# 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 $$T_+\leq S$$.

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 $$0<\varepsilon. The functions $$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.

• I think your statement that $T_+^{1/n}\to\text{Id}_H$, as $n\to\infty$, does not hold. For example, if $T=\pmatrix {1 & 0 \cr 0 & -1}$, then $$T_+=\pmatrix {1 & 0 \cr 0 & 0} = T_+^{1/n}.$$ – Black Oct 17 at 18:19
• @Black you are right, it was a stupid mistake of mine. I have edited my post. – JustDroppedIn Oct 17 at 22:50
• Two points: when $T_+$ is invertible then $T_-=0$, so $T$ is already positive and my conjecture becomes tautological. Secondly, your $S$ is not positive so it doesn't give a counter example. – Black Oct 18 at 0:28
• your first point is exactly what I am saying. About your second point, I didn't notice you wanted $S$ to be positive. I am editing again. – JustDroppedIn Oct 18 at 1:31
• @Black I think we are done here – JustDroppedIn Oct 18 at 1:37