9
$\begingroup$

Let $X$ be a separable Banach space and $\mathcal L$ the collection of bounded linear operators on $X$. The strong operator topology has the sub-basis $\{B_{x,y,\epsilon}\colon x,y\in X,\epsilon>0\}$, where $B_{x,y,\epsilon}=\{T\colon \|Tx-y\|<\epsilon\}$. The Borel $\sigma$-algebra generated by this topology is called the strongly measurable $\sigma$-algebra on $\mathcal L$.

Now the question (which has arisen in a study of the multiplicative ergodic theorem on Banach spaces):

Let $K=\{T\in\mathcal L\colon \text{ker}(T)\ne\{0\}\}$. Is this set strongly measurable?

Thanks for any information.

Improve this question
$\endgroup$
12
  • 2
    $\begingroup$ Initial notes: (1) the unit ball $B$ of $\mathcal{L}$ with the strong operator topology is Polish; (2) it suffices to show $B \cap K$ is Borel in $B$; (3) the set $K$ is clearly analytic in $B$, so it would suffice to show it is coanalytic. $\endgroup$
    – Nate Eldredge
    Jun 13, 2017 at 21:16
  • $\begingroup$ Thanks @NateEldredge: I switched operator algebras for descriptive set theory. $\endgroup$
    – Anthony Quas
    Jun 14, 2017 at 5:44
  • 3
    $\begingroup$ I can answer this as deputy. Let $H$ be a separable Hilbert space. Then for a contraction $T \in B(H)$ and polynomials $p_n(t):=(1-t)^n$, the sequence $(p_n(T^*T))_n$ monotonically converges in SOT to the orthogonal projection onto $\ker T$. Hence for a dense sequence $(v_m)_m$ in the unit ball of $H$, one has $\ker T = 0$ iff $\forall m$ $\exists n$ $\|p_n(T^*T)v_m\|<1/2$. Since each $p_n$ is SOT*-continuous in $T$ and SOT* and SOT measurabilities are the same, this proves SOT measurability of $\{ T:\ker T=0\}$. I don't know a reference nor the case for Banach spaces. $\endgroup$
    – Narutaka OZAWA
    Jun 17, 2017 at 21:45
  • 1
    $\begingroup$ @MichaelGreinecker: So I believe (after scratching my head for a while/having it explained to me by my grad student) that the homeomorphism that Kadets produces isn't linear. This is quite bad, as a linear operator on $X$ would then be conjugate to a non-linear operator on $H$. $\endgroup$
    – Anthony Quas
    Jun 19, 2017 at 23:57
  • 1
    $\begingroup$ One can do it when $X$ is separable and reflexive (or more generally when $X$ is a separable dual space and $T$ is weak*-continuous). For a dense sequence $(x_m)_m$ in the unit ball of $X$, one has $\ker T \neq 0$ iff $\exists m$ $\forall K$ $\exists n$ such that $\| x_m - x_n \| < \| x_m \|/2$ and $\| T x_n \| < 1/K$. $\endgroup$
    – Narutaka OZAWA
    Jun 20, 2017 at 13:33

1 Answer 1

Reset to default
6
$\begingroup$

This is true at least when $X$ is a separable and reflexive. Take a dense sequence $(x_n)_n$ in the unit sphere of $X$. Then, for any $T \in B(X)$, one has $\ker T \neq 0$ iff $\exists m$ $\forall k$ $\exists n$ such that $\| x_m - x_n \| < 1/2$ and $\| Tx_n \| < 1/k$. Indeed, since the closed unit ball of $X$ is weakly compact and any $T \in B(X)$ is weak-weak continuous, if the latter condition holds, then any weak limit point $x$ of the subsequence $(x_{n(k)})_k$ satisfies $\| x \| \geq \| x_m \| - \| x_m - x \| \geq 1/2$ and $Tx = 0$. This proves the "if" direction. The "only if" direction is easy.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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