When the global optimal is outside of the constraint set, what will be the demand?

Question

$u:\mathbb R^n\to\mathbb R$ is a quasi-concave utility function so the indifference curves are convex.

$a,b\in\mathbb R^n$ are two points. Our budget set is the (one-dimensional) segment $[a,b]$ that connects $a$ and $b$.

Given: $$x^*=\arg\max_{x\in[a,b]}u(x)$$

Let $b'$ be a point in the segment $[a,x^*]$. That is: $b'=\lambda a+(1-\lambda)x^*$ for any $\lambda\in[0,1]$.

Prove that:

$$b'=\arg\max_{x\in[a,b']}u(x)$$

Graphically this result is very straight forward but I don't know how to mathematically prove it.

I think we could start of proving that $u(\lambda a+(1-\lambda) x^*)$ is monotonically decreasing with $\lambda$.

Are there named theory related?

By $\mathbb R^n$ did you simply mean the real line $\mathbb R$? The notation $[a,b]$ implies $b>a$, but if we take $n=2$ and the budget set be the line segment connecting $(0,1)$ and $(1,0)$, then it's not obvious which point should be $a\in\mathbb R^2$ and which be $b\in\mathbb R^2$. — Herr K., Commented Aug 7, 2020 at 7:05
@HerrK. Either way, the argument holds geometrically, I think. The notation is a bit unconventional I suppose but it is pretty clear that $a,b$ are any $\mathbb R^n$ points. — High GPA, Commented Aug 7, 2020 at 9:39
I know the result is unaffected. I'm just saying that maybe you should be a little more careful with notations. — Herr K., Commented Aug 7, 2020 at 16:25

user28372user28372 · Accepted Answer · 2020-08-06 18:18:11Z

2

Argue that, given your assumptions on the utility function, $x^*$ is the essentially unique (and hence global) maximum. (You need this because there may be local maxima when the assumptions on the utility function is relaxed - this will violate the proposition you're trying to prove).
Now simply use the definition of global optima: for any $x\leq x^*$, $u(x)\leq u(x^*)$. This should be enough to give you the result.

answered Aug 6, 2020 at 18:18

user28372

$\begingroup$ Upvoted but if the optima are not unique but a convex set (in this case a connected segment), then the proposition seems still hold. If the indifference curve is convex instead of strictly convex, then the optima can be a convex set. $\endgroup$
– High GPA
Commented Aug 6, 2020 at 20:03
1

$\begingroup$ Thats why I mentioned essentially unique. Because all the optima in the convex set will be payoff-equivalent. $\endgroup$
– user28372
Commented Aug 6, 2020 at 20:50
$\begingroup$ Ok I see. I am not sure how to go from $u(x)\leq u(x^*)$ to $u(x)\leq u(b)$. Could you please help with it? $\endgroup$
– High GPA
Commented Aug 6, 2020 at 21:09
$\begingroup$ Assume part 1 has been proved. Then pick any $y\in[a,b^{\prime}]$. Then we know. that $y\leq b^{\prime}$ and $b^{\prime}\leq x^*$ (by definition of $b^{\prime}$). Since $u(.)$ is increasing in the interval $[a,x^*]$, $y\leq b^{\prime} \leq x^* \implies u(y)\leq u(b^{\prime})\leq u(x^*)$. Thus $u(b^{\prime})\geq u(y)$ for all $y\in[1,b^{\prime}]$. $\endgroup$
– user28372
Commented Aug 6, 2020 at 22:57
1

$\begingroup$ Sorry for the late reply. For a line segment, yeah. Easy way to do that is just convert the utility to a single dimension: Define the preference relation $\succsim_R$ represented by the utility function $U:[0,1]\rightarrow\mathbb{R}$ with $U(\lambda) = u\big(a\lambda+b(1-\lambda)\big)$. $\endgroup$
– user28372
Commented Aug 10, 2020 at 15:14

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