# Prove that a function of $n$ variables is concave if and only if the set below its graph in $\mathbb{R}^{\mathbf{n+1}}$ is a convex set.

I know that an analogous result for convex function is proved here. But my question is about using a specific method to proceed. Here is the question:

## Poblem

Prove that a function of $$n$$ variables is concave if and only if the set below its graph in $$\mathbb{R}^{\mathbf{n+1}}$$ is a convex set. Prove this statement for functions of one variable first. Then apply the following theorem for the case of $$n$$ variables.

Theorem$$\quad$$ Let $$f$$ be a function defined on a convex subset $$U$$ of $$\mathbb{R}^{\mathbf{n}}$$. Then, $$f$$ is concave if and only if its restriction to every line segment in $$U$$ is a concave function of one variable.

## My Question

I do not have problem proving the statement for functions of one variable (see My Attempt below). But I am not sure how to apply the above theorem for the case of functions of $$n$$ variables. Could someone please help me with this? Basically, how to formally and rigorously extend the proof of the statement from functions of one variable to $$n$$ variables? Thanks a lot in advance!

## My Attempt

Here is what I have so far:

Proof$$\quad$$ We first prove that a function of one variable $$f$$ is concave if and only if the set on or below its graph in $$\mathbb{R}^2$$ is a convex set. Suppose that $$f$$ is concave, and that $$(x_1,y_1)$$ and $$(x_2,y_2)$$ lie in the set on or below its graph $$G$$; that is, $$f(x_1) \geq y_1$$ and $$f(x_2) \geq y_2$$. Any point on the line segment $$L$$ joining these two points can be represented by $$(tx_2 + (1-t)x_1, ty_2 + (1-t)y_1)$$, where $$t \in [0,1]$$. Since $$f$$ is concave, we have \begin{align*} f(tx_2 + (1-t)x_1) \geq tf(x_2) + (1-t)f(x_1) \geq ty_2 + (1-t)y_1. \end{align*} Thus, the segment $$L$$ lies in the set on or below $$G$$. Hence, the set on or below $$G$$ is convex.

Conversely, suppose the set on or below $$G$$ is convex, and that $$(x_1,f(x_1))$$ and $$(x_2,f(x_2))$$ are in this set. Then, for all $$t \in [0,1]$$, the point $$(tx_1 + (1-t)x_2, tf(x_1) + (1-t)f(x_2))$$ is also in this set. Thus, \begin{align*} tf(x_1) + (1-t)f(x_2) \leq f(tx_1 + (1-t)x_2). \end{align*} So, $$f$$ is concave.

Now, let $$f$$ be a function defined on a convex subset $$U$$ of $$\mathbb{R}^{\mathbf{n}}$$. $$\dots$$ (This is where I got stuck.)

• Good work so far! Are you absolutely totally stuck, as in you have no idea what you could do? The first step is "Suppose $f$ is concave, and let $(\mathbf x_1, y_1)$ and $(\mathbf x_2, y_2)$ lie on or below the graph of $f$. Let $L$ be the line segment joining $\mathbf x_1$ and $\mathbf x_2$ in $\Bbb R^n$ and let $L'$ be the line segment joining $(\mathbf x_1, y_1)$ and $(\mathbf x_2, y_2)$. Consider the function $f|_L$ and the strip $L \times \Bbb R$, and the graph of $f|_L$ in this strip...". If you're getting symbol overload try to visualise why you believe it's true in $\Bbb R^3$! Oct 14, 2023 at 17:15
• @IzaakvanDongen Thank you so much for your comment! I think I get your point. I added an answer below. Could you please take some time and check if this is correct or not? I really appreciate it! Oct 14, 2023 at 21:23

Proof$$\space$$ Cont'd$$\quad$$ Let $$f$$ be a function defined on a convex subset $$U$$ of $$\mathbb{R}^{\mathbf{n}}$$. Let $$(\mathbf{x_1},y_1)$$ and $$(\mathbf{x_2},y_2)$$ lie on or below the graph of $$f$$. Let $$L$$ be the line segment joining $$\mathbf{x_1}$$ and $$\mathbf{x_2}$$ in $$\mathbb{R}^{\mathbf{n}}$$. Let $$L'$$ be the line segment joining $$(\mathbf{x_1},y_1)$$ and $$(\mathbf{x_2},y_2)$$ in $$\mathbb{R}^{\mathbf{n+1}}$$. Consider the function $$f|_L$$, the strip $$L \times \mathbb{R}$$, and the graph of $$f|_L$$ in this strip.
By the Theorem above and by our proof of the statement for the case of functions of one variable, we have that $$f$$ is concave if and only if $$f|_L$$ is a concave function of one variable if and only if the set on or below the graph of $$f|_L$$ is a convex set if and only if $$L'$$ is in the set on or below the graph of $$f|_L$$ if and only if $$L'$$ lies in the set on or below the graph of $$f$$ if and only if the set on or below the graph of $$f$$ is convex.
• This looks good! I would be a bit careful with your "if and only if"s. "$f$ is concave if and only if $f|_L$ is a concave function" isn't true - remember $L$ is a single fixed line segment we're talking about. I would separate the two directions of the "iff" into two different paragraphs. Oct 16, 2023 at 12:48
• @IzaakvanDongen Thank you so much! I just want to confirm my understanding of your comment is correct: Would it be better if I have written "$f$ is concave if and only if the restriction of $f$ to the line segment $L$ in $U$ is a concave function of one variable" (by the Theorem in the Problem)? Oct 16, 2023 at 16:25
• So it's true that $f$ being concave implies $f|_L$ is concave, but it's not true that $f|_L$ being concave implies $f$ is concave - rather "$f|_L$ is concave for all line segments $L$" would imply $f$ is concave. Since you're working with a single line segment $L$ at that point in the proof, I think it's clearer if you don't worry about doing it all in one go with "iff"s, but just do the two directions separately - really only one direction needs this difficult argument, while the other direction is straightforward. This is a bit of a nitpick - I think it's clear that you have the right idea! Oct 16, 2023 at 16:38