Convex implies not subadditive

Question

A function $f: [a,b]\to \mathbb R$ is called convex if for all $x,y \in [a,b], t \in [0,1]$:

$$ f(tx + (1-t)y) \le tf(x) + (1-t)f(y)$$

A function is called subadditive if $f(x+y) \le f(x) + f(y)$.

Is it true that if $f$ is convex then $f$ is not subadditive?

Context: I thought of this question when I read that a concave function with $f(0) \ge 0$ is subadditive.

Answer 1

$e^{-x}$ is both convex and subadditive on $[0,\infty)$:

$$ e^{-x-y} = e^{-x}e^{-y} \leq e^{-x}+e^{-x}e^{-y} \leq e^{-x}+ e^{-y}. $$

EDIT: in fact, any linear function $f:\mathbb{R}\to\mathbb{R}$ is subadditive, superadditive, convex, and concave:

$$\begin{align*} f(tx + (1-t)y) &= tf(x)+(1-t)f(y) \\f(x + y) &= f(x)+f(y) \end{align*}$$