Spectral projection at $\{0\}$

Question

Let $x$ be a positive operator on a Hilbert space $\mathcal H.$ Let $\chi_{(0,\infty)}(x)$ be a spectral projection of $x$. Is it true that $\chi_{(0,\infty)}(x)$ is the orthogonal projection onto closure or range of $x$?

Answer 1

Let $r$ be the range projection of $x$ and $p=\chi_{(0,\infty)}(x)$. Since $rx=x$, it follows that $rf(x)=f(x)$, first for polynomials, then for continuous functions, and then for Borel functions. So $rp=p$. We also have $px=x$ (from $\chi_{(0,\infty)}(t)\,t=t$ on $\sigma (x)$). So $px\xi=x\xi$ for all $\xi\in H$, which means that $pr=r$. Then $$ p=p^*=(rp)^*=pr=r. $$