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This comes from a paper that I am reading:

For $i=1,2$, suppose that $F_i(\cdot,\cdot)$ satisfies the assumption:

$F_i(K_i,L_i)$ is defined for all $K_i\geq 0$, $L_i\geq 0$. $F_i(0,0)=0$. $F_i(K_i,L_i)>0$ for some $(K_i,L_i)>(0,0)$. $F_i(K_i,L_i)$ is homogeneous of degree $1$.

Then, $$ P\equiv\{(X_1,X_2,K,L)\in\mathbb{R}^4|\exists K_1,L_1,K_2,L_2,\text{ and }K_1+K_2\leq K,\\ L_1+L_2\leq L, X_i\leq F_i(K_i,L_i)\}. $$ The author claims that $P$ is a convex cone. I can see that each $F_i$ being homogeneous of degree $1$ implies that $P$ is a cone. But how does convexity follow? I looked at a convex combination such as $$ \varphi(X_1^a,X_2^a,K^a,L^a)+(1-\varphi)(X_1^b,X_2^b,K^b,L^b) $$ but I wasn't successful. Thank you for your help.

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I found a flaw in the answer below in the equation $F_1(K^c_1,L^c_1)=\phi F_1(K^a_1,L^a_1)+(1-\phi)F_1(K^a_1,L^a_1)$. As the question pointed out, showingthat $P$ is a cone is straightforward, convexity is not. I have not deleted the answer to see if a fix is possible.

Suppose $(X^a_1,X^a_2,K^a,L^a)\in P$ and $(X^b_1,X^b_2,K^b,L^b)\in P$. Then, by definition of $P$, there exist non-negative $K^a_1,K^a_2,L^a_1,L^a_2,K^b_1,K^b_2,L^b_1,L^b_2$ such that

\begin{align} &&K^a_1+K^a_2 &\le K^a,&L^a_1+L^a_2 &\le L^a,\\ &&K^b_1+K^b_2 &\le K^b,&L^b_1+L^b_2 &\le L^b,\\ &&X^a_1 &\le F_1(K^a_1,L^a_1),&X^a_2 &\le F_2(K^a_2,L^a_2),\\ &&X^b_1 &\le F_1(K^b_1,L^b_1), &\text{and}\quad X^b_2 &\le F_2(K^b_2,L^b_2). \end{align}

Consider a convex combination: \begin{align}(X^c_1,X^c_2,K^c,L^c) = \phi(X^a_1,X^a_2,K^a,L^a)+(1-\phi)(X^b_1,X^b_2,K^b,L^b).\end{align}

Define \begin{align}X^c_1&=\phi X^a_1+(1-\phi)X^b_1,&X^c_2 &=\phi X^a_2+(1-\phi)X^b_2,\\ K^c_1&=\phi K^a_1+(1-\phi)K^b_1, &K^c_2&=\phi K^a_2+(1-\phi)K^b_2$,\\ L^c_1&=\phi L^a_1+(1-\phi)L^b_1,& \text{and}\quad L^c_2&=\phi L^a_2+(1-\phi)L^b_2.\end{align} Then,

\begin{align} F_1(K^c_1,L^c_1)=\phi F_1(K^a_1,L^a_1)+(1-\phi)F_1(K^a_1,L^a_1) \le \phi X^a_1+(1-\phi)X^b_1=X^c_1.\end{align}

Similarly, $F_2(K^c_2,L^c_2)\le X^c_2$. And clearly, $K^c_1+K^c_2 \le K^c$ and $L^c_1+L^c_2 \le L^c$ so $(X^c_1,X^c_2,K^c,L^c)\in P$.

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  • $\begingroup$ Thank you for your time. I'll get back to what you write when I get home tonight. $\endgroup$ – yurnero Oct 8 '14 at 16:01

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