= Solution
For positive smooth $f$ with the stated decay, <integration by parts> gives
$$
\boxed{\int f''\log f\,dv=-\int\frac{|f'|^2}{f}\,dv}
$$
and, using unit <mass>,
$$
\boxed{\int(fv)'\log f\,dv=-\int vf'\,dv=\int f\,dv=1.}
$$
For zero values of $f$, these calculations can be made with the positive unit-mass approximation $(f+\varepsilon\gamma)/(1+\varepsilon)$ and then passed to the limit whenever the quantities are finite. Positive-time solutions also have the usual Gaussian smoothing.
Since $-\log\gamma=v^2/2+\tfrac12\log(2\pi)$, <mass> conservation gives
$$
H(f_t\mid\gamma)=\int f_t\log f_t\,dv+
\frac12\int v^2f_t\,dv+\frac12\log(2\pi).
$$
The derivative of the first term is $\int\partial_tf_t\log f_t$, because $\int\partial_tf_t=0$. The two identities above and the <energy> equation in (b), with $d=M=1$, yield
$$
\frac d{dt}H(f_t\mid\gamma)
=-\int\frac{|f_t'|^2}{f_t}\,dv+1+
\left(1-\int v^2f_t\,dv\right).
$$
Now expand the <relative Fisher information>:
$$
I(f\mid\gamma)=\int\left|\partial_v\log(f/\gamma)\right|^2f\,dv
=\int\left(\frac{f'}f+v\right)^2f\,dv
=\int\frac{|f'|^2}{f}\,dv+\int v^2f\,dv-2,
$$
where $\int vf'=-1$. \b[Thus the <entropy dissipation identity for Ornstein-Uhlenbeck flow> is]
$$
\boxed{\frac d{dt}H(f_t\mid\gamma)=-I(f_t\mid\gamma)\leq0.}
$$
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