How do I show the following result for two commuting bounded idempotent linear operators on a normed space?

Question

I'm trying an old qualifier problem stating that for two commuting idempotent bounded linear operators $A$ and $B$ on a normed linear space, either $A=B$ or the operator norm of $A-B$ is at least one.

I'm at a complete loss as to how to approach this. Playing round with norms doesn't seem to get me anywhere and I don't see what other theorem to employ.

Answer 1

From $A^2=A, B^2=B$ and $AB=BA$ we get

$(A-B)^3=A-B$ . Let $P:=(A-B)^2$. Then $P^2=P$.

Case 1: $P=0$, hence $A+B=2AB$. Multiplying with $A$ gives: $A+AB=2AB$, thus $A=AB$. We have the same result for $B$: $B=AB$. Hence $A=B$

Case 2: $P \ne 0$. Then: $||P||=||P^2|| \le ||P||^2$ and so $||P|| \ge 1$.

Consequence: $||(A-B)^2|| \ge 1. $ This gives

$1 \le ||(A-B)^2|| \le ||A-B||^2$. Therefore $||A-B|| \ge 1$

Answer 2

Note $(A-B)^3 = A-B$ and so $||A-B|| \le ||A-B||^3 \implies A=B$ or $||A-B|| \ge 1$.