= Solution
Enlarge the space if needed to choose $Y_0$ with density $p$, independent of the driving <Brownian motion>, and solve the same equation with the same driver as $X_0=x$. The second moment of $p$ makes the initial difference square integrable. By parts (a) and (b),
$$
\mathbb E f(Y_t)=\int f(y)p(y)\,dy,
$$
while part (c) gives
$$
\mathbb E|X_t-Y_t|^2\leq D_xe^{-kt},\qquad D_x=\int(x-y)^2p(y)\,dy<\infty.
$$
We must not assume that the smooth bounded $f$ is globally Lipschitz. The hypotheses give $C_Y=\sup_t\mathbb EY_t^2<\infty$, hence <uniform tightness> of $(Y_t)$. For any $R$ and $0<\delta\leq1$, define the modulus of continuity of $f$ on $[-R-1,R+1]$ by $\omega_R(\delta)$. Splitting according to $|Y_t|\leq R$ and $|X_t-Y_t|\leq\delta$, and applying <Markov inequality>, gives
$$
\mathbb E|f(X_t)-f(Y_t)|
\leq\omega_R(\delta)+2\|f\|_\infty\left(\frac{C_Y}{R^2}+\frac{D_xe^{-kt}}{\delta^2}\right).
$$
First choose $R$ large, then $\delta$ small using uniform continuity on that compact interval, and finally let $t\to\infty$. The right side can be made arbitrarily small. Together with part (a), this proves
$$
\boxed{u(t,x)\longrightarrow\int_{\mathbb R}f(y)p(y)\,dy\quad\text{for every }x.}
$$
This is <convergence by synchronous coupling for bounded continuous test functions>. It only needs the constant expectation from part (b) and uniform second moments; it does not add an unstated global bound on $f'$ or require a separate stationarity theorem. As explained in part (c), the literal bounded-drift plus strict-contraction assumptions have no global example; the calculation records the intended consequence under compatible dissipative-drift hypotheses as well.
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