Proving that two projections are at most distance 1 apart

Question

I am trying to solve the very first problem in "The Blue Book" by M. Rørdam. The problem is the following:

• Let $p$ and $q$ be projections in a $C^*$-algebra $A$. Prove that $\Vert p - q \Vert \leq 1$.

To me, it seems like this problem should have a simple and elegant solution, but I am not able to produce it. I have managed to reduce the problem slightly as follows.

Put $s = p-pq$ and $ t = q-pq$. By adding and subtracting $pq$ we obtain $\Vert p - q \Vert = \Vert (p-pq) - (q-pq) \Vert = \Vert s-t \Vert.$ It is straightforward to check that $s^*t = 0 = t^*s$ and that $st^* = 0 = ts^*$. This means that if we realize $A$ on a Hilbert space, then the operators $s$ and $t$ have orthogonal range and domain. I would now like to conclude that $\Vert p - q \Vert = \Vert s-t \Vert \overset{?}{\leq} \max\{ \Vert s \Vert, \Vert t \Vert \} = 1$. The last equality holds since $\Vert s \Vert = \Vert p(1_{\tilde A} - q) \Vert \leq 1$ since $p$ and $1_{\tilde A} - q$ are both projections, and similarly for $t$.

So, what I really would like to know is whether the bound above actually holds, and if so, why. Namely:

• Whether two operators $S,T \in \mathcal{B}(H)$ satisfying $S^*T = 0$ and $ST^* = 0$ also satisfy $\Vert S-T \Vert \leq \max\{ \Vert S \Vert, \Vert T \Vert \}$?

Alternative proofs (or mere suggestions) of the originial problem is also appreciated!

Answer 1

To complement Jake's answer, here is the answer to the OP's question:

If $TS^*=0$, then $T^*TS^*S=0$. If follows that the range projections of $T^*T$ and $S^*S$ are orthogonal to each other. Given $x\in H$, we write $x=x_T+x_S+x_0$, where $x_T\in\text{ran}\,T^*T$, $x_S\in\text{ran}\,S^*S$. Using also $S^*T=0$, \begin{align} \|T-S\|^2&=\|(T-S)^*(T-S)\|=\|T^*T+S^*S\|\\ \ \\ &=\sup\{\langle (T^*T+S^*S)x,x\rangle:\ \|x\|=1\}\\ \ \\ &=\sup\{\langle T^*Tx_T,x_T\rangle+\langle S^*Sx_S,x_S\rangle:\ \|x\|=1\}\\ \ \\ &\leq\sup\{\|T\|^2\|x_T\|^2+\|S^2\|\|x_S\|^2:\ \|x\|=1\}\\ \ \\ &\leq\max\{\|T\|,\|S\|\}^2\,\sup\{\|x_T\|^2+\|x_S\|^2:\ \|x\|=1\}\\ \ \\ &\leq\max\{\|T\|,\|S\|\}^2. \end{align}

Answer 2

Let $p,q$ be projections in some C*-algebra $A$. Then $\sigma(p)\subseteq\{0,1\}$. By the functional calculus, $\sigma(\frac{1}{2}-p)\subseteq\{-\frac{1}{2},\frac{1}{2}\}$ and hence $\frac{1}{2}-p$ is self-adjoint (spectral radius equals norm), which implies that $||\frac{1}{2}-p||=\frac{1}{2}$. Thus $||p-q||=||(\frac{1}{2}-p)-(\frac{1}{2}-q)||\leq\frac{1}{2}+\frac{1}{2}=1$

Edit: This assumes $A$ is unital. If not, pass to the unitization. Since the norm on the unitization extends the norm on $A$, the result holds.

Answer 3

A different proof:

$$(p-q)^2=p(1-q)+q(1-p),$$ and then it is straightforward to show with induction (square at each step) that $$(p-q)^{2^k}=(p(1-q))^{2^{k-1}}+(q(1-p))^{2^{k-1}}$$ for each positive integer $k$. By the triangle inequality and the inequality $\|ab\|\leq\|a\|\|b\|$, this implies that $$\|(p-q)^{2^k}\|\leq 2$$ for all positive integers $k$. Because $p-q$ is self-adjoint, $\|(p-q)^{2^k}\|=\|p-q\|^{2^k}$, so the inequality above holding as $k\to \infty$ implies $\|p-q\|\leq 1$.

---

Although this is rather ad hoc and a proof analyzing the spectrum generalizes better, this proof has the property that it uses only the basic definitions. E.g., the fact that $\|a^{2^k}\|=\|a\|^{2^k}$ when $a$ is self-adjoint follows immediately from the C*-identity $\|x^*x\|=\|x\|^2$.

---

Here's a brief recap using the notation in the question. The norm is at most $1$ because

$$\forall k\in\mathbb N, \|p-q\|^{2^k}=\|(p-q)^{2^{k}}\|=\|s^{2^{k-1}}+(t^*)^{2^{k-1}}\|\leq 2.$$