Minimality of the positive part of a self-adjoint operator Given a bounded self-adjoint operator $T$ on a Hilbert space $H$, the Jordan decomposition asserts that there exists a unique pair of bounded positive operators $(T_+,T_-)$ such that
$$
  T = T_+ - T_-, \quad \text{and}\quad
  T_+T_-=0.
  $$
For that reason $T_+$ and $T_-$ are often called the positive and negative part of $T$, respectlvely.
In particular $T\leq T_+$.
Question.  Is $T_+$ the smallest positive operator bigger than $T$?  In other words, if $S$ is a bounded positive operator on $H$, such that $T\leq S$, does it follow that $T_+\leq S$.
 A: EDIT: The following is true in the case that $T_+$ is invertible.
We have that $T_+-T_-\leq S$. Note that since $T_-T_+=T_+T_-=0$, we have that $T_-p(T_+)=p(T_+)T_-=0$ for any polynomial $p(z)$ with constant term $0$. Now since $T_+$ is invertible, $0\not\in\sigma(T_+)$, so the spectrum of $T_+$ is a closed set contained in an interval $[\varepsilon, M]$, where $0<\varepsilon<M<\infty$. The functions $f_n(t)=t^{1/n}$ defined on $\sigma(T_+)$ are continuous and each one can be uniformly approximated by polynomials with constant term $0$, so we have that $f_n(T_+)T_-=T_-f_n(T_+)=0$ for all $n$. But note that since $f_n\to1$ uniformly on $\sigma(T_+)$ we have that $f_n(T_+)\to\text{id}_H$ in norm. This actually proves that $T_-=0$, and the result follows trivially.
In the case that $T_+$ is not invertible, we cannot say anything: As OP commented, consider for example $T=\pmatrix{1&0\\0&-1}$, then $T_+=\pmatrix{1&0\\0&0}$. Consider the matrix $S=\pmatrix{9&-6\\-6&4}$, then we have that
$$S=\pmatrix{3&-2\\0&0}^*\cdot\pmatrix{3&-2\\0&0}\geq0$$
and we have that
$$S-T=\pmatrix{8&-6\\-6&5}\geq0$$
since the eigenvalues of $S-T$ are $\frac{13\pm 3\sqrt{17}}{2}$. On the other hand, we have that
$$S-T_+=\pmatrix{8&-6\\-6&4}$$
and the eigenvalues of $S-T_+$ are $6\pm 2\sqrt{10}$, one of which is negative.
