Is negative of a concave function convex? What is the relationship between convex set and convex function?

Question

Let f be a function of a single variable defined on the interval I.

Then f is:

concave if for all a ∈ I, all b ∈ I, and all λ ∈ (0, 1) we have f((1−λ)a + λb) ≥ (1 − λ)f(a) + λf(b)

convex if for all a ∈ I, all b ∈ I, and all λ ∈ (0, 1) we have f((1−λ)a + λb) ≤ (1 − λ)f(a) + λf(b)

My question is, if we take a concave function like the picture below, and take the negative of it, will it become convex? Since the appearance will be U-shaped just like a convex function. I understand that the inequality for concave function still holds if we negate the function. But I am curious from the visual point of view, why do we still call it concave since it looks like a convex function?

For my second question, the convex set can be visualized by connecting any two points in a set that still entirely belongs to the set. How does this relate to the definition of convex functions?

Answer 1

Regarding the second question:

• The function $f$ is convex if and only if the set $S_f:=\{(x,y): y\ge f(x)\}$ is convex.

Proof of the if-part: take $x_1<x_2\,$ and write. $$ \boldsymbol{v}_1:=\begin{pmatrix}x_1\\f(x_1)\end{pmatrix}\,,\quad \boldsymbol{v}_2:=\begin{pmatrix}x_2\\f(x_2)\end{pmatrix}\,. $$ Because $S_f$ is convex, $a\boldsymbol{v}_1+(1-a)\boldsymbol{v}_2\in S_f$ for all $a\in[0,1]$. We can write this as $$\tag{1} af(x_1)+(1-a)f(x_2)\ge f\Big(ax_1+(1-a)x_2\Big)\, $$ in other words, $f$ is convex.

Proof of the only-if-part: Let $\boldsymbol{v}_1=\begin{pmatrix}\boldsymbol{v}_{11}\\\boldsymbol{v}_{12}\end{pmatrix}$ and $\boldsymbol{v}_2=\begin{pmatrix}\boldsymbol{v}_{21}\\\boldsymbol{v}_{22}\end{pmatrix}$ be two points in $S_f$. This means that $$\tag{2} \boldsymbol{v}_{12}\ge f(\boldsymbol{v}_{11})\,,\quad\boldsymbol{v}_{22}\ge f(\boldsymbol{v}_{21}) $$ Since $f$ is convex we have for any $a\in[0,1]$, $$\tag{3} af(\boldsymbol{v}_{11})+(1-a)f(\boldsymbol{v}_{21})\ge f\Big(a\boldsymbol{v}_{11}+(1-a)\boldsymbol{v}_{21} \Big)\,. $$ From (2) and (3), $$\tag{4} a\boldsymbol{v}_{12}+(1-a)\boldsymbol{v}_{22}\ge f\Big(a\boldsymbol{v}_{11}+(1-a)\boldsymbol{v}_{21} \Big)\,, $$ in other words, $a\boldsymbol{v}_1+(1-a)\boldsymbol{v}_2\in S_f$ and $S_f$ is convex.