Can you show us what you have tried? The basic definition of convexity is that f(at + (1-t)b) is in the codomain of the function (i.e. the values that actually get output from the function), where a and b are just points in the domain of the function and t is a scalar number ranging from 0 to 1 inclusive.
If a function is defined everywhere and takes on every possible value then it will be convex.
I see your point applied to the case of a single variable function.
For a multivariate function, I have followed either of two approaches (not sure if I am right in the conclusion about convexity in case 1.) :
F(G,W,C,M) = G^a ·W^b· C^c· M^d is an homogeneous equation of degree a+b+c+d .
Therefore for a scalar number t:
F (tG,tW,tC,tM) = (t^(a+b+c+d))·(G^a)·(W^b)·(C^c)·(M^d)
If a+b+c+d>1, equation F presents increasing returns to scale (this is the interpretation from an “economic” standpoint). Therefore, we can state that the function is convex.
The equation I have posted gives a+b+c+d =-2,5779+46,622-15,958+37,241 = 65,3271 > 1 shows increasing returns to scale. Thus it is convex.
In reality a more accurate determination of convexity and/or quasi convexity would be given by demonstrating that the hessian of function F is H(F) > 0. This alternative requires the obtention of all the second derivatives of F and the calculation of the determinant of a 4x4 matrix which means a long and tiresome algebraic development. Have not been able to complete such a huge job (at the moment).
Is it correct to conclude convexity following the reasoning in 1. concerning homogeneity of the function and increasing returns ?
The difference between single and multivariable functions is that a and b are vectors, but t is still a scalar in the [0,1] range.
If you want a conclusive proof, then the suggestion of using the Hessian would be a good one. If you can calculate the partial derivatives (just keep one variable varying and the rest are constant) then you can calculate the Hessian and its determinant.