[proofplan] Fix a real number $\alpha$ and prove directly from the definition that the sublevel set $S_\alpha$ is convex. We take two arbitrary points in $S_\alpha$ and an arbitrary coefficient $t \in [0,1]$, use convexity of $C$ to keep the convex combination inside the domain of $f$, and then use convexity of $f$ to bound its value at that convex combination. The defining inequalities $f(x) \leq \alpha$ and $f(y) \leq \alpha$ then show that the convex combination also belongs to $S_\alpha$. [/proofplan] [step:Fix a sublevel set and choose an arbitrary convex combination] Let $\alpha \in \mathbb{R}$ be fixed, and define the set $S_\alpha \subset C$ by \begin{align*} S_\alpha := \{x \in C : f(x) \leq \alpha\}. \end{align*} To prove that $S_\alpha$ is convex, let $x,y \in S_\alpha$ and let $t \in [0,1]$. Since $x,y \in S_\alpha \subset C$ and $C$ is convex, the vector \begin{align*} z := tx + (1-t)y \end{align*} belongs to $C$. [guided] Fix a real number $\alpha \in \mathbb{R}$. The object whose convexity we must prove is the sublevel set \begin{align*} S_\alpha := \{x \in C : f(x) \leq \alpha\}. \end{align*} By definition, proving that $S_\alpha$ is convex means proving the following statement: whenever $x,y \in S_\alpha$ and $t \in [0,1]$, the convex combination $tx+(1-t)y$ also lies in $S_\alpha$. So let $x,y \in S_\alpha$ and let $t \in [0,1]$ be arbitrary. The membership $x,y \in S_\alpha$ implies two things: first, $x,y \in C$; second, $f(x) \leq \alpha$ and $f(y) \leq \alpha$. Before applying $f$ to the convex combination, we must verify that this convex combination is in the domain of $f$. Since $C$ is convex and $x,y \in C$, the vector \begin{align*} z := tx + (1-t)y \end{align*} belongs to $C$. Thus $f(z)$ is defined. [/guided] [/step] [step:Apply convexity of the function and close the sublevel inequality] Since $f: C \to \mathbb{R}$ is convex, and since $x,y,z \in C$ with $z = tx+(1-t)y$ and $t \in [0,1]$, we have \begin{align*} f(z) \leq t f(x) + (1-t)f(y). \end{align*} Because $x,y \in S_\alpha$, we have $f(x) \leq \alpha$ and $f(y) \leq \alpha$. Since $t \geq 0$ and $1-t \geq 0$, multiplying these two inequalities by $t$ and $1-t$, respectively, and adding gives \begin{align*} t f(x) + (1-t)f(y) \leq t\alpha + (1-t)\alpha. \end{align*} The right-hand side satisfies \begin{align*} t\alpha + (1-t)\alpha = \alpha. \end{align*} Therefore $f(z) \leq \alpha$. Since $z \in C$, this means $z \in S_\alpha$. We have shown that every convex combination of two points of $S_\alpha$ lies in $S_\alpha$, so $S_\alpha$ is convex. [/step]