Bounded operator and fourier transform

[Bridgeport]

Could you help?

What to show: $Tf = (m(y) \hat{f}(y) ) \check\ $ is a bounded operator on $L^2(\mathbb{R})$ iff $m(y) \in L^\infty(\mathbb{R})$

Clarifications:
$ \hat\ $ - fourier transform of f
$ \check\ $- inverse fourier transform


Spasibo!

[shwedka]

Давайте по кусочкам. Преобр. Фурье- огр. оператор в $L_2$, умножение- на ограниченную функцию - ограниченный, и обратное Пр,. Фурье ограниченный.
Композиция 3 ограниченных операторов - ограниченный.

[Bridgeport]

Very thoughtful approach!

I guess I now stack with the proof that Fourier transform is a bounded operator. ( I guess we need to use Fubini theorem and Schwartz inequality, but I don't see how)

[shwedka]

No, you have to use Bessel inequality