$$
\operatorname{cov}(R, L)=\frac{V_{0}^{\prime}}{b} \operatorname{cov}\left(R, R^{\prime}\right)
$$
( $R$ still corresponds to an arbitrary trading strategy). Hence can be rewritten as
$$
\bar{R}-r=-\frac{V_{0}^{\prime}}{b} \operatorname{cov}\left(R, R^{\prime}\right)
$$
In particular, in the special case where you choose $H=H^{\prime}, says that
$$
\bar{R}^{\prime}-r=-\frac{V_{0}^{\prime}}{b} \operatorname{cov}\left(R^{\prime}, R^{\prime}\right)=-\frac{V_{0}^{\prime}}{b} \operatorname{var}\left(R^{\prime}\right)
$$
Using this to substitute for $V_{0}^{\prime} / b$ in where now we are back to an arbitrary trading strategy $H$, we obtain the following:
6CCM388A COURSE NOTES ：
(recall $E\left[C_{1} L / B_{1}\right]=E_{Q}\left[C_{1} / B_{1}\right]$ ). With suitable assumptions about the utility function $u$ to ensure the optimal solution of $(2.22)$ will feature strictly positive consumption values, the following first order necessary conditions must be satisfied:
$$
u^{\prime}\left(C_{0}\right)=\lambda \quad \text { and } \quad u^{\prime}\left(C_{1}(\omega)\right)=\lambda L / B_{1}
$$
Hence
$$
C_{0}=I(\lambda) \text { and } C_{1}(\omega)=I\left(\lambda L / B_{1}\right)
$$