)\chi_{[r<|y|<s]}f(x-y)dy$$
with the Fourier transformation of the truncated kernel as :
$$m_{r,s}(\xi) = \int e^{-2\pi ix\cdot \xi}\chi_{[r<|y|<s]}K(x)dx$$
Prove $\|T_{r,s}\|_{2\to 2}<\infty$  if and only if $m_{r,s} \in L^\infty$
Just as the post here, we can easily show $\|T_{r,s}\|_{2\to 2} \le \|m_{r,s}\|_\infty$
, I have no idea how to show the other direction?
here is a post with same question ,but only prove one direction?What about the other direction?
 A: Let $m$ be a measurable function, and $M$ be the multiplier operator
$$
\widehat{Mf}(p) = m(p)\widehat{f}(p).
$$
We will see that $\|M\|_{L^2\to L^2} = \|m\|_{L^\infty}$.  You already know how to show that $\|M\|_{L^2\to L^2}\leq \|m\|_{L^\infty}$, so we will demonstrate the opposite direction.
First we consider the case that $m\not\in L^\infty$, and check that $M$ is unbounded.  Let $E_N$ be the set on which $|m(p)|\geq N$, which has nonzero measure for every $N>0$.  In particular $E_N\cap B_R$ has nonzero measure for some radius $R=R(N)>0$.  Let $g_N\in L^2(\mathbb{R}^n)$ be the function
whose Fourier transform is the indicator function for this set, so that $\widehat{g_N} = \mathbb{1}_{E_N\cap B_R}$.   Then
$$
\|M g_N\|_{L^2}^2 = \int_{E_N\cap B_R} |m(p)|^2 \,dp
\geq N^2 \int_{E_N\cap B_R}\,dp = N^2 \|g_N\|_{L^2}^2.
$$
This shows that $\|Mg_N\|_{L^2} \geq N\|g_N\|_{L^2}$ for every $N$, and therefore $M$ is not bounded on $L^2$.
Now suppose that $m\in L^\infty$.  Let $E_\delta$ be the set $\{p \mid |m(p)| \geq \|m\|_{L^\infty}-\delta\}$.  For every $\delta>0$, $E_\delta$ has positive measure and for some $R=R(\delta)$, $E_\delta \cap B_R$ has nonzero but finite measure.  Let $g_\delta$ be the function satisfying
$$
\widehat{g_\delta} = \mathbb{1}_{E_\delta \cap B_R}.
$$
Then the same computation as above shows that
$$
\|Mg_\delta\|_{L^2} \geq (\|m\|_{L^\infty}-\delta)\|g_\delta\|_{L^2}.
$$
Therefore, for every $\delta\geq 0$ we have
$$
\|M\|_{L^2\to L^2} \geq \|m\|_{L^\infty}-\delta,
$$
and this completes the proof.
