This is a problem that I have been stuck for a few months.


Let $X$ be a Hilbert space and $A:B:X\to X$ be two non-commuting semi-positive self-adjoint bounded linear operators. Is it true that
$$\|(I+A+B)^{-1}A\|\le 1.$$
 If it is, can you suggest any reference to me?




---


I was able to show $\|(I+A+B)^{-1}A\|\le \|A\|$ and $\|(I+A)^{-1}A\|\le 1$. However, I don't know how to show the desired result without assuming $\|A\|\le 1$ or $AB=BA$. 


If it helps, I can show $A=C^\*D\_1^\*D\_1C$ and $B=C^\*D\_2^\*D\_2C$ for some linear bounded $C,D\_1,D\_2$.



Norm estimation of identity plus two non-commuting self-adjoint operators
The claim is **false.** Consider the following matrix argument.


\begin{eqnarray\*}
 \|(I+A+B)^{-1}A\| \le 1\quad\Leftrightarrow\quad \begin{bmatrix}I & (I+A+B)^{-1}A \\ A(I+A+B)^{-1} & I \end{bmatrix} \ge 0. 
\end{eqnarray\*}
The latter inequality amounts to showing $I \ge (I+A+B)^{-1}A^2(I+A+B)^{-1}$, or in other words, we must show that $(I+A+B)^2 \ge A^2$. Since the map $x \mapsto x^2$ is not operator monotone, suggets that it should be possible to make this inequality fail.


And indeed, here is an explicit counterexample:
\begin{equation\*}
A = \begin{bmatrix}1 &5\\ 5 &61 \end{bmatrix},\quad 
B = \begin{bmatrix}9 &-9\\ -9 & 13\end{bmatrix},\quad 
C = (I+A+B)^2-A^2 = \begin{bmatrix}111 &-654\\ -654& 1895 \end{bmatrix}.
\end{equation\*}
The matrix $C$ has a negative eigenvalue: $1003 - 2\sqrt{305845} < 0$.


