Algebraic approach towards convexity

Question

I have a function: $ u(x) = x_{1} + x_{2} + \min\{x_{1}, x_{2}\}$. How do we algebraically show if it's convex or not? Also, what would be the general way to show if any given function is convex.

Answer 1

This is an edit

By definition,for a convex subset $C\subseteq \mathbb{R}^n$ a convex function $f:C \to \mathbb{R}$ satisfy

$$ f((1-t)x+ty)\leq (1-t)f(x)+tf(y)$$

for every $x,y\in C$ and $t \in (0,1)$.

We can see if we let $x=(1,3)$ and $y=(3,1)$ and $t=\frac{1}{2}$ that

$$ u((1-t)x+ty)=u(2,2)=6 \not \leq 5=\frac{u(1,3)}{2}+\frac{u(3,1)}{2}=(1-t)u(x)+tu(y)$$

meaning your function is not convex. As you mentioned in the comments, the fact that the $\min$ function is not differentiable around $x_1=x_2$ gave you a false answer, since the function is piecewise linear.

The rest of my original answer is irrelevant to the question now, but is still usefull to know.

More generally it helps to know the subspace of convex functions over $\mathbb{R}^n$ is a cone in the vectorspace of functions over $\mathbb{R}^n$.

Specifically, let $C$ be the set of convex functions over $\mathbb{R}^n$, i.e. $\mathbb{R}^n \to \mathbb{R}$.

If $f,g\in C$ and $a,b\in \mathbb{R}^+$ then $a\cdot f+b\cdot g \in C$

This means that you can start to see if "complicated" functions are convex, if they are a conic combination of known convex functions.

Answer 2

I assume you meant $u(x_1,x_2)=x_1+x_2+\min\{x_1,x_2\}$. To check if a function of two or more variables is convex, you need to check if you can differentiate it twice, calculate its Hessian and check if it is positive semi-definite. On top of that the domain needs to be a convex set but since we talk about utility in economics - you don't need to worry about that too much - Can A Utility Function Take On Negative Values?

In your example: $$\mathbf{H}=\begin{bmatrix} \frac{\partial^2u}{\partial x_1^2} & \frac{\partial^2u}{\partial x_1x_2} \\ \frac{\partial^2u}{\partial x_1x_2} & \frac{\partial^2u}{\partial x_2^2} \\ \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$$

so it is?

The approach described above is a general way of showing a function being convex but as pointed in the comment section, you need to be careful - make sure that all the assumptions are met before jumping to conclusions. (Or else, you end up being mistaken as me right now.)

EDIT.

Minimum function mislead me, I agree.

Let's get back to basics:

$u(x_1,x_2)=x_1+x_2+\min(x_1,x_2)=x_1+x_2 + \frac{x_1+x_2-|x_1-x_2|}{2}=\begin{cases}x_1+2x_2 \;\; \mathrm{for} \;\; x_1 \geq x_2\\ 2x_1+x_2 \;\; \mathrm{for} \;\; x_1<x_2 \end{cases}$

It looks like that (on horizontal axis $x_1$, on vertical axis $x_2$:

It is neither convex, nor concave. Due to the absolute value function that is in a core of minimum function: Why is the absolute value function not differentiable at x=0?

I thought that eliminating point where $x_1=x_2$ would solve the issue but as shown in the other answer - it does not.

EDIT. 2 (final)

It is, in fact, concave as the user @afreelunch was telling me. You can see it directly from the definition. Or (what I think is even easier) from the way our utility function is built.

We have $u(x_1,x_2) = \min\{2x_1+x_2,x_1+2x_2\}$. Now, let's take it apart: $f_1(x_1) = 2x_1$ and $g_1(x_2) = x_2$, they are both linear Is linear function convex or concave? - hence, concave (and convex but we care about concavity here) and therefore their sum (property of concave function) $h_1(x_1,x_2)=2x_1+x_2$ is also concave. Similarly, $f_2(x_1) = x_1$ and $g_2(x_2)=2x_2$ and $h_2(x_1,x_2)=x_1+2x_2$ is concave as well. Knowing that, we get $\min\{h_1(x_1,x_2),h_2(x_1,x_2)\}$ - minimum of two concave functions - which is then concave (also a well known fact).

In short, minimum preserves concavity (and maximum preserves convexity).