If $A:L^2([0,2\pi])\to L^2([0,2\pi])$ is translation invariant, how to show that it is a multiplier with $\ell^\infty$ symbol? 
Let $A:L^2([0,2\pi])\to L^2([0,2\pi])$ be continuous and linear and satisfy $AT_a=T_aA$ for all $a\in \mathbb{R}$. Show there is a bounded sequence $\{x_n\} $ such that $\widehat{Af}(n)=x_n\hat{f}(n)$ for all $f\in L^2([0,2\pi])$ 

My work so far: 
$$\widehat{Af}(n) = \frac{1}{2\pi}\int_0^{2\pi} Af(x)e^{-inx}dx
 = \frac{1}{2\pi}\int_0^{2\pi} Af(x)e^{-in(x+a)}e^{ina}dx \\= \frac{1}{2\pi}\int_0^{2\pi} AT_af(x+a)e^{-in(x+a)}e^{ina}dx = e^{ina}\frac{1}{2\pi}\int_0^{2\pi} T_aAf(x+a)e^{-in(x+a)}dx = e^{ina}\langle T_aAf,e^{inx}\rangle = e^{ina}\langle Af, T_{-a}e^{inx}\rangle = e^{ina}\langle Af,e^{in(x+a)}\rangle $$
I'm thinking $x_n=e^{ina}$ but I'm unsure how to show the last is equal to $\hat{f}(n)$. Any help?
I have also shown $\langle Ae^{inx},e^{ikx}\rangle=0$ for $k\neq n$.
 A: 
I have shown $\langle Ae^{inx},e^{ikx}\rangle=0$ for $k\neq n$.

The notation $\langle Ae^{inx},e^{ikx}\rangle$ is a little dangerous, and might be better written as $\langle A(e^{in\bullet})(x),e^{ikx}\rangle$ or $\int A(e^{in\bullet})(x)e^{-ikx}dx$, but we will continue with the notation of the OP.
Let $f(x)= \sum_k \hat f(k) e^{ikx}$ be an arbitrary $L^2$ function. Then since $A$ is bounded and linear,
$$\widehat{ Af}(n)= \langle Af ,e^{inx}\rangle =  \sum_k^{\phantom{1^1}} \hat f(k) \langle A e^{ikx} ,e^{inx}\rangle = \hat f(n)\langle A e^{inx} ,e^{inx}\rangle,  $$
so it remains to show that $x_n := \langle A e^{inx} ,e^{inx}\rangle$ is bounded.
But by Cauchy-Schwarz,
$$  |x_n| \le \|Ae^{inx}\|_{L^2} \|e^{inx}\|_{L^2} \le \|A\|_{L^2\to L^2} \|e^{inx}\|^2_{L^2},$$
and $\|e^{inx}\|^2_{L^2}<C<\infty$ for a constant $C$ that does not depend on $n$ . Therefore, $(x_n)_{n\ge 0} \in \ell^\infty$.

For completeness here is a proof of the statement in the quotes. Note
$$ \langle Af ,e^{inx}\rangle = \langle T_{-a}AT_af ,e^{inx}\rangle = \langle AT_af ,T_ae^{inx}\rangle = e^{ina}\langle AT_a f, e^{inx}\rangle.$$Specialising to $f(x) = e^{ikx}$, linearity gives $A(T_a e^{ik\bullet})=A(e^{ik\bullet}e^{-ika}) = e^{-ika}A e^{ik\bullet} $. So we end up with
$$  \langle Ae^{ikx} ,e^{inx}\rangle =e^{ia(n-k)}\langle Ae^{ikx} ,e^{inx}\rangle$$
Choosing $a$ so that $e^{ia(n-k)} \neq 1$, possible since $n\neq k$, we get that
$$ (1-e^{ia(n-k)})\langle Ae^{ikx} ,e^{inx}\rangle = 0 $$
and hence  $\langle Ae^{ikx} ,e^{inx}\rangle=0$.
