Fix a bounded complex sequence of numbers and define a Hilbert space operator by

We say that is a **Fourier multiplier operator**, and we refer to as the **multiplier sequence**.

It remains for us to verify that is, indeed, a bounded operator. I claim that . It is easy to see that , since

which has -norm . For the reverse inequality, we apply Parseval’s identity to obtain

Fourier multiplier operators have the nice property of commuting with translations: if we define , then

To see this property, let . Then, for given,

Perhaps more interesting is that the converse also holds. Namely, if is a bounded operator which commutes with translations, then is a Fourier multiplier operator. Let be the function , for . For and ,

where . Thus,

For each , we can choose an such that and therefore . Thus, we see that and therefore . Since is bounded and therefore continuous, we conclude that

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