I woud like to show that the generator of the OU process is given by
Af(x)=\frac{1}{2} \sigma^2f''(x)-\alpha x f'(x)

i know generator of brownian motion Af=1/2f''(x) and it should be something like this:

first proof that
P_{t}^{X}f(x)=P^{W}_{g(t)}f(e^{-\alpha t}x) with P_{t}^{X}, P_{t}^{W} transitions functions of OU process and BM, g(t)=\sigma^2(1-e^{-2\alpha t})/2\alpha
and then (i guess) something like
\frac{d}{dt}P_{t}f=AP_{t}f

any help/hints would be appreciated!
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edit:
OU process is defined as X_t=e^{-\alpha t}(X_0+W_{\frac{\sigma ^2}{2\alpha}(e^{2\alpha t}-1)