If $-A \le B\le A$, can be say that the support $s(B)$ of $B$ is smaller than $s(A)$?

Question

For a positive bounded linear operaotr $A$ and a self-adjoint bounded linear operator $B$ on a Hilbert space $H$ with $-A \le B\le A$, can be say that the support $s(B)$ of $B$ is smaller than $s(A)$?

Answer 1

Case 1. $B\geq 0$.

Since $ 0\leq \langle Bh,h\rangle\leq \langle Ah,h\rangle$, $ \langle Bh,h\rangle=0$ on $\ker A$,i.e., $B^{\frac{1}{2}}=0$ on $\ker A$. Hence, $\ker A\subset \ker B^{\frac{1}{2}}\subset \ker B$.

Case 2.

Since $0\leq A+B\leq 2A$, we have $\ker 2A\subset \ker (A+B)$ by Case 1, which is equivalent to $\ker A\subset \ker B$.

Answer 2

I found a weaker version. There must be some easier proof.

Let $A,B$ be finite-rank operators. If $-A\le B\le A$, then $ -s(B)As(B)\le B \le s(B)As(B) $. I can show that $ s(s(B)A s(B)) =s(B)$. Since $ s(s(B)A s(B)) \le s(B)$ is of course true, it suffice to prove $ s(s(B)A s(B)) \ge s(B)$.

For any $\varepsilon >0$, $s(B)As(B)+\varepsilon$ is vertible. Then, $$- s(B)As(B)-\varepsilon \le -s(B)As(B)\le B \le s(B)As(B) \le s(B)As(B)+\varepsilon . $$ Hence, $$-1 \le (s(B)As(B)+\varepsilon)^{-1/2} B (s(B)As(B)+\varepsilon)^{-1/2} \le 1 . $$ By theorem 1.18 in Hiai, we obtain that for every $t>0$, $$\int_0^t \log \mu(t;B)dt \le \int_0^t \log ((s(B)As(B)+\varepsilon)^{1/2} (s(B)As(B)+\varepsilon)^{-1/2} B (s(B)As(B)+\varepsilon)^{-1/2} (s(B)As(B)+\varepsilon)^{1/2} \le \int_0^t \log \mu(t;s(B)As(B)+\varepsilon).$$ Since $\varepsilon$ is arbitrary and the measure of $\{\mu(s(B)As(B))\ne 0\}$ is the standard trace of $s(s(B)As(B))$, it follows that $ s(s(B)A s(B)) \ge s(B)$.