Almost isometric linear maps

Question

Say that a linear map $\varphi : B(\mathcal H) \rightarrow B(\mathcal H)$ is $\epsilon$-almost isometric if $$ 1 - \epsilon \leq \lVert\varphi(a)\rVert \leq 1+\epsilon, \quad \forall a\in B(\mathcal H), \text{s.t. }\lVert a\rVert=1, $$ 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 $\lVert\varphi - \psi\rVert < C\epsilon$?

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

Answer 1

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.

Answer 2

I finally found a version of this question that gives specific bounds at the cost of more conditions.

Theorem 3.4 in Erik Christensen, "Perturbations of operator algebras", Invent. Math. 43 (1977), 1-13.

Let $\Phi$ be a surjective completely positive linear map between $B(H_1)$ and $B(H_2)$ (the algebras can be more general than this). If $t \in [0,1/84]$ such that $\|\Phi\| \leq 1$ and $\|\Phi^{-1}\| \leq 1 + t$, then there exists a $*$-isomorphism $\Psi$ such that $$\| \Phi -\Psi\| \leq 8.5\sqrt t + 7t$$

Answer 3

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.