On operator ranges in Hilbert & Banach spaces

Question

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 $f(Kg) = \mathbb Efg$.

Answer 1

(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

Answer 2

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.

Answer 3

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.

Answer 4

In Embry's 1973 and Douglas' 1966 papers the Hilbert space condition $AA^* < \lambda^2 BB^*$ about the adjoints is replaced by the following equivalent Banach type of condition: $\|A\| < \lambda \|B\|$, so that the whole exotic question of Banach space adjoints $A^* : X \to X$ analogous to Hilbert space adjoints is avoided.

On the other hand, if $A:X \to Y$, the Banach space adjoint defined as $A^* : Y^* \to X^*$ using duality theory does play an important role in the results of Douglas and Embry. Embry's is accessible on the free internet and is easy to read.