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The gradient bound makes $f$ a Lipschitz function, with $|f(x)|\le |f(0)|+K\Vert x\Vert$, so its value at a Gaussian random variable is integrable and square integrable. The future Brownian increment $X_1-X_t$ is independent of ${\mathcal{F}}_t$ and has distribution $N(0,(1-t)I_d)$. Conditioning on the current location therefore givesThis is the Markov property expressed through the Brownian transition semigroup. In particular $M_0=\mu$ and $M_1=f(X_1)$.

Solving the geometric Brownian motion equation over $[t,T]$ givesThe Brownian increment is independent of ${\mathcal{F}}_t$ and has a normal distribution with variance $T-t$. Its exponential moment-generating function therefore givesThe required deterministic price function isThis includes $s=0$, $v=0$ and $t=T$. For $s>0$ and $t<T$, it is continuous and strictly decreasing in $v\ge 0$, with range $(0,\sqrt{s}]$.

If $T\le s$, then $\varphi (W_T)K$ is ${\mathcal{F}}_s$-measurable. Independence and centring of the future Brownian increment make the desired left side zero; the time multiplier on the right is zero as well.

Suppose $T>s$. Conditional on ${\mathcal{F}}_s$, write $W_T=W_s+U$ and $W_t-W_s=V$. The pair $(U,V)$ is jointly normal and independent of ${\mathcal{F}}_s$, withApply the supplied Gaussian integration by parts formula to $z\mapsto \varphi (W_s+z)$, treating the known $W_s$ as its parameter. This givesMultiply by the bounded ${\mathcal{F}}_s$-measurable $K$ and use the defining property of conditional expectation. Thus the required expectation identity holds for every $s<t$, including intervals crossing or lying after $T$.

The diffusion generator is $\mathcal{L}=\mu {\partial }_x+\frac{1}{2}{\sigma }^2{\partial }_{xx}$. By the Feynman-Kac formula, the required conditional characteristic function solvesUse separation of variables in the form $F(t,x)=e^{iux}h(t)$. Since $F_x=iuF$ and $F_{xx}=-u^{2}F$, the temporal factor satisfies the ordinary differential equationThus $h(t)=\exp [(iu\mu -\frac{1}{2}{\sigma }^2{u}^2)(T-t)]$, andThis is bounded in absolute value by one, so it lies in the uniqueness class of the Feynman-Kac formula. As an independent check, integrating the stochastic differential equation gives $X_T=x+\mu (T-t)+\sigma (B_T-B_t)$. The independent Brownian increments have the normal distribution with mean zero and variance $T-t$, yielding exactly the same characteristic function. For $\sigma =0$, the conditional distribution is a point mass at $x+\mu (T-t)$, and the same boxed expression still applies.

For $a<0$, set $v_{\infty }=-{\sigma }^2/(2a)>0$. The candidate invariant probability law of a Markov process isTo prove invariance rather than merely guess a limiting probability distribution, initialize $X_t\sim \pi$ independently of future Brownian increments. Integrating the conditional characteristic function from part (i) against this centered normal distribution givesbecause $v_{\Delta }=v_{\infty }(1-e^{2a\Delta })$. The uniqueness theorem for characteristic functions implies $X_{t+\Delta }\sim \pi$ for every $\Delta \ge 0$. This proves $\pi P_{\Delta }=\pi$, as required. The time-homogeneous Markov property then also makes the process with this Ornstein-Uhlenbeck stationary initialization a strictly stationary process.