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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.

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    $\begingroup$ 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 '17 at 21:16
  • $\begingroup$ Thanks @NateEldredge: I switched operator algebras for descriptive set theory. $\endgroup$ – Anthony Quas Jun 14 '17 at 5:44
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    $\begingroup$ 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 '17 at 21:45
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    $\begingroup$ @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 '17 at 23:57
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    $\begingroup$ 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 '17 at 13:33
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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.

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