Commuting idempotents repel each other

Question

From a Banach algebra $A$ there are elements $e,f$ satisfying $e^2=e,f^2=f,ef=fe,e\ne f$, prove that $\|e-f\|\ge1$;

I can prove$\|e+f|\ge 1$by$\|e-f\|=\|e^2-f^2\|=\|(e-f)(e+f\|\le\|e-f\|\cdot\|e+f\|$, how to substitute$- $for$+$?

Answer 1

Let $E=e(1-f)$ and $F=(1-e)f$; then $E^2=E$, $F^2=F$, $EF=FE$ and $E\ne F$. So, by the same proof as in the question, $1\le\|E+F\|$. Then $$1\le\|e(1-f)+f(1-e)\|=\|e+f-2ef\|=\|(e-f)^2\|\le\|e-f\|^2$$

Answer 2

Hint: Suppose $\|e-f\| <1$. Then $\sum_n (e-f)^{2n+1} $ is convergent. Compute the power using the fact that $e^{k}=e$ and $f^{k}=f$ for all $k \geq 1$. Can you finish?

Answer 3

Letting $$ e'=e-ef, \quad\text{and}\quad f'=f-ef, $$ we have that $e'$ and $f'$ are idempotent elements satisfying $$ e'f'=f'e'=0, \quad\text{and}\quad \|e'-f'\|=\|e-f\|. $$ We may therefore suppose, without loss of generality, that the originally given $e$ and $f$ satisfy $ef=fe=0$.

In this case observe that $$ B=\text{span}\{e, f\} $$ is a closed unital sub-algebra of $A$, with unit $1_B=e+f$. Letting $u=e-f$, we then have that $u^2=1_B$, so the spectrum of $u$ (relative to $B$) is contained in $\{1, -1\}$. Therefore $$ \|e-f\| = \|u\| \geq \sup\big\{|\lambda |: \lambda \in \sigma_B(u)\big\} = 1. $$