Example

The mapping $\Psi(z)f\colon=f(zw)$ is a unitary representation of $S^1=\UU(1)$ in the space $L_2(\C,\g)$ with the gaussian measure $\g(dw)=(2\pi)^{-1}e^{-|w|^2/2}\,dw$. 2. Verify by means of exam that for smooth functions $f$: $$ \ttdl t0\Psi(e^{it})f(w)=iw\pa_wf(w)-i\bar w\pa_{\bar w}f(w) $$

We will show that $\Psi(z)$ is unitary for all $z\in S^1 = U(1)$. The adjoint operator of $\Psi(z)$ is given by $\Psi(z)^* = \Psi(z^{-1}) = \Psi(\bar z)$ because for all $f,g\in L_2(\C,\gamma)$ we have \begin{eqnarray*} \langle \Psi(z)f,g \rangle&=& \int_{\C} f(zw)\overline{g(w)}(2\pi)^{-1}e^{-|w|^2/2}\, dw\\ &=&\int_{\C} f(zw)\overline{g(\bar{z}zw)}(2\pi)^{-1}e^{-|zw|^2/2}\, dw\\ &=&\int_{\C} f(w)\overline{g(\bar{z}w)}(2\pi)^{-1}e^{-|w|^2/2}\, dw = \langle f,\Psi(\bar z)g \rangle. \end{eqnarray*} Here we used the fact that $w\mapsto zw$ is a rotation for $|z|=1$, so by the rotational invariance of the Lebesgue measure we can replace $zw$ by $w$ in the integral. Now for all $f\in L_2(\C,\gamma)$ we have $\Psi(z)^*\Psi(z)f = f(z\bar{z}w) = f(w)$, so $\Psi(z)^*\Psi(z) = 1$ (and analogously $\Psi(z)\Psi(z)^* = 1$, which implies that $\Psi(z)$ is not only injective, but also surjective). Therefore, $\Psi(z)$ is a unitary operator on $L_2(\C,\gamma)$ for all $z\in U(1)$, and since $\Psi$ is clearly a group homomorphism, $\Psi$ is a unitary representation of $U(1)$ in $L_2(\C,\gamma)$.
Using exam, we get \begin{eqnarray*} \frac{d}{dt}[\Psi(e^{it})f(w)]&=&\frac{d}{dt}f(e^{it}w) = (\partial_w f)(e^{it}w)\cdot \frac{d}{dt}(e^{it}w) + (\partial_{\bar w} f)(e^{it}w)\cdot \frac{d}{dt}(e^{-it}\bar w)\\ &=&(\partial_w f)(e^{it}w)\cdot ie^{it}w + (\partial_{\bar w}f)(e^{it}w)\cdot (-i)e^{-it}\bar w. \end{eqnarray*} Therefore, $$ \ttdl t0\Psi(e^{it})f(w) = iw\partial_w f(w) - i\bar{w}\partial_{\bar w}f(w). $$