Proving $(A+B)C=AC+BC$

Question

Let $A,B,C$ be linear operators on a Banach space $X$. I want to prove $(A+B)C=AC+BC$.

For all $x\in X$, we have $(A+B)x=Ax+Bx$ and $(A+B)Cx=z\implies\exists y$ such that $Cx=y$ and $(A+B)y=z$.

Furthermore, for all $x\in X$, we have $(AC+BC)x=ACx+BCx=v\implies\exists u$ such that $Cx=u$ and $(A+B)u=v$.

How can I conclude that $v=z$?

Answer 1

With $y:=Cx$, we have $$(A+B)Cx=(A+B)y=Ay+By=ACx+BCx=(AC+BC)x $$