The Dink Network

Three hours is quite fast, for those questions!

After I noticed:

h(x) = h(ax) + h(bx) ==>
either h(ax) < ah(x) and h(bx) > bh(x),
or h(ax) > ah(x) and h(bx) < bh(x)
or h(ax) = ah(x) and h(ax) = bh(x)

then it became obvious that:
g(x) = ag(ax) + bg(bx) ==>

either g(ax) < g(x) and g(bx) > g(x)
or g(ax) > g(x) and g(bx) < g(x)
or g(ax) = g(bx) = g(x).

From here we find (for positive x and y):
g(x) > g(y) for all y < x ==> g(ax) and g(bx) < g(x) for all suitable a,b, a contradiction by the above.
g(x) < g(y) for all y < x ==> g(ax) and g(bx) > g(x) for all suitable a,b, a contradiction (again) by the above.
Say we have positive x, y with g(x) < g(y). Let:
S={z > 0|g(z) >= g(y)}
Say Y=inf S and Y>0. Then we can find positive Y'<Y such that g(Y')>=g(Y)>=g(y), so that Y' is in S and Y'<Y, contrary to Y' being the inf (greatest lower bound) of S. Therefore inf S = 0. By a similar argument, if:
T={z > 0|g(z) <= g(x)}
then inf T = 0. Thus the function is discontinuous at 0, a contradiction.

Really, I didn't directly use the xg(x) thing in the proof, but I never would have figured out how to do the problem if I hadn't thought of it.