Norm estimation of identity plus two non-commuting self-adjoint operators

Question

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?

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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 $C,D_1,D_2$.

Answer 1

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: $1003 - 2\sqrt{305845} < 0$.

Answer 2

(Unless I messed up the arithmetic, here's a counterexample.)

Let $B=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ and $A=\begin{bmatrix} 1 & 0 \\ 0 & R \end{bmatrix}$ where $R$ is a positive number that we'll choose at the end. Then $(1+A+B)^{-1}=\begin{bmatrix} 3 & 1 \\ 1 & R+2 \end{bmatrix}^{-1}=1/(3R+5)\begin{bmatrix} R+2 & -1 \\ -1 & 3 \end{bmatrix}.$

Then $(1+A+B)^{-1}A=1/(3R+5)\begin{bmatrix} R+2 & -R \\ -1 & 3R \end{bmatrix}.$

Then $||(1+A+B)^{-1}A ||\geq 1/(3R+5)|| \begin{bmatrix} 0 & -R \\ 0 & 3R \end{bmatrix} ||=\frac{\sqrt{10}R}{3R+5}$

Then just take $R$ big enough.

Answer 3

Though two answers are already posted, let me explain how to understand that it is false from general reasoning. The inequality $\|(I+A+B)^{-1}A\|\leqslant 1$ is equivalent to $\|(I+A+B)^{-1}A x\|\leqslant \|x\|$ for any vector $x$. Take $x=A^{-1}(I+A+B)y$, the inequality rewrites as $\|y\|\leqslant \|y+A^{-1}(I+B)y\|$. We are given only that $A,B$ are positive definite, thus for given vector $z$, $A^{-1}z$ may be any vector which forms an acute angle with $z$. For $z=(I+B)y$ this in general allows the vector to form an obtuse angle with $y$ (unless $y$ is an eigenvector of $B$.) So, if the direction of $A^{-1}(I+B)y$ forms an obtuse angle with $y$ and $A^{-1}$ is small enough, the inequality $\|y\|\leqslant \|y+A^{-1}(I+B)y\|$ fails.