Function linear over convex combinations once again

Question

I asked recently here about functions $L:X\rightarrow X$ satisfying
$$ L(ax+by)=aL(x)+bL(y)$$ for $a,b$ such that $a+b=1$. I know now that it also holds $$ L(\sum_{i=1}^n a_i x_i)=\sum_{i=1}^n a_i L(x_i)$$ for $\sum_{i=1}^n a_i =1$ and $n>2$. I'm wondering now if the definition implies also $$ L(ax)=aL(x)$$ for every $a$, not necessarily $a=1$?

Answer 1

Consider the function $L(x)=x+v$ for some constant non-zero vector $v\in X$.

Then $L(0x)=L(0)=0+v=v\neq 0=0L(x)$.