3
$\begingroup$

In the case of Hilbert spaces $\mathcal H$, $\mathcal G$ it is known that the Banach space of all bounded operators $\mathcal B(\mathcal H,\mathcal G)$ is norm-separable if and only if $\operatorname{dim}(\mathcal H)<\infty$ and $\mathcal G$ is separable (or vice versa). In particular if $\operatorname{dim}(\mathcal H)=\infty$ then $\mathcal B(\mathcal H)$ is never norm-separable (regardless of whether $\mathcal H$ is separable or not) as is usually shown via a diagonal argument which isometrically embeds the non-separable space $\ell^\infty$ into $\mathcal B(\mathcal H)$. The overall question I am interested in is:

Given infinite-dimensional normed spaces $X,Y$ (where at least one is not a Hilbert space) is $\mathcal B(X,Y)$ non-separable? Can an affirmative answer be given for some special cases (such as $X=Y$, or $X$ Banach, or $X$ having a Schauder basis, etc.)? If so what is the idea of the proof?

I would like to sketch the Hilbert space argument in more detail to point out where I see issues arising from passing over to more general spaces.

Given an infinite-dimensional Hilbert space $\mathcal H$ (and assuming the axiom of choice) we can find an orthonormal basis $\{x_i\}_{i\in I}$ of $\mathcal H$ (w.l.o.g. $\mathbb N\subset I$). This yields an orthonormal system $\{x_n\}_{n\in\mathbb N}$ in $\mathcal H$ which---for any $z\in\ell^\infty$---lets us define $T_{z,0}(x_n):=z_nx_n$ for all $n\in\mathbb N$ and $T_{z,0}(x_i):=0$ for all $i\in I\setminus\mathbb N$ together with the linear extension onto all of $\mathcal H_0:=\operatorname{span}\{x_i\}_{i\in I}$. Thus $T\in\mathcal L(\mathcal H_0)$ and, moreover $$ \|T_{z,0}\|=\sup_{x\in\mathcal H,\|x\|=1}\|T_{z,0}x\|\geq\sup_{n\in\mathbb N}\|T_{z,0}x_n\|=\sup_{n\in\mathbb N}|z_n|=\|z\|\,. $$ For the converse let any unit vector $x\in\mathcal H_0$ be given so one finds coefficients $\alpha_k$ such that $x=\sum_{k=1}^m \alpha_k x_{n_k}+\sum_{k=1}^{m'}\alpha_k' x_{i_k}$ (where $n_1,\ldots,n_m\in\mathbb N$ and $i_1,\ldots,i_k\in I\setminus\mathbb N$). Using the Pythagorean theorem this yields $$ \|T_{z,0}x\|^2=\Big\|\sum\nolimits_{k=1}^m \alpha_k z_{n_k}x_{n_k}\Big\|^2=\sum\nolimits_{k=1}^m|\alpha_k|^2|z_{n_k}|^2\leq \|z\|^2\sum\nolimits_{k=1}^m|\alpha_k|^2\leq \|z\|^2\|x\|^2\tag{1} $$ Thus in total $\|T_{z,0}\|=\|z\|$ and $T_{z_0}\in\mathcal B(\mathcal H_0)\subset\mathcal B(\mathcal H_0,\mathcal H)$. Because $\overline{\mathcal H_0}=\mathcal H$ (orthonormal basis property) there exists a unique extension $T_z\in\mathcal B(\mathcal H)$ with $\|T_z\|=\|T_{z_0}\|$ so the map $$ \iota:\ell^\infty\to\mathcal B(\mathcal H)\qquad z\mapsto T_z $$ is well-defined, linear and an isometry. But with this the standard proof that $\ell^\infty$ is not separable transfers onto $\mathcal B(\mathcal H)$ showing non-separability of the latter.

$$\boxed{\text{So far, so standard.}}$$

(1) The easiest generalization one might think of is the case of $\mathcal B(X)$ where $X$ is a Banach space which has a Schauder basis $\{x_n\}_{n\in\mathbb N}$ (w.l.o.g. $\|x_n\|=1$). Pursuing the Hilbert space idea given $z\in\ell^\infty$ one can define an operator $T_{z,0}\in\mathcal L(X_0)$ via $T_{z,0}x_n:=z_nx_n$ on $X_0:=\operatorname{span}\{x_n\}_{n\in\mathbb N}$ which satisfies $\|T_{z,0}\|\geq \|z\|$.

Problem 1: Why should $T_{z,0}$ be bounded? (After all (1) crucially relies on the Pythagorean theorem and there to my knowledge is no general connection between the coefficients $\alpha_k$ and $\|\sum_{k=1}^m\alpha_k x_{n_k}\|_X$)

Assuming we can surmount these difficulties we get a unique extension (because of the Schauder basis property and because $X$ is a Banach space) $T_z\in\mathcal B(X)$ so the map $\iota:\ell^\infty\to\mathcal B(X)$, $z\mapsto T_z$ is well-defined, linear and satisfies $\|T_z\|=\|T_{z_0}\|\geq \|z\|$. This is enough to carry over the non-separabilty argument of $\ell^\infty$ because $\|T_z-T_{\tilde z}\|=\|T_{z-\tilde z}\|\geq\|z-\tilde z\|$.

(2) The next step would ask about $\mathcal B(X,Y)$ where $X,Y$ are Banach spaces ($X\neq Y$) and $X$ has a Schauder basis. Same procedure as before but now take a set $\{y_n\}_{n\in\mathbb N}$ of linearly independent unit vectors in $Y$ and define $T_{z,0}x_n:=z_ny_n$. Then $T_{z,0}\in\mathcal L(X_0,Y)$ with $\|T_{z,0}\|\geq\|z\|$.

Problem 2: Why should $T_{z,0}$ be bounded? (Following Problem 1 we additionally have to deal with the problem of $\|\cdot\|_Y$ being independent of $\|\cdot\|_X$, i.e. we would have to relate the coefficients $\alpha_k$ of $\sum_{k=1}^m\alpha_kx_{n_k}$ with the norm of the image $\sum_{k=1}^m\alpha_k\boxed{y_{n_k}}$)

Solving this problem would enable us to finish the argument as before.

(3) Assuming things did not break down until here the most general (and also the most unpleasant) case to consider is the one of the domain being an arbitrary Banach space $X$. As before one could find an infinite set $\{x_n\}_{n\in\mathbb N}$ of linearly independent unit vectors in $X$ (and similarly for $Y$) to define $T_{z,0}:X_0\to Y$ via $T_{z,0}x_n:=z_ny_n$ where $X_0:=\operatorname{span}\{x_n\}_{n\in\mathbb N}$. Again $\|T_{z,0}\|\geq\|z\|$.

Problem 3: Even if one could prove that $T_{z,0}$ is bounded how would one extend $T_{z,0}$ to a bounded operator on all of $X$? (We assume that $X$ does not have a Schauder basis anymore so $X_0$ is not dense in $X$ no matter how we choose our linearly independent unit vectors. Thus the image of $X\setminus \overline{X_0}$ has to be decided on beyond the usual continuity argument)

Such an extension seems possible if $T_{z,0}$ is of finite rank or if $Y$ is an injective space but seems dangerous in general.

These are most of the thoughts I had on this topic up until now. I of course am aware that there might be a more general argument (beyond this diagonal one) which settles (non-)separability of $\mathcal B(X,Y)$ for general normed spaces $X,Y$. Either way I am thankful for any comments and or ideas!

$\endgroup$
4
$\begingroup$

There are infinite-dimensional separable Banach spaces $X$, $Y$ such that all bounded linear operators from $X$ to $Y$ are compact. For example, by a result of Phillips (Phillips, R. S. "On Linear Transformations." Transactions of the American Mathematical Society 48, no. 3 (1940): 516-41) this is the case if $X = c_0$ and $Y$ is a separable conjugate space. If in addition $Y$ satisfies the approximation property (in particular if it has a Schauder basis), then the finite-rank operators are dense in the compact operators from $X$ to $Y$, and this should make ${\mathcal B}(X,Y)$ separable.

$\endgroup$
2
  • $\begingroup$ As I feared for general Banach spaces the question of separability becomes a lot messier. Thank you very much! $\endgroup$ Jan 2 '20 at 18:11
  • 1
    $\begingroup$ Maybe I have missed something in the cited paper, but how about the case $X=Y$ (see the OP)? $\endgroup$
    – Skeeve
    Jan 3 '20 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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