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Take $2n$-tuples of bounded positive operators $x_1,\dots x_n$ and $a_1,\dots a_n$ on a Hilbert space $H$ which have zero kernel and dense image and which satisfy the condition that (1) $$ x_1^* x_1+\dots+x_n^* x_n+a_1^* a_1+\dots+a_n^* a_n $$ is invertible and (2) that the image of $x_i^*x_i+a_i^*a_i$ and $x_i^*$ have dense intersection for each $1\le i\le n$ and (3) that we have the equality (used to define a function $T$ of the $2n$-tuple) $$ T:=x_1(x_1^*x_1+a_1^*a_1)^{-1}x_1^*=\dots=x_n(x_n^*x_n+a_n^*a_n)^{-1}x_n^*\in B(H). $$ The problem is to show that $T$ varies continuously as the $2n$-tuple (subject to the conditions) varies continuously in the norm topology.

[Edit: Each formula for $T$ is a densely defined operator of norm $\le 1$, and so it extends to a bounded operator on $H$. The problem is that as far as I can see no individual formula can show the continuity of $T$ as the $2n$-tuple varies...]

This question may seem strange, but it comes from representing rings of fractions in a Hilbert space. If anyone can say anything about it I would be very grateful!

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    $\begingroup$ Will you please clarify: 1. Are the $x_j$ and $a_j$ assumed to be bounded operators? 2. Are $x^*_jx_j+a_j^*a_j$ assumed to be invertible for all $j=1\dots n$? 3. What are exactly the domain and codomain of $T$? 4. Given that $T=x_1(x^*_1x_1+a^*_1a_1)^{-1}x_1^*$, why do you need continuity w.r.to other objects than just the pair $(x_1,a_1)$? $\endgroup$ Commented Nov 18, 2016 at 14:04
  • $\begingroup$ Why doesn't this follow immediately from the fact that operator addition, multiplication, adjoint and inverse are all continuous with respect to the norm topology? $\endgroup$ Commented Nov 18, 2016 at 14:17
  • $\begingroup$ Bounded - yes. The pairs are not assumed to be invertible, just the 2n sum. I think that the domain of $T$ can be taken to be all of $H$, as it is norm $\le 1$ on a dense subset. I think that if we restrict to a pair we get $T$ bounded but not continuously varying - I think! $\endgroup$ Commented Nov 18, 2016 at 14:35
  • $\begingroup$ Because the contents of $()$ are not invertible. $\endgroup$ Commented Nov 18, 2016 at 14:35

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