# Projections in Banach spaces

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.

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Are you sure that there is an affirmative answer in the Hilbert space case with a family of orthogonal projections? Can't you take the Voltera nest of subspaces and for $Q$ any compact operator which leaves these subspaces invariant? – Bill Johnson Nov 5 '12 at 17:19
@Bill: there are not that many compact operators that leave the Voltera nest invariant... – Mikael de la Salle Nov 5 '12 at 20:13
I had the idea on the orthogonality with $P_0=0, P_1=I$: $\|Qx\|^2=\sum_i\|(P_{s_i}−P_{s_{i+1}})Qx\|^2=\sum_i \|Q(P_{s_i}−P_{s_{i+1}})^2x\|^2$ $\leq \sum_i \|Q(P_{s_i}−P_{s_{i+1}})\|^2\|(P_{s_i}−P_{s_{i+1}})x\|^2 \leq \varepsilon \sum_i \|(P_{s_i}−P_{s_{i+1}})x\|^2= \varepsilon \|x\|^2$, where $\varepsilon$ can be arbitrarily small if we choose well enough $s_i$ point; that is, $Qx=0$ for any vector $x$, i.e. $Q=0$ must hold. – zoltan.leka Nov 5 '12 at 21:38
@Mikael: Bill's suggestion does sound good to me. If the kernel $k(x,y)$ of an integral operator on $L_2([0,1])$ is "upper triangular" (i.e,, supported on $\{(x,y)\mid x\geq y\}$) then it has the desired property. And $k(x,y)$ is an $L_2$ function on the square then the operator is Hilbert-Schmidt. – Leonel Robert Nov 5 '12 at 22:05
Ok, I see where the misunderstanding comes from: of course there are compact operators that leave the Voltera nest invariant (ie such that $P_s Q P_s = Q P_s$), but there are no non-zero compact operators such that $P_s Q = Q P_s$ for all $s$. And Zoltan's argument works. – Mikael de la Salle Nov 6 '12 at 12:30

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.

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@Mikael: I got the discrete case - thanks. But it seems to me that the ultraproduct construction gives a very special projection family: $P_0 = 0$ and $P_s = I$ for every $0 < s \leq 1,$ so $s \mapsto P_sQ$ cannot be continuous at $0$ unless $Q=0.$ For me, it is not obvious the continuous case. – zoltan.leka Nov 8 '12 at 14:35
@Zoltan. I do not follow you, $P_s$ is never equal to the identity. Indeed, none of the $P_n$'s is the identity, so that for every $k$ there is a norm $1$ vector $x_k \in X$ such that $P_s^{(k)} x_k = 0$. If $x$ is the class of $(x_k)_{k}$ in $X^{\mathcal U}$ then $x$ is a norm one vector and $P_s x$ is the class of $(P_s^{(k)}(x_k))_k = (0)_k$, so that $P_s x=0$. – Mikael de la Salle Nov 8 '12 at 15:40
@Mikael: Oh, okay, okay... you are right. – zoltan.leka Nov 8 '12 at 15:55
@Zoltan: you accepted my answer, thanks! But does this mean that you managed to show that for some correct choice of $X$, $(P_n)$ and $Q$, the map $s \mapsto P_s Q_{\mathcal U}$ is continuous? If so I would also be interested to see the argument. – Mikael de la Salle Nov 8 '12 at 16:16
I accepted your answer because I like your sketch and the basic idea seems to be nice. However, the details have not been worked out yet... But somehow I expected a different answer, so probably there are non-constant examples among $L^p$-spaces $(p \neq 2)$ as well. Hmm, that's surprising for me. – zoltan.leka Nov 8 '12 at 16:24

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 $Q(f_{1},f_{2}) = (0,f_{1})$ .

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I'm probably missing something: why is $P_tQ$ norm-continuous? – fedja Nov 18 '12 at 12:19
@fedja : Because $\|(P_{t+\varepsilon}-P_{t})Q(f_{1},f_{2})\| \leq \varepsilon \|f_{1}\|_{\infty} \leq \varepsilon \|(f_{1},f_{2})\|$ – jjcale Nov 18 '12 at 12:32
Very nice, jjcale. Basically the same argument works for $L_p \oplus L_q$ as long as $p\not= q$, so you also get an example for $L_p$ when $1<p\not= 2 <\infty$ because $L_2$ is isomorphic to a complemented subspace of $L_p$. – Bill Johnson Nov 19 '12 at 1:42
@jjcale Thanks! This is very cute :) – fedja Nov 21 '12 at 11:54

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 $$N_k(x)=\sum_j 2^{kj}|x_j|.$$

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This is very nice. – Mikael de la Salle Nov 10 '12 at 20:23
Thank you for your answer! That norm is quite tricky... – zoltan.leka Nov 10 '12 at 20:25
I tried to show how one can guess things like that rather than just providing a formal argument. Yes, the final formula is somewhat convoluted, but each step is fairly natural, isn't it? – fedja Nov 10 '12 at 21:13