Here is the original paper. There are a few steps (practically all of them) I don't understand:

  1. If $A=BC$, then $AA^* = BCC^*B^* = \|C\|^2BB^* - B(\|C\|^2\mathbb I - CC^*)B^* \le \|C\|^2BB^* $

Why the $\le$ sign? In general, the notation $X \le Y$ is meaningful if the operator $Y-X$ is positive, namely $Y-X \ge 0$. In this case, $B(\|C\|^2\mathbb I - CC^*)B^* \ge 0$?

  1. If we suppose ${\rm ran}A \subset {\rm ran}B$, then we can define an operator $C_1$ on $\scr H$ as follows: for $f \in \scr H \dots $ there exists $h \in \{\ker B \}^\perp$ for which $Bh = Af$. Set $C_1 f = h.$

Why should such an $h$ exists? I mean what does $\{\ker B \}^\perp \equiv \overline {{\rm ran } B^*}$ have to do with $ { \rm ran}B$?

  1. $\dots$ because $\ker B$ is closed, it follows that $h \in \{\ker B \}^\perp$ so that $Bh = Af$. Hence $C_1$ has been shown to be bounded.

The purpose was to show the graph of $C_1$ is closed. No idea how he proved it this way.

  1. Define a mapping $D : {\rm ran}B^* \to {\rm ran}A^*$ so that $D(B^*f)=A^*f$. Then $D$ is well defined since $\|D(B^*f)\|^2 \le \dots \le \lambda^2\|B^*f\|^2$.

Is an operator well-defined just because it is bounded?

Remark: this more recent article starts exactly with Douglas lemma, except it uses $A^*A$ and $B^*B$ instead of $AA^*$ and $BB^*$ of the original paper. These are clearly not the same operators, since $A$ and $B$ are not supposed to be normal in the first place, just bounded.

So? Which one is true?


The two papers differ in the order of $B$ and $C$, which is exactly equivalent to using $A^*$ and $B^*$ instead of $A$ and $B$.

  1. Yes. You have $CC^*\leq\|C\|^2\,I$, and so $BCC^*B^*\leq \|C\|^2\,BB^*$.

  2. You have $Af=BCf$. Write $Cf=x+h$ with $x\in \ker B$ and $h\in(\ker B)^\perp$. Then $Af=Bh$.

  3. The Closed Graph Theorem gives you that if the graph of $C_1$ is closed, then $C_1$ is bounded.

  4. Yes. If $B^*f=B^*g$, then $$\|D(B^*f)-D(B^*g)\|=\|D(B^*(f-g))\|\leq\lambda\|B^*(f-g)\|=\lambda\|B^*f-B^*g\|=0. $$So $D(B^*g)=D(B^*f)$.

  • $\begingroup$ If I use $A^*$ and $B^*$ instead of $A$ and $B$ the decomposition becomes $A^*=B^*C$, hence $A=A^{**}= C^*B \neq CB$, right? $\endgroup$ – ric.san Jun 11 at 7:53
  • 1
    $\begingroup$ Obviously, the $C$ is not the same between both formulations. One will be the adjoint of the other. $\endgroup$ – Martin Argerami Jun 11 at 13:14

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