Given two projections $p,q\in B(H)$, it is well-known that if $\|p-q\|<1$, then there exists a unitary $u\in B(H)$ with $q=upu^\*$. 


The proof that immediately occurs to me uses comparison of projections (the inequality forces $p\sim q$ and also $p^\perp\sim q^\perp$, and then one constructs the unitary by adding the corresponding partial isometries), and implies that one can even replace $B(H)$ with any von Neumann algebra that contains both $p$ and $q$. 



> 
> Questions:
> 
> 
> 1. Does anyone know of a more direct proof? Comparison of projections is a very basic result in von Neumann algebras, but somehow here it feels like overkill.
> 2. Does this property hold in a C$^\*$-algebra? Proof, or counterexample?
> 
> 
> 



If two projections are close, then they are unitarily equivalent
**Edit:** I'm leaving the old post below, but before I want to write the proof as suggested by Bruce from his book, which uses the ideas in a more efficient way.


Assume that $\|p-q\|<1$, with $p,q\in A$, a unital C$^\*$-algebra. Let $x=pq+(1-p)(1-q)$. Then, as $2p-1$ is a unitary, 
$$
\|1-x\|=\|(2p-1)(p-q)\|=\|p-q\|<1.
$$
So $x$ is invertible. Now let $x=uz$ be the polar decomposition, $z=(x^\*x)^{1/2}\in A$. Then $u=xz^{-1}\in A$. Also, $px=pq=xq$, and $qx^\*x=qpq$, so $qx^\*x=x^\*xq$, and then $qz=zq$. Then
$$
pu=pxz^{-1}=xqz^{-1}=uzqz^{-1}=uqzz^{-1}=uq.
$$
So $q=u^\*pu$.


=============================================
*(the old post starts here)*


*(A good friend pointed me to the ideas in this answer, so I'm sharing them here)*


The result holds in any unital C$^\*$-algebra. So assume that $\|p-q\|<1$, with $p,q$ in a unital C$^\*$-algebra $A\subset B(H)$. 



> 
> **Claim 1:** There is a continuous path of projections joining $p$ and $q$.
> 
> 
> 


*Proof.* Let $\delta\in(0,1)$ with $\|p-q\|<\delta$. For each $t\in[0,1]$, let $x\_t=tp+(1-t)q$. Then
$$
\|x\_t-p\|=\|(1-t)(p-q)\|<\delta(1-t),
$$
$$
\|x\_t-q\|=\|t(p-q)\|<\delta t.
$$
This, together with the fact that $x\_t$ is selfadjoint, implies that $\sigma(x\_t)\subset K=[-\delta/2,\delta/2]\cup[1-\delta/2,1+\delta/2]$ (since $\min\{t,1-t\}\leq1/2$). Now let $f$ be the continuous function on $K$ defined as $0$ on $[-\delta/2,\delta/2]$ and $1$ on $[1-\delta/2,1+\delta/2]$. Then, for all $t\in[0,1]$, $f(x\_t)\in A$ is a projection. And
$$
t\to x\_t\to f(x\_t)
$$
is continuous, completing the proof of the claim. **Edit:** years later, I posted [this answer](https://math.stackexchange.com/a/1810356/22857) to a question on MSE that proves the continuity.



> 
> **Claim 2:** We may assume without loss of generality that $\|p-q\|<1/2$.
> 
> 
> 


This is simply a compacity argument, using that each projection in the path $f(x\_t)$ is very near another projection in the path. The compacity allows us to make the number of steps finite, and so if we find projectons $p=p\_0,p\_1,\ldots,p\_n=q$ and unitaries with $u\_kp\_ku\_k^\*=p\_{k+1}$, we can multiply the unitaries to get the unitary that achieves $q=upu^\*$. 



> 
> **Claim 3:** If $\|p-q\|<1/2$, there exists a unitary $u\in A$ with $q=upu^\*$. 
> 
> 
> 


Let $x=pq+(1-p)(1-q)$. Then 
$$
\|x-1\|=\|2pq-p-q\|=\|p(q-p)+(p-q)q\|\leq2\|p-q\|<1,
$$
so $x$ is invertible. Let $x=uz$ be the polar decomposition. Then $u$ is a unitary. Note that
$$
qx^\*x=q(qpq+(1-q)(1-p)(1-q))=qpq,
$$
so $q$ commutes with $x^\*x$ and then with $z=(x^\*x)^{1/2}$. Note also that $px=xq$, so $puz=uzq=uqz$. As $z$ is invertible, $pu=uq$, i.e.
$$
q=u^\*pu.
$$
Note that $u=xz^{-1}\in A$. 


