TITLE: Set is Convex regardless of b
QUESTION [5 upvotes]: Let the function $f$ be convex, $f :\Bbb R^n \rightarrow \Bbb R$ and let 
$$S = \{x : f(x) \le b\}$$
The proposition states that the set $S$ is convex regardless of $b$. Can someone explain to me how this proposition holds for all $b$?

REPLY [4 votes]: Fix $b$ any real number, then study $S$ (with $b$ fixed).

If it is empty, it is convex,
Otherwise, if $x,y \in S$, and $\mu \in [0,1]$, then what about $\mu x + (1-\mu) y$ ? Just compute $f(\mu x + (1-\mu) y) <= \mu f(x) + (1-\mu)f(y)$ by convexity of $f$, and then because $x,y\in S$, $f(x),f(y) \leq b$ (by definition). So $\mu f(x) + (1-\mu)f(y) \leq \mu b + (1-\mu) b = b$.

So regardless of the actual value for $b\in\mathbb{R}$, $S$ is convex (but $b$ is fixed).