From the result above and the substitution \( u = s + t \), \[ P_t U_\alpha f = \int_0^\infty e^{-\alpha s} P_{s+t} f \, ds = \int_t^\infty e^{-\alpha (u - t)} P_u f \, du = e^{\alpha t} \int_t^\infty...From the result above and the substitution \( u = s + t \), \[ P_t U_\alpha f = \int_0^\infty e^{-\alpha s} P_{s+t} f \, ds = \int_t^\infty e^{-\alpha (u - t)} P_u f \, du = e^{\alpha t} \int_t^\infty e^{-\alpha u} P_u f \, du \] Hence \[ \frac{P_t U_\alpha f - U_\alpha f}{t} = \frac{1}{t} \left[e^{\alpha t} \int_t^\infty e^{-\alpha u} P_u f \, du - U_\alpha f\right] \] Adding and subtracting \( e^{\alpha u} U_\alpha f \) and combining integrals gives \begin{align*} \frac{P_t U_\alpha f - U_\al…
