Let $S^1$ be a $U(1)$-set, and let $L^2(S^1)$ be the Hilbert space of square-integrable complex-valued functions $f : S^1 → \Bbb C$. It can be shown that the map
$$ T_{\infty} : U(1) → \text{GL}(L^2(S^1)) $$ $$ \hspace{4cm} e^{i\theta} ↦ \left(f ↦ (e^{i\phi} ↦ f\left(\left(e^{i\theta}\right)^{-1}e^{i\phi}\right))\right)$$
is a representation of U$(1)$. I'm trying to determine whether it is unitary. By that I mean, that for all $e^{i\theta} ∈ U(1)$, we have
$$T_{\infty}(e^{i\theta})^* = T_{\infty}(e^{i\theta})^{-1} = T_{\infty}(e^{-i\theta}).$$
To this end, let $f ∈ L^2(S^1)$ and $e^{i\phi} ∈ S^1 $ be arbitrarily given. If I interpret the $^*$ (i.e., the Hermitian adjoint) correctly, that would mean that we have, for all $f ∈ L^2(S^1), e^{i\phi} ∈ S^1$
$$ (T_{\infty}(e^{i\theta})^*f)(e^{i\phi}) = (T_{\infty}(e^{-i\theta})f)(e^{i\phi})= f\left(\left(e^{-i\theta}\right)^{-1}e^{i\phi}\right) = f\left(e^{i(\theta+\phi)}\right).$$
In evaluating the LHS, however, I get confused, as I find several definitions of the adjoint. In some texts, for $f : V → W$, we have $f^* : W → V$, while in others (and on Wikipedia), $f^* : W^* → V^*$.. I would say that in this case, the former actually makes more sense, as we still want to plug in functions from $L^2(S^1))$ not from its dual. Secondly, the texts say for example that "there is at most one function $f^*$ that preserves the inner product" (in the sense that $\langle f(v), w\rangle = \langle v, f^*(w)\rangle$), this function $f^*$ being the adjoint. But they don't actually give a function rule for $f^*$, so how can I evaluate $(T_{\infty}(e^{i\theta})^*f)(e^{i\phi})$..?
(Also, is $T_{\infty}$ generally called that function representation? Or are there other canonical names for it?)
$$\langle F(.+a),H\rangle=\langle F,H(.-a)\rangle$$ so the adjoint of $F\to F(.+a)$ is $H\to H(.-a)$ (and hence those operators are unitary since the adjoint is the inverse).
This works for any $L^2(G)$ where we are using the Haar measure on the (locally compact abelian) group $G$. If $G$ is not abelian it works too with the right Haar measure and $F\to F(a .),H\to H(a^{-1}.)$.