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?

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

1 Answer 1


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.