[proofplan] Fix two points $x,y\in C$ and a coefficient $t\in[0,1]$. Convexity of $C$ keeps the convex combination $z=(1-t)x+ty$ inside $C$, so every function $f_i$ can be evaluated at $z$. Applying convexity to each $f_i$ gives a family of upper bounds, and each upper bound is dominated by $(1-t)f(x)+t f(y)$ because $f(x)$ and $f(y)$ are suprema. Taking the supremum over the index set then gives the desired extended-valued convexity inequality. [/proofplan] [step:Form the convex combination inside the domain] Let $x,y\in C$ and let $t\in[0,1]$. Define $z\in V$ by \begin{align*} z=(1-t)x+ty. \end{align*} Since $C$ is convex and $x,y\in C$, we have $z\in C$. Thus $f_i(z)$ is defined for every $i\in I$, and $f(z)$ is defined as an element of $\mathbb{R}\cup\{+\infty\}$. Because $I$ is nonempty and every $f_i(z)$ is a real number, the supremum \begin{align*} f(z)=\sup_{i\in I} f_i(z) \end{align*} is never $-\infty$; it is either real or $+\infty$. [/step] [step:Bound each member of the family by the same extended-real quantity] Fix $i\in I$. Since $f_i:C\to\mathbb{R}$ is convex and $x,y,z\in C$ with $z=(1-t)x+ty$, we have \begin{align*} f_i(z)\le (1-t)f_i(x)+t f_i(y). \end{align*} By the definition of $f$ as a pointwise supremum, \begin{align*} f_i(x)\le f(x) \end{align*} and \begin{align*} f_i(y)\le f(y). \end{align*} Since $1-t\ge 0$ and $t\ge 0$, monotonicity of multiplication by nonnegative scalars in $\mathbb{R}\cup\{+\infty\}$ gives \begin{align*} (1-t)f_i(x)\le (1-t)f(x) \end{align*} and \begin{align*} t f_i(y)\le t f(y), \end{align*} with the convention $0\cdot(+\infty)=0$ when $t=0$ or $t=1$. Adding these two inequalities in the extended real line yields \begin{align*} (1-t)f_i(x)+t f_i(y)\le (1-t)f(x)+t f(y). \end{align*} Combining the last inequality with convexity of $f_i$ gives \begin{align*} f_i(z)\le (1-t)f(x)+t f(y). \end{align*} [guided] Fix an index $i\in I$. The goal at this stage is not yet to estimate the supremum directly; instead, we estimate one member of the family and make the estimate independent of $i$. Because $f_i:C\to\mathbb{R}$ is convex, and because the previous step showed that $z=(1-t)x+ty$ belongs to $C$, the defining convexity inequality for $f_i$ applies to the pair $x,y\in C$ and the coefficient $t\in[0,1]$. Hence \begin{align*} f_i(z)\le (1-t)f_i(x)+t f_i(y). \end{align*} Now we replace the two values $f_i(x)$ and $f_i(y)$ by the larger quantities $f(x)$ and $f(y)$. By definition, \begin{align*} f(x)=\sup_{j\in I} f_j(x) \end{align*} and \begin{align*} f(y)=\sup_{j\in I} f_j(y). \end{align*} Since $i$ is one of the indices in $I$, the supremum at $x$ dominates the particular value $f_i(x)$, and the supremum at $y$ dominates the particular value $f_i(y)$. Thus \begin{align*} f_i(x)\le f(x) \end{align*} and \begin{align*} f_i(y)\le f(y). \end{align*} The coefficients in the convex combination are nonnegative: $1-t\ge 0$ and $t\ge 0$. Therefore multiplying the two inequalities by these coefficients preserves their direction in the extended real line: \begin{align*} (1-t)f_i(x)\le (1-t)f(x) \end{align*} and \begin{align*} t f_i(y)\le t f(y). \end{align*} At the endpoints $t=0$ and $t=1$, this uses the standard extended-valued convexity convention $0\cdot(+\infty)=0$, so no undefined expression appears. Adding the two inequalities gives \begin{align*} (1-t)f_i(x)+t f_i(y)\le (1-t)f(x)+t f(y). \end{align*} Combining this with the convexity estimate for $f_i$ gives the index-independent bound \begin{align*} f_i(z)\le (1-t)f(x)+t f(y). \end{align*} This is the key point: the right-hand side no longer depends on $i$, so it can control the supremum over all indices. [/guided] [/step] [step:Take the supremum over the index set] The previous step shows that every element of the set \begin{align*} \{f_i(z):i\in I\} \end{align*} is bounded above by the same extended-real number $(1-t)f(x)+t f(y)$. Hence its supremum is bounded above by that number: \begin{align*} \sup_{i\in I} f_i(z)\le (1-t)f(x)+t f(y). \end{align*} Using the definition of $f(z)$, this becomes \begin{align*} f(z)\le (1-t)f(x)+t f(y). \end{align*} Since $z=(1-t)x+ty$, we obtain \begin{align*} f((1-t)x+ty)\le (1-t)f(x)+t f(y). \end{align*} Because $x,y\in C$ and $t\in[0,1]$ were arbitrary, $f$ is convex on $C$ in the extended-valued sense. [/step]