Proving a function is quasi-concave

Question

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!

Answer 1

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.

Answer 2

Your approach goes in the right direction. But you should verify $D_2>0$ and not $D_2\geq0$. In theory, one should also verify that $D_1<0$, but this is very easy to see.

As for strictly quasi-concave, you should not use the theorem about bordered Hessian matrix because it does not discuss this case (I would still conjecture that the same result can also be shown for strictly quasi-convex functions, but I do not have a proof). Instead, you could try to use the definition.

Answer 3

Eigenvalues of the Hessian matrix of the function $U=x_1+\mathrm{ln}(x_2) $ turns out to be $0$ and $-\frac{1}{{x_{2}}^2}$ which implies that $U$ is concave in $\mathbb{R}^2_{++}.$

Every function that is concave is also Quasi Concave too.

Answer 4

You need to construct the bordered Hessian you mention. I don't see that Osborne gives a proper citation, but this construction is given by Arrow-Enthoven (1961). But the condition you should be thinking about is whether this object is negative semi-definite or not. The "alternating sign test of determinants of principal minors" is just a way to establish definiteness.

A possibly more sensible way to proceed in these days when computation is cheap is simply to calculate the eigenvalues of the bordered Hessian evaluated at a point $x$. Provided the function in question is continuously twice differentiable then the bordered Hessian will be symmetric, and you'll simply be interested in whether or not all the eigenvalues are non-negative or not.

Finally, the necessary and sufficient condition for quasi-concavity is that the property of negative-semi-definiteness holds for all $x$ in the domain of interest.