Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question

3 Answers 3

up vote 4 down vote accepted

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).

share|improve this answer

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 $y_n\in T(\mathrm{span}\{y_1,\ldots,y_n\})$.

share|improve this answer
I think I understand. Just to be clear, the range of $T$ is contained in the closure of the span of $y_n$'s, and, since $Y$ is closed, it must be contained in $Y$. It looks that the condition $Y$ closed is important, right? –  Markus Aug 25 '13 at 18:14
@Markus: Exactly what I had in mind. I don't know how to answer if $Y$ isn't closed. I'll edit to make it a little more explicit. –  Jonas Meyer Aug 25 '13 at 18:16
Yes, in fact the closure of the range equals the closure of span of $y_n$'s. I suspect the general case is still open. Thank you. I will wait a few days, in case someone knows the general case, otherwise I will accept your answer. –  Markus Aug 25 '13 at 18:32

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.

share|improve this answer
Excellent example! –  Markus Aug 14 at 21:07
I was wondering, in your example can a closed infinite dimensional and codimensional $Y$, be the closure of the range of a bounded operator? By the same reasoning, $T$ would have to be compact. –  Markus Aug 14 at 23:38
@Markus, Let $X$ and $Y$ be separable Banach spaces. I claim that there is an injective nuclear operator $T\colon X\to Y$ with dense range. Let $(y_n)_{n=1}^\infty$ be a sequence that is dense in the unit sphere of $Y$. Choose a sequence $(f_n)_{n=1}^\infty$ in the unit ball of $X^*$ that is total. Let $Tx = \sum_{n=1}^\infty \tfrac{1}{2^n} \langle x, f_n\rangle y_n$. Clearly $T$ is injective as $(f_n)_{n=1}^\infty$ is total. The range of $T$ is dense in $Y$. –  Tomek Kania Aug 14 at 23:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.