Write $P_f=\operatorname{prox}_f$ and $P_h=\operatorname{prox}_h$. Since $w_k=x_k+z_{k-1}$,
$$
y_k=P_h(w_k),
\qquad z_k=w_k-P_h(w_k),
$$
and therefore
$$
\boxed{w_{k+1}=T(w_k),
\qquad T=I-P_h+P_f(2P_h-I).}
$$
With reflected proximal maps $R_f=2P_f-I$ and $R_h=2P_h-I$,
$$
T=\frac12(I+R_fR_h).
$$
Firm nonexpansiveness of each proximal map is equivalent to nonexpansiveness of its reflection. Thus $R_fR_h$ is nonexpansive, and its average with the identity is firmly nonexpansive. This is the <Douglas–Rachford method>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 339 / 3 / b / Solution

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For two proximal maps $P_f,P_h$, define reflected maps $R_f=2P_f-I$ and $R_h=2P_h-I$. The Douglas--Rachford fixed-point map is
$$
T=\frac12(I+R_fR_h)=I-P_h+P_f(2P_h-I).
$$
Because reflected proximal maps are nonexpansive, $T$ is firmly nonexpansive.


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 Incoming links: Douglas–Rachford method