Let $A$ and $B$ be positive semidefinite matrices such that $0 \leq A,B \leq I$ and suppose that $A + B >0$. Does it follow that $\|(I-A)(I-B)\|_2 < 1$?
This is true for the scalar case--let $0\leq a,b\leq 1$ and $a+b = \alpha > 0.$ Then
$(1-a)(1-b) = e^{\log(1-a)(1-b)} = e^{\log(1-a)+\log(1-b)} \leq e^{-a-b} = e^{-\alpha} < 1.$
Unfortunately there's not really a matrix analog to this argument.
I've tried using the typical eigenvalue inequalities and messed around with complex representations using holomorphic functional calculus, but I can't get anything to work so far.
I've also done some computing and the result seems to be true; however, I could just have easily looked over a simple counterexample.
Yes. Since both $A$ and $B$ are between $0$ and $I$ in positive semidefinite partial ordering, both $\|I-A\|_2$ and $\|I-B\|_2$ are bounded above by $1$. Therefore, for any unit vector $x$, we have $$ \|(I-A)(I-B)x\|_2\le\|I-A\|_2\|(I-B)x\|_2\le\|I-A\|_2\|(I-B)\|_2\|x\|_2\le1.\tag{1} $$ If $\|(I-A)(I-B)\|_2=1$, then $\|(I-A)(I-B)x\|_2=1$ for some unit vector $x$. Therefore by $(1)$ we must have $\|(I-B)x\|_2=\|I-B\|_2=1$. Since $I-B$ is positive semidefinite, $\max_{\|u\|_2=1}\|(I-B)u\|_2$ is attained only at unit eigenvectors corresponding to the maximum eigenvalue of $I-B$. Therefore $(I-B)x=x$, i.e. $Bx=0$. But then the condition $\|(I-A)(I-B)x\|_2=1$ implies that $\|(I-A)x\|_2=1$. Therefore $Ax=0$ too. Hence $(A+B)x=0$, which is a contradiction to the assumption that $A+B>0$.