# Measurability of $T \to \Pi_{ker T}$ w.r.t. SOT

I came across the following technical question, to which I could not - after some time of thinking - find an answer:

Let $\mathcal{U},\mathcal{H}$ be two real (in general infinite dimensional) separable Hilbert spaces. For some linear subspace $\bar{U} \subseteq \mathcal{U}$, let $\Pi_U$ denote the orthogonal projection on this subspace. The question is:

Is the mapping $T \mapsto \Pi_{\text{ker}T}$ measurable from $L(\mathcal{U},\mathcal{H})$ to $L(\mathcal{U})$ when both spaces are endowed with the strong operator topology (and measurability is meant w.r.t. the Borel-$\sigma$-algebra of the SOT on both sides)?

Any hints and thoughts on this are more than appreciated!

• can you write the projection as $lim 1 - (T^*T)^{\frac 1 n}$, and therefore the limit of continuous funtions ?
– user83457
Mar 13, 2018 at 15:34
• Could you please elaborate on this? I can't work your argument out. Would appreciate it! Mar 15, 2018 at 9:40
• It looks like the answer below does that
– user83457
Mar 15, 2018 at 11:09
• I (partially) agree. However, how do you deal with the two questions I added below? Mar 15, 2018 at 12:35

It is true if you consider the unit ball. (Which should be sufficient because $\Pi_{\ker(\lambda T)} = \Pi_{\ker(T)}$ for all $\lambda\neq 0$.)

Also note that $\ker(T) = \ker(T^\ast T)$. I will therefore restrict my attention to normal operators $T$. That may or may not be sufficient for the application you have in mind.

Then we have $\Pi_{\ker(T)} = 1_{\{0\}}(T)$, where we use the measurable functional calculus.

If we define $f_n(x) = \begin{cases} (1-|x|)^n & |x|\leq 1 \\ 0 & |x|\geq 1\end{cases}$ we have a sequence continuous, uniformly bounded functions on $\mathbb{C}$ with $f_n\to 1_{\{0\}}$ pointwise. Therefore $f_n(T) \xrightarrow[n\to\infty]{s.o.t} 1_{\{0\}}(T)$.

It is therefore sufficient to prove that $B_1(L(U))_{sot} \to L(U)_{sot}, T\mapsto f(T)$ is continuous for all continuous $f:\mathbb{R}\to[0,1]$. If $T_i\xrightarrow{s.o.t} T$ and $x\in U$ arbitrary, then approximate $f$ uniformly on the unit disk $\mathbb{D}=\{z\in\mathbb{C} : |z|\leq1\}$ by some polynomial $p$. Now remember that $p(X)-p(Y) = \tilde{p}(X,Y)(X-Y)$ for some polynomial $\tilde{p}$ of degree $\deg(p)-1$ and that $\mathbb{D}$ contains $\sigma(T)$ for all $T\in B_1(L(U))$ so that:

\begin{align*} \|f(T)x-f(S)x\| &\leq \|p(T)x-p(S)x\| + 2\epsilon\|x\| \\ &\leq \|\tilde{p}(T,S)\cdot (T-S)x\| + \epsilon\|x\| \\ &\leq const\|(T-S)x\| + 2\epsilon\|x\| \end{align*}

Now if $T\xrightarrow{s.o.t}S$, then this proves $\forall\epsilon>0: \limsup_{T\to S} \|f(T)x-f(S)x\|\leq 2\epsilon\|x\|$ to that $f(T)\xrightarrow{s.o.t}f(S)$.

• How does non-measurability follow from that? Mar 13, 2018 at 14:56
• It doesn't. I thought you asked about continuity. Mar 13, 2018 at 15:22
• I have edited the answer. Mar 13, 2018 at 16:04
• Thank you very much for your answer! I have no experience with measurable functional calculus, so right now I cannot follow all of your arguments. However, I see your general line of argument on how to obtain the measurability. But intuitively: Concerning the reduction to the unit ball, isn't it a problem that the constant $\lambda$ can certainly not be chosen independent of $T$? Wouldn't you need to be able to push an arbitrary open set of $L(U,H)$ into the unit ball? Mar 14, 2018 at 12:19
• Why do the weak and strong operator topology coincide on bounded sets of $L(U,H)$, $L(H,U)$ respectively? I could neither prove this myself nor find a source, which states this statement. Can you help me out on this? Mar 15, 2018 at 9:07