[proofplan] We prove both implications directly from the definitions. If $C$ is convex, then the only non-automatic case of the extended-valued convexity inequality occurs when the right-hand side is finite; in that case the relevant convex combination is forced to lie in $C$. Conversely, if $\iota_C$ is convex and $x,y\in C$, then the convexity inequality gives $\iota_C((1-t)x+ty)\le 0$, which is possible only when $(1-t)x+ty\in C$. [/proofplan]

[step:Assume $C$ is convex and prove the extended-valued convexity inequality]Assume that $C$ is convex. Let $x,y\in V$ and let $t\in[0,1]$. We must prove \begin{align*} \iota_C((1-t)x+ty)\le (1-t)\iota_C(x)+t\iota_C(y). \end{align*} We use the standard extended-real convention that $a(+\infty)=+\infty$ for $a>0$, $0(+\infty)=0$, and $r+(+\infty)=+\infty$ for every $r\in\mathbb{R}\cup\{+\infty\}$. If the right-hand side is $+\infty$, the inequality holds because $\iota_C((1-t)x+ty)\in\{0,+\infty\}$. It remains to consider the case where \begin{align*} (1-t)\iota_C(x)+t\iota_C(y)<+\infty. \end{align*} Since each term is nonnegative, this implies that $(1-t)\iota_C(x)<+\infty$ and $t\iota_C(y)<+\infty$. Hence either $t=1$ or $x\in C$, and either $t=0$ or $y\in C$. If $0<t<1$, then $x\in C$ and $y\in C$, so convexity of $C$ gives $(1-t)x+ty\in C$. Therefore \begin{align*} \iota_C((1-t)x+ty)=0=(1-t)\iota_C(x)+t\iota_C(y). \end{align*} If $t=0$, then the finite right-hand side condition gives $x\in C$, and $(1-t)x+ty=x\in C$. If $t=1$, then the finite right-hand side condition gives $y\in C$, and $(1-t)x+ty=y\in C$. In both endpoint cases the same equality holds. Thus the convexity inequality holds for all $x,y\in V$ and all $t\in[0,1]$, so $\iota_C$ is extended-valued convex.[/step]

[guided]Assume that $C$ is convex. To prove that $\iota_C$ is convex, we must verify the defining inequality for arbitrary $x,y\in V$ and arbitrary $t\in[0,1]$: \begin{align*} \iota_C((1-t)x+ty)\le (1-t)\iota_C(x)+t\iota_C(y). \end{align*} The point of using the extended-valued indicator is that most cases are automatic. The value of $\iota_C$ is always either $0$ or $+\infty$. We use the standard extended-real arithmetic convention: $a(+\infty)=+\infty$ for $a>0$, $0(+\infty)=0$, and adding $+\infty$ to any extended-real value gives $+\infty$. If \begin{align*} (1-t)\iota_C(x)+t\iota_C(y)=+\infty, \end{align*} then the desired inequality is immediate, because the left-hand side is either $0$ or $+\infty$, and both are at most $+\infty$ in the extended order. So suppose instead that \begin{align*} (1-t)\iota_C(x)+t\iota_C(y)<+\infty. \end{align*} Since both summands are nonnegative, neither summand can be $+\infty$. Thus $(1-t)\iota_C(x)<+\infty$ and $t\iota_C(y)<+\infty$. The first condition means that either $1-t=0$, so $t=1$, or else $\iota_C(x)=0$, so $x\in C$. Similarly, the second condition means that either $t=0$ or $y\in C$. Now separate the interior and endpoint cases. If $0<t<1$, then neither coefficient is zero, so the preceding paragraph gives $x\in C$ and $y\in C$. Since $C$ is convex, the convex combination $(1-t)x+ty$ belongs to $C$. Therefore \begin{align*} \iota_C((1-t)x+ty)=0. \end{align*} Also $\iota_C(x)=0$ and $\iota_C(y)=0$, so \begin{align*} (1-t)\iota_C(x)+t\iota_C(y)=0. \end{align*} Hence the required inequality holds. If $t=0$, then the combination is $x$. The finite right-hand side condition forces $x\in C$, so $\iota_C(x)=0$ and the inequality holds with both sides equal to $0$. If $t=1$, then the combination is $y$. The finite right-hand side condition forces $y\in C$, so again both sides are equal to $0$. Therefore the extended-valued convexity inequality holds in every case, and $\iota_C$ is convex.[/guided]

[step:Assume $\iota_C$ is convex and show $C$ is closed under convex combinations] Assume that $\iota_C$ is extended-valued convex. Let $x,y\in C$ and let $t\in[0,1]$. Since $x,y\in C$, we have $\iota_C(x)=0$ and $\iota_C(y)=0$. By convexity of $\iota_C$, \begin{align*} \iota_C((1-t)x+ty)\le (1-t)\iota_C(x)+t\iota_C(y). \end{align*} The right-hand side is \begin{align*} (1-t)\iota_C(x)+t\iota_C(y)=0. \end{align*} Thus \begin{align*} \iota_C((1-t)x+ty)\le 0. \end{align*} Since $\iota_C$ only takes the values $0$ and $+\infty$, this forces \begin{align*} \iota_C((1-t)x+ty)=0. \end{align*} By the definition of $\iota_C$, we obtain $(1-t)x+ty\in C$. Hence $C$ is convex. [/step]

[step:Conclude the equivalence] We have shown that convexity of $C$ implies extended-valued convexity of $\iota_C$, and that extended-valued convexity of $\iota_C$ implies convexity of $C$. Therefore $\iota_C$ is extended-valued convex if and only if $C$ is convex. [/step]