[proofplan] We prove the convexity inequality for $g\circ f$ directly. Given two points of $C$ and a coefficient in $[0,1]$, convexity of $C$ puts the corresponding convex combination back in $C$, so convexity of $f$ gives an upper bound for $f$ at that point. The interval property of $J$ ensures that the scalar convex combination of $f(x)$ and $f(y)$ lies in the domain of $g$; then monotonicity of $g$ and convexity of $g$ combine to give the desired inequality. [/proofplan] [step:Fix a convex combination in the domain of the composition] Let $x,y\in C$ and let $t\in[0,1]$. Define the point $z\in V$ by \begin{align*} z=(1-t)x+ty. \end{align*} Since $C$ is convex and $x,y\in C$, we have $z\in C$. Therefore $f(z)$ is defined, and since $f(C)\subset J$, we also have $f(z)\in J$. Similarly, $f(x)\in J$ and $f(y)\in J$. Since $J\subset\mathbb{R}$ is an interval and $t\in[0,1]$, the scalar convex combination \begin{align*} a=(1-t)f(x)+t f(y) \end{align*} belongs to $J$. Thus both $f(z)$ and $a$ lie in the domain of $g$. [/step] [step:Use convexity of $f$ to compare the two arguments of $g$] By convexity of $f:C\to\mathbb{R}$, applied to the points $x,y\in C$ and the coefficient $t\in[0,1]$, we obtain \begin{align*} f((1-t)x+ty)\le (1-t)f(x)+t f(y). \end{align*} With the notation $z=(1-t)x+ty$ and $a=(1-t)f(x)+t f(y)$, this says \begin{align*} f(z)\le a. \end{align*} [guided] The goal is to prove convexity of the composite function, so we need an inequality involving \begin{align*} (g\circ f)((1-t)x+ty). \end{align*} The first function applied to the convex combination is $f$, so the natural first step is to use convexity of $f$. Since $x,y\in C$, $t\in[0,1]$, and $C$ is convex, the point \begin{align*} z=(1-t)x+ty \end{align*} belongs to $C$. Hence the convexity inequality for $f:C\to\mathbb{R}$ is applicable and gives \begin{align*} f(z)=f((1-t)x+ty)\le (1-t)f(x)+t f(y). \end{align*} Define the real number $a$ by \begin{align*} a=(1-t)f(x)+t f(y). \end{align*} Then the inequality becomes \begin{align*} f(z)\le a. \end{align*} This comparison is exactly what monotonicity of $g$ can use: it turns an inequality between inputs of $g$ into an inequality between their images under $g$. [/guided] [/step] [step:Apply monotonicity and convexity of $g$] From the previous step, $f(z)\le a$, and both $f(z)$ and $a$ belong to $J$. Since $g:J\to\mathbb{R}$ is nondecreasing, we have \begin{align*} g(f(z))\le g(a). \end{align*} Since $g$ is convex on $J$, and since $f(x),f(y)\in J$ with $a=(1-t)f(x)+t f(y)\in J$, we also have \begin{align*} g(a)\le (1-t)g(f(x))+t g(f(y)). \end{align*} Combining these inequalities gives \begin{align*} g(f((1-t)x+ty))\le (1-t)g(f(x))+t g(f(y)). \end{align*} Equivalently, \begin{align*} (g\circ f)((1-t)x+ty)\le (1-t)(g\circ f)(x)+t(g\circ f)(y). \end{align*} [/step] [step:Conclude convexity of the composition] The inequality above holds for every $x,y\in C$ and every $t\in[0,1]$. Therefore $g\circ f:C\to\mathbb{R}$ satisfies the defining convexity inequality on $C$. Hence $g\circ f$ is convex. [/step]