Say that a linear map $\varphi : B(\mathcal H) \rightarrow B(\mathcal H)$ is a $\epsilon$-almost isometric if $$ 1 - \epsilon \leq \|\varphi(a)\| \leq 1+\epsilon, \quad \forall a\in B(\mathcal H), \text{s.t. }\|a\|=1, $$ where $B(\mathcal H)$ denotes the set of bounded operators on a Hilbert space $\mathcal{H}$.

Does there exist a constant $C$ such that for every $\epsilon>0$ and every $\epsilon$-almost isometric linear map $\phi$, there exists an isometric linear map $\psi : B(\mathcal H) \rightarrow B(\mathcal H)$ such that $\|\varphi - \psi\| < C\epsilon$?

If that is too hard or false, what about in finite dimension?

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    $\begingroup$ You probably mean that "there exists an $\epsilon \in (0,1)$ such that...", right? Otherwise, every bounded linear operator is "almost isometric". $\endgroup$ – Jochen Glueck Jun 23 '18 at 11:21
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    $\begingroup$ @WlodAA But it's the third word in the title and the sixth word in the question. It would be better English for me to have written "almost isometric, linear map". $\endgroup$ – Chris Ramsey Jun 23 '18 at 21:59
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    $\begingroup$ Can this statement hold for all Banach spaces (perhaps for invertible $\varphi$ only)? In the answer it is proven for Hilbert spaces, which does not say much since there is many isometries on a Hilbert space, but that's a start. Or perhaps to the contrary, can this be a characterization of having a lot of isometries? $\endgroup$ – erz Jun 23 '18 at 22:46
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    $\begingroup$ You've probably already looked at this, Chris, but is there some extra juice to be squeezed from Kadison's original paper on isometries between C*-algebras? Those results are usually stated for surjective isometries but some of the lemmas have info about e.g. extreme points of unit ball of B(H) $\endgroup$ – Yemon Choi Jun 24 '18 at 0:33
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    $\begingroup$ Also: if you're prepared to weaken $C\epsilon$ to "continuous function of $\epsilon$ which vanishes at zero" then maybe one could try to run an ultraproduct argument? Jarosz has exploited this in some old work which boosts isometric results to "almost implies near" results $\endgroup$ – Yemon Choi Jun 24 '18 at 0:35

The answer is yes if you assume $\varphi$ is surjective, and you're only looking for a function $f(\epsilon)$ which tends to zero as $\epsilon$ tends to zero.

Let's call a function $\varphi$ satisfying your given condition an $\epsilon$-isometric, linear map.

Theorem: For every $\epsilon > 0$ there is a $\delta > 0$ such that for every $\mathcal{H}$ and every linear, surjective, $\delta$-isometric map $\varphi : B(\mathcal{H})\to B(\mathcal{H})$, there is a linear isometry $\psi : B(\mathcal{H})\to B(\mathcal{H})$ such that $\|\varphi - \psi\| \le \epsilon$.

Proof: Suppose otherwise and fix $1/k$-isometric, linear, surjective maps $\varphi_k : B(\mathcal{H}_k)\to B(\mathcal{H}_k)$ which remain $\epsilon$-far away from any isometry. Then we may define a map $$ \Phi : \prod B(\mathcal{H}_k) / \bigoplus B(\mathcal{H}_k) \to \prod B(\mathcal{H}_k) / \bigoplus B(\mathcal{H}_k) $$ in the obvious way. Note that $\Phi$ is surjective, linear and isometric. By a theorem of Kadison, then, $\Phi$ has the form $\Phi(x) = u \rho(x)$, where $u$ is a unitary and $\rho$ is a Jordan $^*$-isomorphism. (Recall that a Jordan $^*$-isomorphism is a linear, $^*$-preserving map $\rho$ satisfying $\rho(ab + ba) = \rho(a)\rho(b) + \rho(b)\rho(a)$.) We may find unitaries $U_k\in B(\mathcal{H_k})$ such that the sequence $(U_k)$ represents $u$. Then, multiplying by $u^*$ on the left, we may assume that $\Phi = \rho$. Then $\varphi_k$ is an $\epsilon_k$-approximate Jordan $^*$-isomorphism in the sense that $$ \| \varphi_k(ab + ba) - \varphi_k(a)\varphi_k(b) - \varphi_k(b)\varphi_k(a)\| \le \epsilon_k \|a\|\|b\| $$ where $\epsilon_k\to 0$. By a result of Ilišević and Turnšek, there is an actual Jordan $^*$-isomorphism $\psi_k : B(\mathcal{H}_k)\to B(\mathcal{H}_k)$ such that $\|\psi_k - \varphi_k\|\to 0$, and since Jordan $^*$-isomorphisms are isometries, this is a contradiction.

  • $\begingroup$ This is a nice argument. Ultimately, for me, it would be great to get some bound on the convergence when one is only dealing with finite-dimensional C*-algebras. $\endgroup$ – Chris Ramsey Jun 26 '18 at 14:50
  • $\begingroup$ The result of Ilišević and Turnšek that I mentioned has an explicit bound that looks like $K\epsilon^{1/2} (1 + o(1))$. You might be able to construct a function $g$ such that every $\epsilon$-isometric, linear surjection is a $g(\epsilon)$-approximate Jordan $^*$-isomorphism, but that would probably involve following through the proof of Kadison's theorem. $\endgroup$ – Paul McKenney Jun 26 '18 at 19:25
  • $\begingroup$ Also, note that the function $f$ above is uniform over the dimension of $H$, so it's got that going for it. $\endgroup$ – Paul McKenney Jun 26 '18 at 19:27

By the polar decomposition we have $\phi = AU$ for some positive-semidefinite self-adjoint $A$ and some partial isometry $U$. Since $\phi$ has no kernel $U$ must be a (not necessarily surjective) isometry. Note that $A$ also satisfies the hypothesis $1-\epsilon \leq \|Ax\| \leq 1 + \epsilon$ for $\|x\|=1$. We claim that $\|A - I\| \leq \epsilon$. This follows from the spectral theorem, though presumably there is also a more elementary proof.

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    $\begingroup$ Note that the operator $\phi$ in the question is not an operator on $\mathcal{H}$ but on $B(\mathcal{H})$. Thus, there is no polar decomposition for $\phi$, in general. $\endgroup$ – Jochen Glueck Jun 23 '18 at 10:27
  • $\begingroup$ Ah, sorry, I overlooked that. $\endgroup$ – Sean Eberhard Jun 23 '18 at 10:43
  • $\begingroup$ Thanks for your reply. Anyway, I'm still interested in your answer for operators $\phi$ on Hilbert spaces. In case that $U$ is not surjective, how do you obtain the estimate for $\|Ax\|$? $\endgroup$ – Jochen Glueck Jun 23 '18 at 10:51
  • $\begingroup$ I should have said that estimate holds for all $x \in \operatorname{im} U$ such that $\|x\|=1$, and then it follows that $\|A|_U - I|_U\|\leq \epsilon$. $\endgroup$ – Sean Eberhard Jun 23 '18 at 11:03
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    $\begingroup$ @SeanEberhard Thanks for the response but ultimately this won't lead to an answer to my question for the reasons pointed out. $\endgroup$ – Chris Ramsey Jun 23 '18 at 12:49

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