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A follow up to my previous question:

Given $u(x) = (x_1-b_1)^\alpha (x_2-b_2)^\beta(x_3-b_3)^\gamma$

and ending up with from a maximization that

$$x(p,w) = (b_1, b_2, b_3) + (w - (b \cdot p))\left(\frac{\alpha}{p_1},\frac{\beta}{p_2},\frac{\gamma}{p_3}\right)$$

How do I show that $x(p, w)$ is convex? I already showed that it is homogeneous of degree zero and satisfies Walras' Law. The book from Mas-Colell says that the solution is obvious, but I have not gotten anywhere with it. My setup for showing convexity is that I want to show:

$$x(tp + (1-t)p', tw + (1-t)w') < tx(p,w) + (1-t)x(p',w')$$

But I'm not finding anyway to simplify this expression to show it is true. Any help would be appreciated, as I feel like I am missing something very obvious.

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I am not sure whether the question asks about the function being convex, or simply the demand correspondence/function being a convex set. The latter is often asked and if that is the case here is an answer.

This is a general answer to your question which may be more useful to you and future readers than a specific one.

The following theorem holds in general:

Theorem: If preferences are convex, then the marshallian demand x(p,w) is a convex set for every p>>0 and $w \ge 0$. Note: p>>0 here means that every element of the price vector is strictly positive.

Proof:

The Budget set $B(p,w) \equiv \{ x \in \mathbb{R}_+^n:px \le w \} $ is a convex set.

If two points $ x, x' \in x(p,w )$, then x ~ x' (the consumer must be indifferent between the two as they are both part of x(p,w), i.e. his best choice. If one of them would be better than the other, then only the best one would be part of x(p,w) as the marshallian demand is derived from maximization.)

For all $\alpha \in [0,1]$ we have that $\alpha x + (1-\alpha)x' \in B(p,w)$ by convexity of B(p,w). Furthermore, $\alpha x + (1-\alpha)x'\succeq x$ by the convexity of preferences. Therefore $\alpha x + (1-\alpha)x'\in x(p,w)$ Q.E.D.

Your preference relation is convex, so the theorem applies and your set x(p,w) is convex. Note that a preference relation is (strictly) convex if and only if the utility function is (strictly) quasiconcave. Your utility function as far as I can see is concave, which implies it is quasiconcave, which implies convex preferences.

Further note that if preferences are strictly convex, then x(p,w) is single-valued for every p>>0 and $w \ge 0$. Hence in that case x(p,w) would be a convex set as a point is a convex set.

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  • $\begingroup$ I ended up using a general proof as well, even though I am still a little unsure whether my setup would be feasible as well. Thank you. $\endgroup$ – Kitsune Cavalry Oct 10 '15 at 20:05

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