## Torus Fourier Multiplier Operators

Fix a bounded complex sequence of numbers $(\lambda_{n})_{n=1}^{\infty}$ and define a Hilbert space operator $T: L^{2}([-\pi,\pi]) \rightarrow L^{2}([-\pi,\pi])$ by

$\displaystyle Tf(x) := \sum_{n \in \mathbb{Z}}\lambda_{n}a_{n}e^{inx}, \indent \forall f = \sum_{n \in \mathbb{Z}}a_{n}e^{inx} \in L^{2}([-\pi,\pi])$

We say that $T$ is a Fourier multiplier operator, and we refer to $(\lambda_{n})_{n=1}^{\infty}$ as the multiplier sequence.

It remains for us to verify that $T$ is, indeed, a bounded operator. I claim that $\left\|T\right\| = \sup_{n}\left|\lambda_{n}\right|$. It is easy to see that $\left\|T\right\| \geq \sup_{n}\left|\lambda_{n}\right|$, since

$\displaystyle Te^{inx} = \lambda_{n}e^{inx}, \indent \forall n \in \mathbb{Z}$

which has $L^{2}$-norm $\left|\lambda_{n}\right|$. For the reverse inequality, we apply Parseval’s identity to obtain

$\displaystyle \left\|Tf\right\|_{L^{2}} = \left(\sum_{n \in \mathbb{Z}}\left|\lambda_{n}a_{n}\right|^{2}\right)^{\frac{1}{2}} \leq \sup_{n}\left|\lambda_{n}\right|\left(\sum_{n \in \mathbb{Z}}\left|a_{n}\right|^{2}\right)^{\frac{1}{2}}= \sup_{n}\left|\lambda_{n}\right|\left\|f\right\|_{L^{2}}$

Fourier multiplier operators have the nice property of commuting with translations: if we define $\tau_{h}f(x) = f(x-h)$, then

$\displaystyle T \circ \tau_{h} = \tau_{h} \circ T, \indent \forall h \in \mathbb{R}$

To see this property, let $f = \sum_{n \in \mathbb{Z}}a_{n}e^{inx} \in L^{2}([-\pi,\pi])$. Then, for $h \in \mathbb{R}$ given,

$\displaystyle T \circ \tau_{h}f(x) = \sum_{n\in \mathbb{Z}}\lambda_{n}a_{n}e^{in(x-h)} = \tau_{h}\left(\sum_{n \in \mathbb{Z}}\lambda_{n}a_{n}e^{inx}\right) = \tau_{h}(Tf)(x) = \tau_{h}\circ Tf(x)$

Perhaps more interesting is that the converse also holds. Namely, if $T: L^{2}([-\pi,\pi])$ is a bounded operator which commutes with translations, then $T$ is a Fourier multiplier operator. Let $f \in L^{2}([-\pi,\pi])$ be the function $f(x)=e^{inx}$, for $n \in \mathbb{Z}$. For $h \in \mathbb{R}$ and $x \in [-\pi,\pi]$,

$\displaystyle \begin{array}{lcl}\sum_{m \in \mathbb{Z}}\lambda_{m}e^{im(x-h)} = \tau_{h} \circ Tf(x)= T \circ \tau_{h}f(x)= T(e^{in(x-h)}) &=& e^{-inh}Tf(x)\\&=& e^{-inh}\sum_{m \in \mathbb{Z}}\lambda_{m}e^{imx}\end{array}$

where $(\lambda_{m})_{m \in \mathbb{Z}} \in \ell^{2}$. Thus,

$\displaystyle \sum_{m \in \mathbb{Z}}\lambda_{m}(e^{-imh} - e^{-inh})e^{imx} = 0 \Longrightarrow \lambda_{m}(e^{-imh}-e^{-inh}) = 0, \indent \forall m \in \mathbb{Z}$

For each $m \neq n$, we can choose an $h \in \mathbb{R}$ such that $e^{-imh} \neq e^{-inh}$ and therefore $\lambda_{m} = 0$. Thus, we see that $Tf(x) = \lambda_{n}e^{inx}$ and therefore $\lambda_{n} = Tf(0)$. Since $T$ is bounded and therefore continuous, we conclude that

$\displaystyle Tf(x) = T\left(\lim_{N \rightarrow \infty}\sum_{n=-N}^{N}a_{n}e^{inx}\right) = \lim_{N\rightarrow\infty}\sum_{n=-N}^{N}a_{n}Te^{inx} = \sum_{n \in \mathbb{Z}}\lambda_{n}a_{n}e^{inx}$