The situation described in this question is probably too complex to come up with any reasonably simply and accurate reply.
If however we simplify the question into "How is lift related to airspeed if angle of attack is constant?" then the answer is a bit easier.
The lift equations is L = Cl 1/2 Rho Vsquared, where V is the TAS.
If angle of attack is constant then Cl is constant. And if altitude, temperature and humidity are constant then so is Rho (if we ignore compressibility).
Taking the (now constant Cl, and 1/2Rho) out of the equation gives:
Lift is proportional to TAS squared.
So if we double the TAS we get 4 times the lift. Or to put it another way if lift equals weight at a given speed, then lift will be 1/4 of the weight at 1/2 of that speed.
The above argument is of course a simplification in that it ignores the complications of such things as upwash, downwash and effective angle of attack.