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Lemma 1 from Anderson & Trapp's Shorted Operators, II is

Let $A$ and $B$ be bounded operators on the Hilbert space $\mathcal H$. The following statements are equivalent:

(1) ran($A$) $\subset$ ran($B$).

(2) $AA^\* \le \lambda^2 BB^\*$ for some $\lambda \ge 0$.

(3) There exists a bounded operator $C$ such that $A = BC$.

Moreover, if (1), (2) and (3) are satisfied, there exists a unique operator $C$ so that ker($A$) = ker($C)$ and ran($C$) $\subset$ closure(ran($B^\*$)).

They follow this with the statement, "the lemmas and the original proofs remain valid for operators between two Hilbert spaces."

Question:I would like to know if there is a similar statement for more general Banach spaces, and if so, where I might find it.

My context: I am considering the Banach space $\Omega = C(U_1) \times C(U_2)$ of continuous functions over two domains. I have a covariance operator $$K : \Omega^\* \to \Omega$$ which is decomposed as $$K = \binom{K_{11} ~ K_{12}}{K_{21} ~ K_{22}}.$$ I want to apply the above lemma to $A = K_{21}$ and $B = K_{22}^{1/2}$.

Edit: If we have a probability measure $\mathbb P$ on $\Omega$, then continuous linear functionals $\Omega^\*$ are random variables. Thus the expectation $\mathbb Efg$ for $f, g \in \Omega^\*$ is well-defined. The covariance operator is the bilinear form defined by $f(Kg) = \mathbb Efg$.

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What is the norm on $C(U_1)\times C(U_2)$? Just the maximum, so that we really get $C(U_1 \cup U_2)$, the continuous functions on the disjoint union of $U_1$ and $U_2$? Then $\Omega^*$ is the $\ell_1$ sum of $M(U_1)$ with $M(U_2)$. So I don't understand your "edit"... –  Matthew Daws Feb 4 '10 at 21:11
    
@Matt, yes. I'm looking at continuous functions over a union $C(U \cup V)$ where $U$ and $V$ are compact planar regions, and considering the sup norm. –  Tom LaGatta Feb 4 '10 at 22:19

3 Answers 3

up vote 4 down vote accepted

(1) does not generally imply (3) for bounded operators between Banach spaces. The first example I have a reference for was due to Douglas and was included in "Factorization of operators on Banach space" by Embry in 1973. That paper has much more that might interest you, such as the fact that a factorization holds when you have the reverse range inclusion of the adjoints.

See also: http://www.jstor.org/stable/2043114

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+1 because that reference might plug in as part of background to something I've been toying with on and off for the last 5 years :) –  Yemon Choi Feb 4 '10 at 21:36
    
Jonas, thank you for the reference. This answers my question in the negative. Now I need to figure a way to make it work for a positive operator K. –  Tom LaGatta Feb 4 '10 at 23:01
1  
You're welcome. Funny thing about the second link: The preview page is the entire article, but if you click on it without a subscription to JSTOR you are invited to buy it for 24 USD. –  Jonas Meyer Feb 10 '10 at 15:23

One way to reformulate (1) as a factorization result is this:

Suppose $S:X\to Z$ and $T:Y\to Z$ are bounded linear operators and $SX\subset TX$. Then $S$ factors through the map $T$ induces from $Y/T^{-1}(0)$ into $Z$. To see this, WLOG $T$ and $S$ are one to one, and just observe that by e.g. the closed graph theorem $SX\subset TX$ implies that for some $a$, $SB_X \subset aTB_Y$.

Of course, this implies that if $T^{-1}(0)$ is complemented in $Y$, then $S$ factors through $T$ itself.

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Slightly off-topic: but this is vaguely reminiscent of some of the setup in Waelbroeck's "Quotient Banach spaces" series of papers - which I believe you referred to on some other question? –  Yemon Choi Feb 4 '10 at 23:11
    
IIRC Waelbroeck's set up was more complicated and he added morphisms to make the categorical aspects nicer. –  Bill Johnson Feb 4 '10 at 23:19
    
That sounds right - it was his version of something done by Guy Noel, I think, & meant to be more to the taste of functional analysts... –  Yemon Choi Feb 4 '10 at 23:57

My initial impression is that for what you want, you're going to need a notion of $A^*: E\to E$ when $A:E\to E$ is an operator on a Banach space. I don't know much about this, but some years ago did see this short paper

MR2053349 (2005a:46045)
Gill, Tepper L.(1-HWRD-EE); Basu, Sudeshna(1-HWRD); Zachary, Woodford W.(1-HWRD-EE); Steadman, V.(1-DC)
Adjoint for operators in Banach spaces. Proc. Amer. Math. Soc. 132 (2004), no. 5, 1429--1434

which requires a choice of Hilbert space rigging $H_1 \hookrightarrow E \hookrightarrow H_2$.

One thing that might go wrong with $(1) \implies (3)$ in general Banach spaces is the non-existence, in general, of projections from $E$ onto a closed subspace. However, that doesn't rule out the possiblity that something like $(1)\implies(3)$ does indeed hold; I'd need to think about this a bit more.

Edit: ah, I see that in your setting the operators go from one Banach space to another, rather than from the space to itself. That might make a difference: and indeed, since you're mapping into a $C(K)$-space and not just an arbitrary one, more tools might be available.

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@Yemon, thanks for the fast response. In this context, K is an operator from Omega* to Omega. I neglected to write that because I'm still trying to wrap my head around this functional analytic point of view! If I do everything by the abstract Wiener space construction, then there is the Hilbert-Schmidt subspace to take advantage of, so the Hilbert space rigging actually arises quick naturally. (1) => (3) is in fact the direction I am trying to prove. Thanks! –  Tom LaGatta Feb 4 '10 at 21:01
    
(quite not quick) –  Tom LaGatta Feb 4 '10 at 21:01
    
I meant Cameron-Martin, not Hilbert-Schmidt. You can tell I was in a rush when I wrote that comment. @Yemon, that's a really enlightening paper. Thanks for the reference! –  Tom LaGatta Feb 5 '10 at 17:14

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