Let $M$ ba a purely infinite von Neumann algebra. Suppose that $X$ (may not be bounded) is a positive self-adjoint affiliated with $M$. We know that the spectral projection of $X$ lies in $M$.
My question : Suppose that there exits $t>0$ such that $X\geq 1+t$, can we find $c>0$ such that the projection $\chi_{(c,\infty)}(X)=0$?
If $\chi_{(c,\infty)}(X)=0$ then $X$ is bounded. So any positive unbounded operator with spectrum inside $(1+t,\infty)$ will be a counterexample.
For instance let $M$ be the diagonal masa in $B(\ell^2(\mathbb N))$. That is, let $\{e_n\}$ be the canonical basis and $\{E_k\}$ the corresponding orthogonal projections. Then $M=\{E_k\}''$. Consider $$D(X)=\operatorname{span}\{e_k:\ k\}$$ and let $X:D(X)\to D(X)$ be the linear operator induced by $$ Xe_n=(1+t+n)e_n. $$ Then $X$ is positive, unbounded, $X\geq 1+t$, and $\sigma(X)=\{1+t+n:\ n\in\mathbb N\}$. In particular $\chi_{(c,\infty)}(X)\ne0$ for all $c>0$.
In the example $M$ is not purely infinite, but as MaoWao notices we can take a type III factor $N$ and work on $M\otimes N$ and consider the operator $X\otimes I_N$.