I was reading through Dynamic Programming by Richard Bellman today, and I got to exercise 7 in chapter one. You are asked to prove a theorem, but I feel like the theorem itself is... well, not quite wrong, but extremely misleading.
The theorem states the following:
"Consider the case where $\phi (x)$ is a monotonically increasing function which is strictly concave. Show that the solution of the corresponding functional equation,
$$f_N(c) = \max_{0\leq y \leq c} [\phi(y) + f_{N-1}(c-y)], N\geq2,$$ $$f_1(c) = \phi(c),$$
has the form $$y_N = 0, 0 \leq c\leq c_N$$ $$y_N = z_N, c>c_N$$
where $z_N$ is the unique solution of $$\phi '(y) = f_{N-1}(c-y)$$ for $N\geq 2$, and show how to determine the sequence $\{c_N\}$."
The thing is, it seems pretty obvious to me that the solution to this problem should be interior. If we take $N=2$ for example, $$f_2 = \max_{0\leq y\leq c}[\phi(y)+f_1(c-y)] = \max_{0\leq y\leq c}[\phi(y)+\phi(c-y)]$$
It's pretty obvious that $$\phi'(0) > \phi'(c)$$ so the solution should be in the interior where $$\phi'(y)=\phi'(c-y)$$ which given the assumption of strict concavity implies that $y = {c\over{2}}$.
A similar, more tedious calculation shows that, since $$\phi'(y_2) = \phi'(y_1)$$ $\phi'(y_3)$ also is equal to $\phi'(y_1)$.
Inductively, it seems like $y_i = {c\over N}$ should be accurate. I suppose that the theorem would still be "accurate", since we can just say that $c_N = 0$ for all $N$, but I feel like it's more likely that I'm wrong.
Either that or Bellman the Master is wrong, which could be the case given that this book never had a revision.