Dear All, 


I am absolutely lost in the following problem:


Let $P\_s, \: s \in [0,1],$ be a uniformly bounded family of projections (idempotents) in a Banach space $X$ such that $P\_s P\_t = P\_{{\rm min}(s,t)}$. Let $Q$ be a bounded linear operator on $X$ such that $QP\_s = P\_sQ$ for every $s \in [0,1]$ and the function $$s \mapsto P\_s Q $$ is continuous in operator norm. Does it follow that the function must be constant (i.e. $P\_sQ \equiv P\_0Q$)? 


For simplicity, one can also assume that the function $s \mapsto P\_s$ is strongly continuous. I can give an affirmative answer only in some trivial situations (finite dimensional case, Hilbert spaces with family of orthogonal projections) but nothing more. 



Projections in Banach spaces
I guess that the answer is no in general. More precisely what I consider as the discrete version of your question has a negative answer. I guess that one should be able to find a couterexample to your question by an ultraproduct argument, but I did not check the details.


By discrete version of your question I mean: if $(P\_n)\_{n \in \mathbb N}$ is a family of uniformly bounded projections on $X$ such that $P\_0=0$ and $P\_n P\_m = P\_{\min(n,m)}$, and $Q$ is a (not necessaily bounded) linear map on $X$ that commutes with the $P\_n$'s and is such that $\sup\_n \|P\_n Q -P\_{n-1} Q\|<\infty$. Does it follow that $\sup\_n \|P\_n Q\|<\infty$?


If $(e\_n)$ is a Schauder basis in $X$, and take $P\_n$ the map $x = \sum\_k x\_k e\_k \mapsto P\_n(x) = \sum\_{k \leq n} x\_k e\_k$. Then $P\_n$ satisfies your assumption. For a sequence $z\_n \in \{{-1,1\}}$, define $Q x\_n = z\_n x\_n$ so that $\|P\_n Q - P\_{n-1} Q\| = \|P\_n - P\_{n-1}\|$. Then $\sup\_n \|P\_n Q\| <\infty$ for every such $Q$ is equivalent to the basis $(e\_n)$ being unconditional. It is well known that there exist bases that are not unconditional. This answers negatively the discrete question.


Here is how I would try to deduce a continuous example from a discrete example $X,P\_n,Q$~: for any $k$, define $P\_s^{(k)} = P\_{[2^k s]}$ and $Q\_k = P\_{2^k}Q/\|Q P\_{2^k}\|$. Then consider a non principal ultrafilter on $\mathbb N$, and construct the following operators on $X^{\mathcal U}$, and the operators $P\_s = (P\_s^{(k)})\_{k \in \mathbb N}$ and $Q\_{\mathcal U} = (Q\_k)\_{k \in \mathbb N}$. Then $\|Q\_{\mathcal U}\|=1$, so that $P\_0 Q\_{\mathcal U} = 0 \neq Q\_{\mathcal U}= P\_1 Q\_{\mathcal U}$. One probably needs to assume something more on $Q$ to prove that $s \mapsto P\_s Q$ is continuous. I leave this to you.


