Question About an Application of Implicit Function Theorem

nonmathtype

Hi everyone,

I do economics but am very poor at Math. I had a specific and perhaps silly question about the implicit function theorem, but will be grateful for an urgent response.

Suppose we have a function, U(x, y).
x and another variable z are linearly related so the function can also be specified as U(z,y) by substituting z for x.

It can be shown by using the implicit function theorem that y = f(x), and also separately that y = g(z) such that U(x,y)=0 and U(z,y)= 0 respectively.

Is it them possible to conclude that y = h(x,z) exists ?

Thank you in advance !

chiro

Hey nonmathtype and welcome to the forums.

You haven't specified what the function h is not explicitly or implicitly as a relationship to your other functions. What properties of h did you have in mind?

nonmathtype

Hi Chiro, thank you so much for your response.

Actually I am not interested in any specific functional form or properties of h. All I am looking for is its existence.

So it is clear that both f and g will exist by the implicit function theorem as mentioned above (because the partial derivative of U with respect to y is not zero, and U is continuously differentiable by assumption. These are both functions with a single argument (single variable functions) i.e y = f(x) and y = g(z).

Does this imply that a function that contains both x and z in the domain will exist as well i.e y = h(x,z) ?

chiro

You should be able to do that both for explicit and implicit representations of the functions.

For example if you can't get z = r(x) (i.e. an explicit function of x) then provided you have a function for y = f(x) and y = g(z), then you can add them together and divide by 2 to get y = 1/2(f(x) + g(z)) = h(x,z).

I'm assuming your assumptions where you have existence of y = f(x) and y = g(z) and I'm assuming they point to the same variable y.

Also since you didn't mention the functional form, I assume you just want to find any function with h(z,x) so the one provided which is a linear combination of the two solutions does satisfy the requirement at least.

nonmathtype

Thanks a lot chiro, it is much appreciated :-) Yes your assumptions about what I was trying to say are correct...both y = f(x) and y = g(z) point to the same y.

I did not realize that it could be so obvious ! I was worried regarding the fact that the domain of both f and g is R1, whereas the domain of the proposed function h is R2. So I did not think that the existence of h on the basis of existence of f and g would follow in such a straightforward way.

Once again, thanks a lot !