Is it true that the cost of a closed walk is nonnegative if all its cycles have nonnegative costs? Consider a directed graph $G$ and a closed (not necessarily directed) walk $\omega$ constructed from some edges of $G$. The term "not necessarily directed" means here that $\omega$ is a walk of the underlying undirected graph of $G$.
Let $\kappa^+(\omega)=\sum\limits_{i=1}^n(v_i + S_i u_i)$ and $\kappa^-(\omega)=\sum\limits_{i=1}^n(v_i-S_i u_i)$ be the cost and anti-cost of $\omega$, respectively, where $u_i$ and $v_i$ are two nonnegative parameters for the $i$th edge within $\omega$ and $S_i$ is a sign for the $i$th edge. Here $S_i=1$, if the $i$th edge points in the direction of the endpoint of $\omega$, otherwise $S_i=-1$.
My question is whether it is true that $\kappa^\pm(\omega) \ge 0$ if $\kappa^\pm(\chi) \ge 0$ for all $\chi \subseteq \omega$ (not necessarily directed) cycles.
If no, then what other $\psi \subseteq \omega$ subwalks should satisfy $\kappa^\pm(\psi) \ge 0$ to achieve $\kappa^\pm(\omega) \ge 0$?
 A: We can prove this by strong induction on the length of the walk. (When $\omega$ is a closed walk of length $0$, its cost is of course nonnegative.)
In general, suppose $\omega$ is a closed walk of length $n$ visiting vertices $v_0, v_1, v_2, v_3, \dots, v_n = v_0$. I am not requiring these vertices to be distinct, of course. Now pick the pair $(v_i, v_j)$ with $v_i = v_j$, $i<j$, and $j$ as small as possible: the first time that $\omega$ revisits a vertex.
The sub-walk $\chi$ visiting vertices $v_i, v_{i+1}, \dots, v_j$ is a cycle, because none of its vertices can be repeats, and so we know that the total cost of its edges $v_i v_{i+1}, \dots, v_{j-1} v_j$ is nonnegative. Deleting it from $\omega$, we get the sub-walk $\omega'$ visiting vertices $v_0, v_1, \dots, v_i, v_{j+1}, \dots, v_n$ which is also a closed walk. In fact, all cycles contained in $\omega'$ are nonnegative, because they were also contained in $\omega$.
So the total cost of $\omega'$ is nonnegative, by the inductive hypothesis (it's shorter). Since the total cost of $\omega$ is the sum of the costs of $\chi$ and $\omega'$, we know that the total cost of $\omega$ is nonnegative, proving the result we wanted.
We did not assume anything about the costs beyond the fact that edges have them, so this works for your cost and anti-cost equally well.

One caveat: you might want to allow your walks to backtrack (going from a vertex $v$ to another vertex $w$ back to the same vertex $v$) but not allow the same for a cycle. In that case, the argument above no longer quite works: the sub-walk $\chi$ might not be a cycle in that case.
In this case, $\chi$ will have length $2$ and take edge $vw$ followed by edge $wv$. Your formulas for $\kappa^+$ and $\kappa^-$ have a "constant" cost and a "direction-dependent" cost; let these be $c$ and $d$ for the edge in question. Then $\kappa^+(\chi)$ and $\kappa^-(\chi)$ will both have cost $(c+d) + (c-d) = (c-d) + (c+d) = 2c$ for traversing this edge once in each direction. As long as the "constant" cost $c$ is always nonnegative, which is assumed, the cost of $\chi$ will still be nonnegative.
So in this case, the result still holds, though we have to use a bit more information about the details of the costs.
