Bounded operators with prescribed range - part II

Question

This is a continuation of the question bellow, in a more particular case.

Bounded operators with prescribed range

If $X$ is a separable Banach space and $Y$ is a closed, infinite dimensional subspace of $X$, can one always find an bounded operator $T$ on $X$ with $\rm{Range}(T)=Y$?

Even weaker version: Can one always find an bounded operator $T$ with infinite dimensional range such that $\rm{Range}(T)\subseteq Y$?

The last question feels much weaker and I suspect it is true. Intuitively, it would be strange if a large 'chunk' of Banach space is not 'visited' by any bounded operator (defined on the entire space).

Answer 1

The answer is no in general. Recall that if $T:X\longrightarrow Y$ is a surjective operator, then $X/{\rm ker}(T)$ is isomorphic to $Y$. The question may thus be equivalently stated: Let $X$ be a separable Banach and $Y$ a subspace of $X$. Does there exist a subspace $Z\subseteq X$ such that $X/Z$ is isomorphic to $Y$?

The answer is of course yes if $X$ is isomorphic to $\ell_2$. It is also yes if $X$ contains a complemented subspace isomorphic to $\ell_1$, because $\ell_1$ is surjectively universal for the class of separable Banach spaces (that is, $\ell_1$ admits a surjective operator onto any separable Banach space, and in particular onto any of its subspaces).

For a counterexample, let us look at the opposite phenomenon to the surjective universality property enjoyed by $\ell_1$; more precisely, let us consider a space that is injectively universal for the class of separable Banach spaces: every separable Banach space is (isometrically) isomorphic to a subspace of $C([0,1])$ (the Banach-Mazur theorem). So, if the OP's question were to have an affirmative answer, it would have to be the case that every separable Banach space is isomorphic to a quotient of $C([0,1])$; this is false.

Non-reflexive counterexamples: Let $Y$ be a nonreflexive subspace of $C([0,1])$ such that $c_0$ is not isomorphic to a subspace of $Y$ (e.g., $\ell_1$, the James space, and many others). Then there is no surjective operator from $C([0,1])$ onto $Y$. Indeed, a classical result of Pelczynski (see, e.g., p.119 of the book Topics in Banach space theory by Albiac and Kalton) tells us that non-weakly compact operators from $C(K)$ spaces fix a copy of $c_0$, and the claimed counterexample follows since a surjective operator onto a non-reflexive space is non-weakly compact.

Reflexive counterexamples: Every reflexive quotient of a $C(K)$ space is super-reflexive; this theorem is due to H.P. Rosenthal, who showed that a reflexive quotient of a $C(K)$ space is isomorphic to a quotient of $L_q(\mu)$ for some probability measure $\mu$ and $2\leq q<\infty$ (Corollary 11 of On subspaces of $L_p$, Ann. Math. 97 (1973), p.344-373) (Remark: it is true more generally that every reflexive quotient (in the Banach space sense) of a $C^\ast$-algebra is super-reflexive; this is due to Jarchow, On weakly compact operators on $C^\ast$-algebras, Math. Ann. 273 (1986), p.341-343). So, take $Y$ to be any reflexive subspace of $C([0,1])$ that is not super-reflexive. Then there is no surjective operator onto $Y$. For example, take $Y$ to be a subspace of $C([0,1])$ isomorphic to $(\bigoplus_{n=1}^\infty\ell_q^n)_{\ell_p}$, where $1<p<\infty$ and $q\in\{ 1,\infty\}$.

More reflexive counterexamples: let $1\leq p<\infty$. Then $\ell_p$ is isomorphic to a quotient of $C([0,1])$ if and only if $p\geq2$. Thus any subspace $Y$ of $C([0,1])$ such that $Y$ is isomorphic to $\ell_p$, for some $1<p<2$, is not a quotient of $C([0,1])$. (The fact that every operator from $C(K)$ to $\ell_p$ is compact whenever $1\leq p<\infty$ is Exercise 6.10 in the aforementioned book of Albiac and Kalton. Since $C([0,1])^{\ast\ast}$ is isomorphic to a space $L_\infty(\mu)$ for a suitable measure $\mu$, and since $\ell_p$ is reflexive for $1<p<2$, for the case $1<p<2$ it suffices to show that every operator from an $L_\infty(\mu)$ space to $\ell_p$ is compact whenever $1<p<2$; for some discussion of this see Remark 2 on p.211 of H.P. Rosenthal, On quasi-complemented subspaces of Banach spaces, with an appendix on compactness of operators from $L^p(\mu)$ to $L_r(\nu)$, J. Funct. Anal. 4 (1969), p. 176-214).

Answer 2

I can only answer the weaker version of the question. Using Hahn–Banach one can "inductively" construct sequences $(y_n)_{n=1}^\infty$ in $Y$ and $(\phi_n)_{n=1}^\infty$ in $X^*$ such that $\phi_n(y_n)=1$ and $\phi_n(y_m)=0$ if $m<n$. Let $T_n:X\to X$ be the rank one operator defined by $T_n(x)=\phi_n(x)y_n$. Let $T=\sum_{n=1}^\infty a_nT_n$, with $0\neq a_n\in\mathbb C$ small enough to make the sum convergent, e.g. $a_n=\dfrac{1}{\|T_n\|2^n}$. Then the range of the nuclear operator $T$ is contained in $Y$ (because $Y$ is closed) and contains the infinite dimensional span of $\{y_n\}_{n=1}^\infty$ (it is straightforward to show by induction that $y_n\in T(X)$ for each $n$, in fact $y_n\in T(\mathrm{span}\{y_1,\ldots,y_n\})$.

Answer 3

Argyros and Haydon constructed a separable Banach space $X$ such that each bounded linear operator $T\colon X\to X$ is of the form $T=cI_X + K$ where $K$ is compact.

Let $Y$ be an infinite-dimensional subspace of $X$ that has infinite codimension. Then $Y$ is not the range of any operator on $X$. Indeed, if $T\colon X\to X$ were an operator with ${\rm im}\,T=Y$, then $T$ is not a Fredholm operator by the very definition of a Fredholm operator. For each $c\neq 0$ and $K$ compact $cI_X + K$ is a Fredholm operator of index 0. Thus $T$ must be compact but the ball of $Y$ cannot be contained in the range of $T$ as it is not compact.

This example tells us that there exist spaces whose only finite-codimensional and finite-dimensional subspaces are operator ranges.