This is from economics, but I think there's a lot of math involved and I want to make sure I didn't mess anything up. There is a utility function U = $x_1$ + $\ln(x_2)$ such that $x_1$ and $x_2$ are nonnegative.
How do I prove this function is quasi-concave?
I was looking online for any notes about this, and it tells me I need to create this
https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/qcc/t
"bordered Hessian matrix"
As in for this problem it seems to me I need to calculate
\begin{bmatrix}
0 & f'_{1}(x) & f'_{2}(x)\\
f'_{1}(x) & f''_{11}(x) & f''_{12}(x) \\
f'_{2}(x) & f''_{21}(x) & f''_{22}(x) \\
\end{bmatrix}
So I do that for the above function and I get
\begin{bmatrix}
0 & 1 & \frac{1}{x_2} \\
1 & 0 & 0 \\
\frac{1}{x_2} & 0 & -\frac{1}{x_{2}^{2}} \\
\end{bmatrix}
Then it says I need to find the determinant and if n = even, I need $D_{n} \geq 0$. So here $D_{2}$ for the above matrix I got to be $\frac{1}{x_{2}^{2}}$. Since $x_2$ is nonnegative, this means that the determinant is $\geq 0$.
Does this prove this function is quasi-concave? Is my general approach right or am I totally missing the point of this question?
What about strictly quasi-concave? The online notes I found never mentions this, but is that just when it is strictly greater than 0? So in this case, this function is also strictly quasi-concave too? If someone can verify if my guess is correct, that would be great.
Finally, on an unrelated note, how do I prove this function is continuous using math? I know it is when I graph it, but is there a mathematical proof for this?
Thank you so much for the help!
Best Answer
I actually am not familiar with the method you're using, so I'm using a different method.
The function is actually concave on the (natural) domain $\Bbb{R} \times (0, \infty)$. Note that the function $(x_1, x_2) \mapsto \ln(x_2)$ is concave, because the function $\ln$ is concave (check its second derivative). The function $(x_1, x_2) \mapsto x_1$ is an affine function, and hence is concave (and convex). Summing two concave functions produces a concave function, and every concave function is quasiconcave.
The only question remaining is strictness. Neither of the above functions are strictly (quasi)concave, so we need a separate argument. Strict quasiconcavity means that, for all $(x_1, x_2), (y_1, y_2)$ in the domain, and any $\lambda \in (0, 1)$, we have $$f(\lambda x_1 + (1 - \lambda)y_1, \lambda x_2 + (1 - \lambda)y_2) > \min \{f(x_1, x_2), f(y_1, y_2)\}.$$ Quasiconvexity means the above with $\ge $ substituted for $>$. So, let's suppose that we have equality. Without loss of generality, assume $f(x_1, x_2) \le f(y_1, y_2)$. Then, $$\lambda x_1 + (1 - \lambda)y_1 + \ln(\lambda x_2 + (1 - \lambda)y_2) = x_1 + \ln(x_2).$$ By the strict concavity of $\ln$ (again, examine the second derivative), we have \begin{align*} &x_1 + \ln(x_2) > \lambda x_1 + (1 - \lambda)y_1 + \lambda\ln(x_2) + (1 - \lambda)\ln(y_2) \\ \iff \, &(1 - \lambda)x_1 + (1 - \lambda)\ln(x_2) > (1 - \lambda)y_1 + (1 - \lambda)\ln(y_2) \\ \iff \, &x_1 + \ln(x_2) > y_1 + \ln(y_2) \iff f(x_1, x_2) > f(y_1, y_2), \end{align*} which contradicts $f(x_1, x_2) \le f(y_1, y_2)$. Therefore, $f$ is quasiconcave.