# Proving quasi-concavity for a utility function

I have a utility function, and I want to prove that it is a quasi-concave function: $$u(x_1,x_2)= 2x_1x_2+x_1+2x_2$$ I do this by showing that the set of points where the utility is larger than or equal to any given level forms a convex set, and denote this level of utility by $$U^{x}$$ The set of points where the utility function is larger than or equal to can be represented as: $$S ={ (x_1,x_2)|2x_1x_2+x_1+2x_2 ≥ u^{x}}$$ So I have to prove that the set S is convex for all possible values $$u^{x}$$ I consider two arbitrary points ($$x_1^{a},x_2^{a}$$) and ($$x_1^{b},x_2^{b}$$) in the set S that satisfy: $$[2x_1^{a}x_2^{a}+x_1^{a}+2x_2^{a} ≥ u^{x} ]$$ $$[2x_1^{b}x_2^{b}+x_1^{b}+2x_2^{b} ≥ u^{x} ]$$ Now, if I take a convex combination of those points: $$(x_1,x_2)= \lambda(x_1^{a},x_2^{a})+(1-\lambda)(x_1^{b},x_2^{b}$$ where $$(0≤\lambda≤1)$$. This lies within the line segment connecting ($$x_1^{a},x_2^{a}$$) and ($$x_1^{b},x_2^{b}$$)

Now I evaluate the utility function at the point $$(x_1,x_2)$$: $$u(x_1,x_2)= 2x_1x_2+x_1+2x_2$$

I now substitute the expression for $$(x_1,x_2)$$:

$$u(x_1,x_2)= 2 (\lambda x_1^{a}+(1-\lambda)(x_1^{b})(\lambda x_2^{a}+(1-\lambda)(x_2^{b}))) + \lambda x_1^{a}+(1-\lambda)(x_1^{b} + 2 \lambda x_2^{a}+(1-\lambda)(x_2^{b}))$$

I know that I now have to show that u(x_1,x_2) is larger than or equal to $$u^{x}$$, but I am not sure how to proceed further. I would greatly appreciate any help.

EDIT: Solved!

• For quasi-concavity, you have to show that the set of points where the utility is larger or equal (!) to a certain level is convex. Commented May 6 at 14:33
• Thank you, I have updated my answer.
– Noah
Commented May 6 at 14:40
• Something to consider is that you have three (!) more left parentheses in your final equation than right parentheses. Now I am a man of left wing leanings, but this seems too radical even to me. Commented May 6 at 15:44
• If you solved it, you can post an answer to your own question. Other readers might be interested. Commented May 6 at 20:06

Consider the upper level set $$\begin{eqnarray*}S_a&=&\left\{(x_1,x_2)\in\mathbb{R}^2_+|2x_1x_2+x_1+2x_2\geq a\right\}\\ &=& \left\{(x_1,x_2)\in\mathbb{R}^2_+|2(x_1+1)\left(x_2+\frac{1}{2}\right)\geq a+1\right\}\\ &=& \left\{(x_1,x_2)\in\mathbb{R}^2_+|(x_1+1)\left(x_2+\frac{1}{2}\right)\geq \dfrac{a+1}{2}\right\}\\ &=& \left\{(u,v)-\left(1,\frac{1}{2}\right)\in\mathbb{R}^2_+|uv\geq \dfrac{a+1}{2}\right\} \\ &=& \left\{(u,v)\in[1,\infty)\times[\frac{1}{2},\infty)|uv\geq \dfrac{a+1}{2}\right\} -\left(1,\frac{1}{2}\right) \ \ \ [\text{Element-wise subtraction}]\end{eqnarray*}$$ which is a convex set if and only if $$\left\{(u,v)\in[1,\infty)\times[\frac{1}{2},\infty)|uv\geq \dfrac{a+1}{2}\right\}=\left\{(u,v)\in\mathbb{R}^2_+|uv\geq \dfrac{a+1}{2}\right\}\bigcap [1,\infty)\times[\frac{1}{2},\infty)$$ is a convex set. So, it is sufficient to show

$$T_k=\left\{(u,v)\in\mathbb{R}^2_+|uv\geq k\right\}$$

is a convex set for all $$k\in\mathbb{R}$$.

We'll now show that.

For $$k\leq 0$$, $$T_k=\left\{(u,v)\in\mathbb{R}^2_+|uv\geq k\right\}=\mathbb{R}^2_+$$ is a convex set.

Now consider for any $$k>0$$, the set $$T_k=\left\{(u,v)\in\mathbb{R}^2_+|uv\geq k\right\}$$. To show that $$T_k$$ is convex, consider any $$(u',v')\in T_k$$ and $$(u'',v'')\in T_k$$, and any $$\lambda\in (0,1)$$, we'll show that $$(\lambda u'+(1-\lambda)u'', \lambda v'+(1-\lambda)v'')\in T_k$$

To show this, we'll need the following inequality: Without loss of generality letting $$k\leq u'v'\leq u''v''$$, the following holds: $$u'v''+u''v'=u'v'(\frac{v''}{v'}+\frac{u''}{u'})\geq u'v'(\frac{u'}{u''}+\frac{u''}{u'})\geq 2k$$ [Please note that $$(\frac{u'}{u''}+\frac{u''}{u'})\geq 2$$ because $$\min_{a>0} a + \frac{1}{a} = 2$$]

Therefore, we also have $$u'v''+u''v'\geq 2k$$ along with $$u'v'\geq k$$, $$u''v''\geq k$$.

Now, we'll show that $$(\lambda u'+(1-\lambda)u'', \lambda v'+(1-\lambda)v'')\in T_k$$. Here is the argument: $$\begin{eqnarray*} && (\lambda u'+(1-\lambda)u'')(\lambda v'+(1-\lambda)v'') \\ &=& \lambda^2 u'v'+ \lambda(1-\lambda)(u'v''+u''v')+ (1-\lambda)^2u''v''\\ &\geq & \lambda^2 k+ \lambda(1-\lambda)(2k)+ (1-\lambda)^2k \\ &=& (\lambda^2 + 2\lambda(1-\lambda)+ (1-\lambda)^2)k \\ &=& k \end{eqnarray*}$$ Therefore, $$(\lambda u'+(1-\lambda)u'', \lambda v'+(1-\lambda)v'')\in T_k$$. So, $$T_k=\left\{(u,v)\in\mathbb{R}^2_+|uv\geq k\right\}$$ is a convex set for all $$k\in\mathbb{R}$$.

Alternatively, you could check the bordered hessian: $$\begin{bmatrix} 0 & \frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2}\\ \frac{\partial f}{\partial x_1} & \frac{\partial^2 f}{\partial x_1 \partial x_1} & \frac{\partial^2 f}{\partial x_1 \partial x_2}\\ \frac{\partial f}{\partial x_2} & \frac{\partial f}{\partial x_1 \partial x_2} & \frac{\partial^2 f}{\partial x_2 \partial x_2}\end{bmatrix}$$ The determinant of this matrix is strictly positive for $$x_1, x_2 > 0$$, which shows the function is quasi-concave on $$\mathbb{R}^2_{++}$$.