If a function $L:X\rightarrow X$ has a property that $$L(ax+by)=aL(x)+bL(y)$$ for $x, y \in X$ and $a,b$ such that $a+b=1$ does it imply that $$ L(ax+by+cz)=aL(x)+bL(y)+cL(z)$$ provided that $a+b+c=1$?
The point is that the author of A Course in credibility states that the projection onto an affine subspace $\mathbb Q$ has a property $$ \mathrm {Pro}((a+b)^{-1}(aX+bY)|\mathbb Q)=(a+b)^{-1}(a\mathrm{Pro(X|\mathbb Q)}+b\mathrm{Pro(Y|\mathbb Q)}).$$ It implies that for $a+b=1$ $$ \mathrm {Pro}((aX+bY)|\mathbb Q)=a\mathrm{Pro(X|\mathbb Q)}+b\mathrm{Pro(Y|\mathbb Q)}$$ and I'm wondering if we can generalize this rule, and maybe I'm blind, but I don't see a direct way to prove it :(
It's a useful fact, and it's worth working out the details for yourself one time. You can do it in two steps, using
$$ax + by +cz = ax + (b+c)\frac{by + cz}{b+c}$$
and
$$\frac{by + cz}{b+c} = \frac{b}{b+c}y + \frac{c}{b+c}z$$
This generalizes to an arbitrary number of vectors, and I like to think of it as "you can generate all convex combinations of $N$ vectors by taking successive pairwise convex combinations", where you're allowed to use previously generated vectors in later convex combinations, and "pairwise convex combinations" is just a complicated way of saying "anything lying on the line segment connecting two vectors".