Proving quasi-concavity for a utility function

Question

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!

Answer 1

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 $k\in\mathbb{R}$.

Answer 2

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 $\mathbb{R}^2_{++}$.