$u:\mathbb R^n\to\mathbb R$ 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:


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?

  • $\begingroup$ 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 $b\in\mathbb R^2$. $\endgroup$ – Herr K. Aug 7 '20 at 7:05
  • $\begingroup$ @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. $\endgroup$ – High GPA Aug 7 '20 at 9:39
  • 1
    $\begingroup$ I know the result is unaffected. I'm just saying that maybe you should be a little more careful with notations. $\endgroup$ – Herr K. Aug 7 '20 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.

  • $\begingroup$ 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 Aug 6 '20 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 Aug 6 '20 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 Aug 6 '20 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 Aug 6 '20 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 Aug 10 '20 at 15:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.