. It was asserted at the end of Subsection 8.3.3 and established in Exercise 8.3 that $v_{L .}(x)$ given by (8.3.13) is the only bounded continuous function having a continuous derivative and satisfying the linear complementarity conditions (8.3.18)-(8.3.20). There are, however, unbounded functions that satisfy these conditions. Let $0<L<K$ be given, and assume that
$$
\frac{2 r}{2 r+\sigma^2} K>L \text {. }
$$
(i) Show that, for any constants $A$ and $B$, the function
$$
f(x)=A x^{-\frac{2 r}{\sigma^2}}+B x
$$
satisfies the differential equation
$$
r f(x)-r x f^{\prime}(x)-\frac{1}{2} \sigma^2 x^2 f^{\prime \prime}(x)=0 \text { for all } x \geq 0 .
$$
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8 American Derivative Securities
(ii) Show that the constants $A$ and $B$ can be chosen so that
$$
f(L)=K-L, \quad f^{\prime}(L)=-1 .
$$
(iii) With the constants $A$ and $B$ you chose in (ii), show that $f(x) \geq(K-x)^{+}$ for all $x \geq L$.
(iv) Define
$$
v(x)=\left\{\begin{array}{l}
K-x, 0 \leq x \leq L, \\
f(x), \quad x \geq L .
\end{array}\right.
$$
Show that $v(x)$ satisfies the linear complementarity conditions (8.3.18)$(8.3 .20)$, but $v(x)$ is not the function $v_{L .}(x)$ given by (8.3.13).
(v) Every solution of the differential equation (8.8.7) is of the form (8.8.6). In order to have a bounded solution, we must have $B=0$. Show that in order to have $B=0$, we must have $L=\frac{2 r}{2 r+\sigma^2} K$, and in this case $v(x)$ agrees with the function $v_{L_*}(x)$ of (8.3.13).