Let $X$ be a normed vector space. Let $T,S$ be bounded linear operators such that $T^2=T,S^2=S,ST=TS$. Show either $T=S$ or $\|T-S\|\geq 1$
We consider two cases:
$\ker T \neq \ker S$. Then WLOG we may assume that there exists $x \in \ker S \setminus \ker T$. Then $Tx \neq 0$ while $Sx = 0$. So $$ (T-S)(Tx) = T^2x - STx = Tx - TSx = Tx, $$ and so, $ \| T - S \| \geq 1$.
$\ker T = \ker S$. We note that $\operatorname{im}(1-T) \subseteq \ker T = \ker S$. So, for any $x$, $$ Sx = S((1-T)x + Tx) = S\underbrace{(1-T)x}_{\in \ker S} + STx = STx. $$ By the same argument, we have $Tx = TSx = STx$ for all $x$. Therefore $T = S$.