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.

Question

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.)

Answer 1

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.