Lower bound of trace norm on operator multiplication 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?
 A: 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., $k>\frac1{\kappa\sqrt2}$).

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 
$$\tag3
\text{Tr}(AP_1)<\text{Tr}(AQ).
$$
