The norm of the difference of two projections in a C$^\ast$-algebra

Question

I'm trying to show that if $p$ and $q$ are projections in a $C^*$-algebra (i.e. $p=p^2=p^*$), then the norm of their difference is less than or equal to $1$.

Any help would be greatly appreciated!

Answer 1

Represent your algebra faithfully on some Hilbert space $H$. Then consider for $v\in H$:

$$\|(p-q)v\|^2=\|p(1-q)v-(1-p)qv\|^2=\|p(1-q)v\|^2+\|(1-p)qv\|^2$$ where the right equality holds because $p$ and $(1-p)$ project onto orthogonal subspaces. Now further bound each summand:

$$≤\|p\|^2\,\|(1-q)v\|^2+\|1-p\|^2\,\|qv\|^2≤\|(1-q)v\|^2+\|qv\|^2 =\|v\|^2.$$

It follows that $p-q$ is a contraction, hence has norm $≤1$.

Answer 2

Let $p$ be a projection in $A$. Recall that $\sigma(p)\subset\{0,1\}$. Then in the unitization $A^+$ of $A$, by functional calculus we have $$\left\|p-\frac12\right\|\leq\max\left\{\left|0-\frac12\right|,\left|1-\frac12\right|\right\}=\frac12.$$ Thus, for any projections $p,q\in A$, we have $$\|p-q\|\leq\left\|p-\frac12\right\|+\left\|\frac12-q\right\|\leq1.$$