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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.)

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    $\begingroup$ 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 $\Bbb R^3$! $\endgroup$ Oct 14, 2023 at 17:15
  • $\begingroup$ @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! $\endgroup$
    – Beerus
    Oct 14, 2023 at 21:23

1 Answer 1

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Thanks a lot for @IzaakvanDongen's comment. Based on my understanding of his suggestions, I would like to add an answer. All mistakes are mine.

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.

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  • $\begingroup$ 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. $\endgroup$ Oct 16, 2023 at 12:48
  • $\begingroup$ @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)? $\endgroup$
    – Beerus
    Oct 16, 2023 at 16:25
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    $\begingroup$ 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! $\endgroup$ Oct 16, 2023 at 16:38
  • $\begingroup$ @IzaakvanDongen Oh, right! Thank you so much for pointing it out! I will amend my answer. $\endgroup$
    – Beerus
    Oct 17, 2023 at 15:25

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