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}

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