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

$$u:\mathbb R^n\to\mathbb R$$ 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 $b\in\mathbb R^2$. 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. 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. Aug 7, 2020 at 16:25

1. 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).

2. 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.

• 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. Aug 6, 2020 at 20:03
• Thats why I mentioned essentially unique. Because all the optima in the convex set will be payoff-equivalent.
– user28372
Aug 6, 2020 at 20:50
• 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? Aug 6, 2020 at 21:09
• 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 $y\in[1,b^{\prime}]$.
– user28372
Aug 6, 2020 at 22:57
• 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 $U(\lambda) = u\big(a\lambda+b(1-\lambda)\big)$.
– user28372
Aug 10, 2020 at 15:14