Does $\operatorname{Tr}(e^x(\lambda,\infty) x) =\operatorname{Tr}(px)$ imply that $e^x(\lambda,\infty) \le p \le e^x[\lambda,\infty)$?

Let $$H$$ be a Hilbert space and $$\operatorname{Tr}$$ be the standard trace on $$B(H)$$. Let $$x$$ be a self-adjoint operator in $$B(H)$$. Let $$e = e^x(\lambda,\infty)$$ be the spectral projection. Assume that $$\operatorname{Tr}(e x) =\operatorname{Tr}(px)$$ for a projection $$p$$ with $$\operatorname{Tr}(p)=\operatorname{Tr}(e)$$. Prove that $$e^x(\lambda,\infty) \le p \le e^x[\lambda,\infty).$$

More generally, if $$p$$ is a projection $$\operatorname{Tr}(p)=s$$ and $$\operatorname{Tr}(e )\le \operatorname{Tr}(p) \le \operatorname{Tr}( e^x[\lambda,\infty))$$ and $$\operatorname{Tr}(px)= \operatorname{Tr}(ex) + \lambda (s-\operatorname{Tr}(e) )$$, then do we have $$e \le p \le e^x[\lambda,\infty).$$

If you allow $$\operatorname{Tr}(ex)$$ to be infinite, then the implication is not true. For instance take $$x=I$$, $$\lambda=1/2$$ (so $$e=I$$) and $$p$$ any infinite projection other than the identity.
When $$x$$ is positive and compact, and $$\lambda>0$$, the first implication is true. We have $$x=\sum_j x_j e_j,$$ where $$x_1\geq x_2\geq\cdots$$ and $$\{e_j\}$$ are pairwise orthogonal rank-one projections. Let $$k$$ such that $$x_k=\max\{x_j:\ x_j>\lambda\}$$. Then $$e=\sum_{j\leq k} e_j$$ and $$xe=\sum_{j\geq k} x_je_j$$. Thus $$\operatorname{Tr}(xe)=\sum_{j\leq k} x_j.$$ Using the orthonormal basis associated with the $$\{e_j\}$$ to calculate the other trace, we have $$\operatorname{Tr}(px)=\sum_j x_j\,\operatorname{Tr}(pe_j).$$ Note that $$\sum_j \operatorname{Tr}(pe_j)\leq\operatorname{Tr}(p)=\operatorname{Tr}(e)=k$$. Now \begin{align} 0&=\operatorname{Tr}(xe)-\operatorname{Tr}(px) =\sum_{j=1}^k (1-\operatorname{Tr}(pe_j))\,x_j +\sum_{j>k} x_j\operatorname{Tr}(pe_j). \end{align} Now all the factors in the sums are non-negative, so we have $$\operatorname{Tr}(pe_j)=\begin{cases} 1,&\ 1\leq j\leq k\\ 0,&\ j>k\end{cases}.$$ When $$j\leq k$$, we get $$0=1-1=\operatorname{Tr}(e_j-e_jpe_j).$$ So, being positive, we get that $$e_j-e_jpe_j=0$$; that is, $$e_j(1-p)e_j=0$$. Then $$(1-p)e_j=0$$ and $$pe_j=e_j$$. When $$j>0$$, $$0=\operatorname{Tr}(pe_j)=\operatorname{Tr}(e_jpe_j)$$, so $$e_jp=0$$. So $$p=p_1+\sum_{j=1}^k e_j$$ for some projection $$p_1$$ orthogonal to $$\{e_j\}$$; comparing traces, we get that $$p=\sum_{j=1}^k e_j=e$$.
$$x =\begin{pmatrix}3 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 1\end{pmatrix}$$ with $$e = 1_{(2, \infty)}(x)$$ and $$p = 1 - e$$.
• You are right for the second one. What about the first one with the additional condition that $Tr(e) =Tr(p)$? – user92646 Apr 4 at 23:28