Non-uniqueness of linear function extension with cone positiveness Let $M$ be a proper linear subspace of $\mathbb R^n,$ and let $K$ be a cone in $\mathbb R^n$ without the origin ($K$ may not be convex and for any $x \in K$, $−x \notin K$). Suppose $f:\mathbb R^n \to \mathbb R$ is linear and $f>0$ on $K.$ Fix $x_0 \notin M$. Does there exist a linear $g: R^n \to \mathbb R$ such that $g>0$ on $K,$ $g= f$ on $M,$ and $f(x_0) \ne g(x_0)?$ This is a companion problem of my other question. That problem is my try for this problem. However, my attempt was proved to be false in the post. So I asked this problem here. Could anyone help me?

Counterexample: In $\mathbb R^2,$ let $K=\\{(x,y):y>0\\},$ $M=\\{(x,x): x\in \mathbb R\\}.$ Let $f(x,y)=y.$ Then $f>0$ on $K.$ Suppose $g$ is linear and $g>0$ on $K$ and $g=f$ on $M.$ Because $g>0$ on $K,$ $g(x,y)=cy$ for some $c>0.$ Because $g=f$ on $M,$ we have $g(1,1)=c=f(1,1)=1.$ Therefore $c=1,$ hence $g=f$ everywhere.