How to prove the convexity of the function

Question

A consumer has a utility function of the form $U(x; y) = x^a +y^b$ where both $a$ and $b$ are non-negative. What additional restrictions on the values of the parameters $a$ and $b$ are imposed by each of the following assumptions?
(i) Preferences are quasi-linear, convex, and $x$ is a normal good.
(ii) Preferences are homothetic.
(iii) Preferences are homothetic and convex.
(iv) Goods $x$ and $y$ are perfect substitutes.

Answer:
(i) a = 1 and b is between 0 and 1.
(ii) a = b.
(iii) a = b and a is between 0 and 1.
(iv) a = b = 1.

In the third part of the Question

For homothetic preferences, MRS has to depend on ratio of $x$ and $y$.
Therefore $a=b$.
But I don't understand, the restrictions on the parameter to prove convexity that $a$ has to lie between $0$ and $1$. If we use the bordered hessian method, shouldn't a be greater than $1$.
I used the borderred hessian to compute $fxx$,$fyy$ and $fxy$ and I think it should be that $a >1$.
But the answer is wrong?

please help :(

Answer 1

You have correctly found conditions under which the utility function $U(x,y)$ is convex. However, the question concerns the convexity of preferences, not of the function $U(x,y)$.

Traditionally, economists interpret convexity of preferences as the requirement that the indifference curves be convex; generally, this means that the utility function $U(x,y)$ be quasi-concave. Here, it seems that there is the stronger assumption that $U(x,y)$ is concave.

Answer 2

Preferences are convex if and only if the utility function that represents these preferences is quasi-concave. So you only need to show that $U(x,y)$ is quasi-concave. Simon-Blume, Theorem 21.20 characterizes the quasi-concavity in terms of a single determinant (I assume you have this book) being $>0$ for all $x,y$. In your case, since $U(x,y)= x^a + y^a$, this determinant will be so only if $a<1$.

Intuitively, since $U(x,y)$ is sum of two functions, these functions should be concave in order to have a concave $U(x,y)$, which requires again $a<1$.