Lower bound of trace norm on operator multiplication

Question

Suppose $\mathscr{H}$ is a separable Hilbert space. Let the class of bounded operators and bounded trace operators are given by $\mathfrak{B}(\mathscr{H})$ and $\mathfrak{I}_1(\mathscr{H})$ respectively. Let also the set of all positive operators with trace one is given by $\mathfrak{S}(\mathscr{H})$; $\mathfrak{S} (\mathscr{H})\subseteq \mathfrak{I}_1(\mathscr{H})$. Suppose there exists a self adjoint operator $A \in \mathfrak{B}(\mathscr{H})$ such there exists $p_\ast \in \mathfrak{S}(\mathscr{H})$ satisfies, $$ tr\left(A(p - p_\ast)\right) > 0, \forall p \in \mathfrak{S}(\mathscr{H}) \backslash p_\ast $$ where $tr(A p_\ast)$ is allowed to be zero. Is it possible to have some lower bound relation in the following form: $$ \kappa \left\|p - p_\ast\right\|_1^m \leq tr\left(A(p - p_\ast)\right), , \forall p \in \mathfrak{S}(\mathscr{H}) \backslash p_\ast $$ for some $m \geq 1$ and $\kappa >0$, where $\left\| \cdot \right\|_1$ is the norm for Trace class operator?

Answer 1

Let $H=\ell^2(\mathbb N)$ with $\{E_{kj}\}\subset B(H)$ the corresponding canonical matrix units.

Let $$S(H)=\{E_{11},E_{22},\ldots\},\ \ \ A=\sum_{k=2}^\infty \frac1k\,E_{kk},\ \ \ \ p_*=E_{11}.$$ Then $$ \text{Tr}(A(p-p_*))=\text{Tr}(A(E_{kk}-E_{11})=\frac1k>0. $$ And $$ \|p-p_*\|_1=\text{Tr}\,((E_{kk}-E_{11})^2)^{1/2}=\text{Tr}\,(E_{kk}+E_{11})^{1/2}=\sqrt2. $$

So for all $\kappa>0$ and $m\geq1$ the inequality fails for sufficiently large $k$ (i.e., $k>\frac1{\kappa\sqrt2}$).

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Edit: I think with the interpretation that $S(H)$ consists of all positive operators with trace one, the inequality $\text{Tr}(A(p-p_*))>0$ for all $p$ is impossible.

Write $p_*=\sum_j\mu_jP_j$, where $\mu_j\geq0$, $\sum \mu_j=1$, and $P_j$ is a rank-one projection for all $j$. Assume, without loss of generality, that $\mu_j>0$ for all $j$. Let $p=\mu_1Q_1+\sum_{j\geq2}\mu_jP_j$, with $Q_1\ne P_1$. Then $$ 0<\text{Tr}(A(p-p_*))=\mu_1\,\text{Tr}(A(Q_1-P_1)). $$ So $\text{Tr}(AP_1)<\text{Tr}(AQ_1)$ for all rank-one projections $Q_1$ . In particular, if $P_k\ne P_1$, then $$\tag1\text{Tr}(AP_1)<\text{Tr}(AP_k).$$ Now let $p=\mu_kP_1+\sum_{j\ne k}\mu_jP_j$. Then $$ 0<\text{Tr}(A(p-p_*))=\mu_k\,\text{Tr}(A(P_1-P_k)), $$ and we obtain $$\tag2 \text{Tr}(AP_k)<\text{Tr}(AP_1), $$ contradicting $(1)$. The only way to avoid this is that $p_*$ is a rank-one projection $P$. So now, for every positive trace-one operator $Q$ (in particular, rank-one projections) other than $P_1$ we have $$\tag3 \text{Tr}(AP_1)<\text{Tr}(AQ). $$