From Osher Doctorow

Let's consider a representation of the magnitudes of physical
variables (extended to include geometric ones) in which 1 represents
infinity and 0 is the least magnitude.

I've argued previous that either the speed of light is infinite (or 1
on this scale) or that it is just a finite phase-transition related
quantity with actual velocities extending to infinity (although
possibly on an imaginary scale for "superluminal" phase).

Here let's examine a different possibility, namely that the
superluminal and subluminal domains are equal in magnitude ranges, or
in other words that the speed of light c = 1/2 in the above
notation. Whether the superluminal domain is "purely geometric" or
not won't be argued here.

Now look at the unit square with one corner at the origin and a
diagonally opposite corner at (1, 1) where the coordinates are the
usual planar (x, y).

Consider a function f of x and y:

1) f(x, y)

where f represents in some way a curve or surface over the first
quadrant of the x-y plane and over the unit square described above.
Since the diagonal of the unit square is y = x with dy/dx = 1, if f
rose faster than 1 then it would hit the top horizontal boundary of
the unit square somewhere on the line segment y = 1 with x in (0,
1). However, this means that for the finite x at which the curve/
surface hits the boundary, y is infinite, raising the question of what
prevents the curve/surface from "tunneling through" past the boundary
"above y = infinity". The problem doesn't arise for (1, 1) because
both x and y are infinite, and we wouldn't intuitively expect either x
or y to change further.

So arguably the maximum slope of the infinite curve f or an infinite
curve on surface f is 1, and in the same units of x and y so to speak,
f should be = 1 and = 1 at its maximum.

Now let's start all over and choose f(x, y) = 1 to maximize the curve
or surface but add the condition that it is maximized over the
diagonal line y = x, yielding:

2) max(f(x,y)) = 1 + lambda(y - x) (Lagrange Multiplier method)

It is fairly easy to see that Probable Influence/Causation P(A--B)
is:

3) P(A--B) = 1 + y - x = max(f(x,y)) of (2)

and that it satisfies P(A--B) = 1 on y = x. The partial derivative
of P(A--B) with respect to y is 1, so that P(A--B) like f doesn't
rise faster than 1 but rises with rate 1 with respect to y, and the
partial derivaitve of P(A--B) with respect to x is -1 which has
similar behavior but decreases with x.

Osher Doctorow