Let $m:\Bbb{R}^d \to \Bbb{C}$ be the Fourier multiplier define the cooresponding operator $T_m$ as $$f \mapsto \mathcal{F}^{-1}(m\hat{f})$$ for $f\in \mathcal{S}(\Bbb{R}^d)$
Prove $T_m$ can be extended to $L^2\to L^2$ bounded if and only if $m \in L^\infty$
My attempt:
First use the Plancherel's theorem we have $$\|T_mf\|_2 = \|m\hat{f}\|_2 \le \|m\|_\infty\|f\|_2 $$
Which implies $\|T_m\|_{2\to 2} \le \|m\|_\infty$, I need to prove the other direction, I have no idea how to prove $\|m\|_\infty \le \|T_m\|_{2\to 2}$, maybe we can find $f$ to prove that the inequality above can be equal.