(Local time). Let $W(t)$ be a Brownian motion. The ItôDoeblin formula in differential form says that
$$
d f(W(t))=f^{\prime}(W(t)) d W(t)+\frac{1}{2} f^{\prime \prime}(W(t)) d t .
$$
In integrated form, this formula is
$$
f(W(T))=f(W(0))+\int_0^T f^{\prime}(W(t)) d W(t)+\frac{1}{2} \int_0^T f^{\prime \prime}(W(t)) d t .
$$
The usual statement of this formula assumes that the function $f^{\prime \prime}(x)$ is defined for every $x \in \mathbb{R}$ and is a continuous function of $x$. In fact, the formula still holds if there are finitely many points $x$ where $f^{\prime \prime}(x)$ is undefined, provided that $f^{\prime}(x)$ is defined for every $x \in \mathbb{R}$ and is a continuous function of $x$ (and provided that $\left|f^{\prime \prime}(x)\right|$ is bounded so that the integral $\int_0^T f^{\prime \prime}(W(t)) d t$ is defined). However, if $f^{\prime}(x)$ is not defined at some point, naive application of the Itô-Doeblin formula can give wrong answers, as this problem demonstrates.
(i) Let $K$ be a positive constant, and define $f(x)=(x-K)^{+}$. Compute $f^{\prime}(x)$ and $f^{\prime \prime}(x)$. Note that there is a point $x$ where $f^{\prime}(x)$ is not defined, and note also that $f^{\prime \prime}(x)$ is zero everywhere except at this point, where $f^{\prime \prime}(x)$ is also undefined.
(ii) Substitute the function $f(x)=(x-K)^{+}$into (4.10.42), replacing the term $\frac{1}{2} \int_0^T f^{\prime \prime}(W(t)) d t$ by zero since $f^{\prime \prime}$ is zero everywhere except at one point, where it is not defined. Show that the two sides of this equation cannot be equal by computing their expected values.
(iii) To get some idea of what is going on here, define a sequence of functions $\left\{f_n\right\}_{n=1}^{\infty}$ by the formula
$$
f_n(x)= \begin{cases}0 & \text { if } x \leq K-\frac{1}{2 n}, \\ \frac{n}{2}(x-K)^2+\frac{1}{2}(x-K)+\frac{1}{8 n} & \text { if } K-\frac{1}{2 n} \leq x \leq K+\frac{1}{2 n}, \\ x-K & \text { if } x \geq K+\frac{1}{2 n}\end{cases}
$$