Convex implies not subadditive 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. 

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