II.1.4.4: Composition

Theorem: [Sum of functions] Suppose $h$ is defined on $\mathcal{X}\otimes\mathcal{U}$ and $f(x)=\int_U h(x,u)dF(u)$. Suppose $k$ is defined on $\mathcal{X}\otimes\mathcal{X}\otimes\mathcal{U}$ and satisfies $$h(x,u)=k(x,x,u)=\min_{y\in\mathcal{X}} k(x,y,u)$$ for all $x\in\mathcal{X}$ and $u\in\mathcal{U}$. Then $g$ defined by $$g(x,y)=\int_U k(x,y,u)dF(u)$$ satisfies $$f(x)=\min_{y\in\mathcal{X}} g(x,y)$$

Proof: $h(x,u)=k(x,x,u)\leq k(x,y,u)$. Integrate to get $f(x)=g(x,x)\leq g(x,y)$. QED

Proof: $g(x)\geq v(x,u)\geq\inf_{u} v(x,u)=f(x)$, and because $y\in X(u)$ also $g(y)=v(y,u)=f(y)$. QED

Proof: $g(x)\geq f(x)$ and thus $\gamma(g(x))\geq\gamma(f(x))$. Also $g(y)=f(y)$ and thus $\gamma(g(y))=\gamma(f(y))$. For the second part we have $\gamma(g(x))\geq\eta(g(x))\geq\eta(f(x))$ and $\gamma(g(y))=\eta(g(y))=\eta(f(y))$. QED

Theorem: Suppose $f=\min_{k} f_{k}$ and let $S_{i}$ be the set where $f=f_{i}$. If $y\in S_{i}$ and $g$ majorizes $f_{i}$ at $y$, then $g$ majorizes $f$ at $y$.

Proof: First $g(x)\geq f_{i}(x)\geq\min_{k} f_{k}(x)=f(x)$. Because $y\in S_{i}$ also $g(y)=f_{i}(y)=f(y)$. QED

This implies that if $f=\min_{k} f_{k}$ has a quadratic majorizer at each $y$, if each of the $f_{k}$ has a quadratic majorizer at each $y$.