Projections in Banach spaces - MathOverflow

Projections in Banach spaces

2012-11-05T14:06:02Z

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.

Answer by Mikael de la Salle for Projections in Banach spaces

2012-11-05T21:06:17Z

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.

Answer by fedja for Projections in Banach spaces

2012-11-10T00:29:46Z

I'm not sure if I like the ultrafilters, so I decided to find some elementary construction. I do not have much imagination for chains of commuting projections either, so let us consider the space of all functions $f:[0,1]\to X$ where $X$ is the space of all sequences and let $P_s$ just keep the values to the left of $s$ and kill the values to the right of $s$. So far so good. The operator $Q$ will just act in each layer separately, so it is going to be an operator in $X$ really. The devil is in the choice of the norm. We need to take an advantage of small support somehow, which calls for considering a sequence of $L^{p_k}$-norms on $[0,1]$ with decreasing $p_k>1$. Then we'll have the full strength of Holder backing us. However, there needs to be some penalty for using norms with small $p_k$, so it is natural to pair each $p_k$ with some norm $N_k$ in $X$, which are increasing fast so that the final norm, which is just the infimum of $\sum_k\|N_k(f_k(t))\|_{L^{p_k}}$ over all decompositions $f=\sum_k f_k$ will have to be a hard trade-off rather than a trivial collapse to a single term. Now the first thing we want is $$ N_{k+1}(Qx)\le N_k(x) $$ This will ensure that for the first small $k$ we will fire with Holder to gain on the power of the length of the interval but for the large $k$, when Holder finally betrays us and the reduction of power becomes useless, we will use $$ N_k(Qx)\le 2^{-k} N_k(x) $$ so each faraway term will just take care of itself. The possibilities for the choice of such family of norms and $Q$ are unlimited but, being (sort of) an analyst, I like the backward shift and weighted spaces, so put $$ (Qx)_{j}=2^{-j}x_{j+1}, $$ and $$ N_k(x)=\sum_j 2^{kj}|x_j|. $$

Answer by jjcale for Projections in Banach spaces

2012-11-18T11:41:52Z

Here a simple example :

Let X be the cartesian product of $L^{\infty}$ and $L^{1}$ on the interval $[0,1]$, let $P_{t}$ the canonical projection on the subspace of functions with support $[0,t]$ and choose $Q(f_{1},f_{2}) = (0,f_{1})$ .