# Smooth pasting conditions for optimal investment with transactions costs

I'm reading this paper relating to optimal investment with transaction costs where some value function $$F(x)$$ is optimized. At some boundary $$x=u$$ it will be optimal to pay a proportional cost $$C$$ which gives the boundary condition

$$\begin{equation} F'(u) = -C \end{equation}$$

The author argues that optimality also implies a boundary condition for the second derivative

$$\begin{equation} F''(u) = 0 \end{equation}$$

but I'm struggling to understand why this is the case. Any hints that will help me understand the intuition behind this condition?

• Does $C$ depend on $u$? Jul 6, 2020 at 8:56
• $C>0$ is just some constant. Jul 6, 2020 at 9:00
• I'm confused, doesn't the second equation follow from $F'$ being constant? Jul 6, 2020 at 9:11
• It's only constant at $x=u$. Note that $x$ is the variable and $x=u$ is the boundary. Jul 6, 2020 at 9:13

I think that they are saying that, at special point $$u$$, we have:

$$F'(u) = -\rho.$$

Also, that in a neighborhood from the left, we have:

$$F'(u- dU) = -\rho$$

for any small positive $$dU$$.

Then, with Taylor on the left:

$$F'(u) - F''(u)dU = - \rho$$

Hence: $$F''(u) = 0$$

Basically, if the first derivative of a function is constant in a neighborhood of a point, then its second derivative must be null at that point.

• Why is $F'(u- dU) = -\rho$ true? Jul 6, 2020 at 21:06
• Equation (10) it seems. Setting $v$ to $u-dU$.
– ir7
Jul 6, 2020 at 23:36