# Douglas' Lemma for Bounded operators on Hilbert Space

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 $$D(B^*g)=D(B^*f)$$.
• 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? – ric.san Jun 11 at 7:53
• Obviously, the $C$ is not the same between both formulations. One will be the adjoint of the other. – Martin Argerami Jun 11 at 13:14