A nonlinear mapping on $L^2(S^1)$ that commutes with all translation operators is necessarily measurable?

Question

Let $H:= L^2(S^1)$, where $S^1$ is the circle, and $\tau_a : H \to H$ be the translation operator for each $a \in S^1$: \begin{equation} (\tau_a f)(x):= f(x+a) \end{equation}

Then, it is clear that each $\tau_a$ is a surjective linear isometry on $H$. Now, suppose that $F : H \to H$ is some nonlinear mapping satisfying \begin{equation} F( \tau_a f) = \tau_a F(f) \text{ for all } a \in S^1 \text{ and }f \in H \end{equation}

Then, I wonder if such $F$ is necessarily measurable. The definition of (weak) measurability is presented in wikipedia.

By Pettis Theorem, weak meausrability of $F$ is equivalent to strong measurability since $H$ is separable.

Could anyone please help me? This seems like a highly nontrivial question.

Answer 1

For $f \in H$, call $Mf$ the mean value of $f$ on $\mathbb{S}_1$. Let $\phi : \mathbb{R} \to \mathbb{R}$ be any non-Borel function. Call $\mathbb{1} \in H$ the constant function equal to $1$ everywhere. Then $f \mapsto \phi(Mf)\mathbb{1}$ is a non-measurable map from $H$ to $H$ which commutes with translations.