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Let $A,B,C$ be self-adjoint operators of $L^2(\mathbb{R}^n)$ ($A$ and $B$ unbounded), $A\geq 0$, $B \geq 0$, with $\sqrt{A} C$ and $\sqrt{B} C$ bounded. Is the following inequality true for some constant $c \geq 0$, where $\left| \! \left| \cdot \right| \! \right|$ is the operator norm, \begin{align*} \left| \! \left| \sqrt{A+B} C \right| \! \right| \leq c \left| \! \left| \sqrt{A} C \right| \! \right| + c \left| \! \left| \sqrt{B} C \right| \! \right| ? \end{align*}

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  • $\begingroup$ Should $A$ be $B$ in the last term? $\endgroup$
    – Nate Eldredge
    Commented Jul 8, 2020 at 13:34
  • $\begingroup$ Yes thanks (and sorry) $\endgroup$
    – user140442
    Commented Jul 8, 2020 at 13:36
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    $\begingroup$ Also, do you need some more assumptions to ensure that $A+B$ is self adjoint? $\endgroup$
    – Nate Eldredge
    Commented Jul 8, 2020 at 13:37
  • $\begingroup$ Yes you can assume that $\endgroup$
    – user140442
    Commented Jul 8, 2020 at 13:38

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$$ \|\sqrt{A+B}Cx\|^2=(\sqrt{A+B}Cx,\sqrt{A+B}Cx)=\\ ((A+B)Cx,Cx)=(ACx,Cx)+(BCx,Cx)=\|\sqrt{A}Cx\|^2+\|\sqrt{B}Cx\|^2, $$ taking the supremum over unit vectors $x$ we get $$ \|\sqrt{A+B}C\|^2\leqslant \|\sqrt{A}C\|^2+\|\sqrt{B}C\|^2\leqslant (\|\sqrt{A}C\|+\|\sqrt{B}C\|)^2. $$

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For any unit vector $x\in L^2(\mathbb{R}^n)$, $$\|\sqrt{A+B}\,Cx\|^2=(\sqrt{A+B}\,Cx,\sqrt{A+B}\,Cx) =(ACx,Cx)+(BCx,Cx)=\|\sqrt A\,Cx\|^2+\|\sqrt B\,Cx\|^2 \le(\|\sqrt A\,Cx\|^2+\|\sqrt B\,Cx\|)^2 \le(\|\sqrt A\,C\|+\|\sqrt B\,C\|)^2.$$ So, $$\|\sqrt{A+B}\,C\|\le\|\sqrt A\,C\|+\|\sqrt B\,C\|.$$

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