Strongly convex implies boundedness of the arguments set

Question

On page 459 Section 9.1.6 of the Boyd's book it is said that if $f$ is strongly convex on $S$ then the sublevel sets of $f$ are bounded and so the set $S$ is bounded.

How come ? Take $f(x)=x^2$ is strongly convex on $\mathbb{R}$, which is not bounded. What am I missing ?

Answer 1

How do you define a sublevel set?

For me, the $\alpha$-sublevel set of $f:x\mapsto x^2$ is $\{x\in\mathbb R\mid x^2\le\alpha\}=[-\sqrt\alpha,\sqrt\alpha]$, which is of course bounded.

EDIT: I see your problem. In the book you mistook $S$ with $\mathbb R$, whereas its definition is given before, see Equation (9.3) of the same book.