# Check for concavity, quasi-concavity with given value of function

I am studying for an upcoming exam in microeconomics and I have always struggled with proving concavity, convexity in different exercizes. This one I can't solve for the life of me. It is kind of hard to find the exact point I struggle with, because I'm pretty confused on this one overall.

We are given sets of points for which the value of the function is f(x) = 1, 2, 4.

We are then asked if these curves are consistent or inconsistent with the function being either concave or quasi-concave.

We are given tips:

1. Write down the definition of concavity:

$$f: D -> \mathbb R$$ is concave if for all $$x, x' \in D, t \in (0,1):$$

$$f(tx + (1-t)x') \geq tf(x) + (1-t)f(x')$$

1. Select two points in the domain (1,3) and write down the convex combination for the points.

???

1. Evaluate the value of the function.

2. Evaluate the value of the convex combination for each of the points for this function.

3. Use the defintion to show where the function is concave and/or quasi-concave

What throws me off here is that there is no actual function given, but only values for the function. If I had a function, say $$f(x,y) = x^\beta y^{1-\beta}$$, I could choose a point and check with a certain t against the concavity defintion, right?

I hope this doesn't fall under the no homework-clause. I don't have to turn this in, I'm just lost :)

• "Select two points in the domain (1,3) and write down the convex combination for the points." What exactly do you mean by "???" Seems like this could be explained. Commented Feb 13, 2020 at 12:31
• Hint: You don't know the exact form of $f$, but you do know that its values at points $(1,1)$, $(2,2)$ and $(3,3)$ and $(2,2)$ is a convex combination of $(1,1)$ and $(3,3)$. Commented Feb 13, 2020 at 17:30

First look at the shape of the indifference curves. (I assume that $$f$$ is a utility function here.) The better-sets are convex, so this is consistent with quasi-concavity of the function $$f$$.
Second, consider the convex combination $$(2,2)=\frac{1}{2}(1,1)+\frac{1}{2}(3,3)$$. You have $$f(2,2)=2$$, but $$\frac{1}{2}f(1,1)+\frac{1}{2}f(3,3)=\frac{1}{2}1+\frac{1}{2}4=2.5>2$$, which is inconsistent with the function being concave.