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Let $A : \mathcal{H} \to \mathcal{H}$ and $B : \mathcal{H} \to \mathcal{H}$ be trace-class (hence compact) Hermitian operators on a separable Hilbert space. Assume that $A$ is strictly positive and that $B$ is positive and rank-1. I'm interested in conditions when $A - \epsilon B \ge 0$ for some strictly positive $\epsilon \in \mathbb{R}$ (as usual, $>$ and $\ge$ for operators refers to being positive (semi)definite).

If $A$ is finite dimensional, then $A - \epsilon B \ge 0$ for some $\epsilon > 0$ always. This is because the smallest eigenvalue of $A - \epsilon B$ obeys $\lambda_\min >0$ for $\epsilon=0$ and varies continuously with $\epsilon$.

If $A$ is infinite dimensional and $B=\vert \phi\rangle\langle\phi\vert$ for some eigenvector $\vert \phi\rangle$ of $A$ (with corresponding eigenvalue $\lambda>0$), then it is clear that $A - \epsilon B\ge 0$ for $\epsilon\le \lambda$.

What about when $A$ is infinite dimensional and $B$ does not have the form of $\vert \phi\rangle\langle\phi\vert$?

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    $\begingroup$ It is not always true in finite dimensions, if $A$ can be positive semidefinite. You'd need all of its eigenvalues to be strictly positive. In infinite dimensions you need $\phi$ to belong to the spectral projection $\chi_{[\epsilon,\infty)}(A)$, assuming $\|\phi\|=1$. $\endgroup$ – Nik Weaver Apr 3 '20 at 17:00
  • $\begingroup$ Sorry, I should have been more explicit. $A$ is strictly positive. $\endgroup$ – Artemy Apr 3 '20 at 17:24
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In "On Majorization, Factorization, and Range Inclusion of Operators on Hilbert Space (1966)", R. G. Douglas proved the following result (Theorem 1 in the paper):

Theorem. Let $C$ and $D$ be bounded linear operators on a real or complex Hilbert space $\mathcal{H}$; then the following are equivalent:

(i) $C\mathcal{H} \subseteq D \mathcal{H}$.

(ii) There exists a number $\lambda \in [0,\infty)$ such that $CC^* \le \lambda^2 DD^*$.

(iii) There exists a bounded linear operator $E$ on $\mathcal{H}$ such that $C = DE$.

Now, if you choose $C$ in the theorem as the positive square root $\sqrt{B}$ of $B$ and $D$ in the theorem as the positive square root $\sqrt{A}$ of $A$, you can characterize the property you are interested in by means of a range condition.

More precisely:

Corollary. Write your rank-$1$ operator $B$ as $B = \alpha \vert \phi\rangle\langle\phi\vert$ for a number $\alpha > 0$ and a vector $\vert \phi\rangle \in \mathcal{H}$ of norm $1$ (not necessarily an eigenvector of $A$). Then the following are equivalent:

(i) There exists $\varepsilon > 0$ such that $A \ge \varepsilon B$.

(ii) $\vert \phi\rangle$ is an element of the range of $\sqrt{A}$.

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