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Suppose $u(x,y)$ is a concave and strictly increasing $\mathcal{C}^2$ function (think of a utility function from economics).

Define the one variable function $f(x)=u(x,e^r(K-x))$ for all $x\in [0,K]$, where $r,K\in\mathbb{R}$.

I want to show that $f$ has a maximum and that this maximum is unique.

I know that since $[0,K]$ is a compact set, $f([0,K])=\{ f(x)\in\mathbb{R}| x\in [0,K]\}$ is a closed and bounded set, thus $f$ has a maximum value in $[0,K]$.

My hypothesis is that $f$ is a concave function. If this is true, we can apply the fact that every maximum point of $f$ is a global maximum.

I also know, from the theory of concave functions, that because $u$ is a concave function of two variables $u''_{11},u''_{22}\leq 0$ and $u''_{11}u''_{22}-(u''_{12})^2\geq 0$. With this in mind, I have tried to show that $f''(x)\leq 0$.

Applying the chain rule to $f$, I want to show that

$$f''(x)=u''_{11}(x,e^r(K-x))-2e^ru''_{12}(x,e^r(K-x))+e^{2r}u''_{22}(x,e^r(K-x))$$

is less than or equal to $0$. I have tried without success.

Is there anything wrong in my reasoning? Is my approach bad? Any hints?

bobbo
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1 Answers1

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Without some additional hypothesis your proposition is false.

Let's take $u(x,y) = \ln(x^2+y^2)$ - it's strictly concave, strictly increasing, $\mathcal C^\infty$ on $x>0$, $y>0$.

But the function $F(x) = \ln (x^2+(1-x)^2)$ (think $r=0$, $K=1$) is convex on $[0,1]$.

It would be interesting to study the inverse of your proposition, i.e. whether $f$ is convex for any $u$ satisfying your conditions=)

TZakrevskiy
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  • I appreciate your counterexample! Hmm, but then I wonder how I can show that f has a unique maximum value. – bobbo Feb 08 '14 at 18:25
  • @bobbo you are welcome. Neither you can show uniqueness (again, without additional info on your functions) - take my counterexample, it has two maxima, $x=0$ and $x=1$. – TZakrevskiy Feb 08 '14 at 20:08
  • are you able to provide a counterexample for a function $u(x,y)$ which is defined for all $x,y \geq 0$? Since utility functions has a property called completeness: for every $(x,y)$ where $x,y\geq 0$ are real numbers, $u(x,y)$ is defined. – bobbo Feb 09 '14 at 14:21
  • @bobbo $u(x,y)=\ln(1+x^2+y^2)$ does the trick. If $K=1$ and $r=0$, then $f$ is convex and has two maxima on $[0,1]$. – TZakrevskiy Feb 10 '14 at 02:24