Let $A$ and $B$ be two bounded operators on a Hilbert space $H$.
$1)$ $\exists M>0,$ $\forall x\in H$ : $||A^{\ast}x|| \leq M||B^{\ast}x||,$
$2)$ $\exists C$ bounded operator on $H$ : $A=BC.$
We want to show that $1$ $\Longrightarrow$ $2$.
Assume that $1)$ is true. Now we define $D: R(B^{\ast})\longrightarrow R(A^{\ast})$ by $D(B^{\ast}f)=A^{\ast}f.$
$D$ is fine defined, and since $$||D(B^{\ast}f)||=||A^{\ast}f||\leq M||B^{\ast}f||. $$
$D$ admits a unique extension $\overline{D}:\overline{R(B^{\ast})}\longrightarrow R(A^{\ast}),$
and if we put $D=0$ on $R(B^{\ast})^{\perp},$ we will have $DB^{\ast}=A^{\ast},$ so $A=BD^{\ast}$ with $C=D^{\ast}.$
I don't understand why if we put $D= 0$ on $R(B^{\ast})^{\perp},$ we will have $DB^{\ast}=A^{\ast}?$
I tried to do as follows : if $f\in H$ so $B^{\ast}f\in R(B^{\ast})\subset H=\overline{R(B^{\ast})}\bigoplus R(B^{\ast})^{\perp}.$ So if $B^{\ast}f=B^{\ast}f_{1}+B^{\ast}f_{2}$ then, $DB^{\ast}f=DB^{\ast}f_{1}$ ......
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**if anyone has an idea to help me, thanks !!!! **
The equality $DB^*=A^*$ holds irrespective of whatever $D$ does on $(R(B^*)^\perp$. Indeed, for any $x\in H$ you have $B^*x\in R(B^*)$, $$ DB^*x=\overline DB^*x=A^*x. $$ As this holds for any $x$, $DB^*=A^*$. The choice to make $D=0$ on $R(B^*)^\perp$ is the simplest one, there is no meaningful argument to justify another choice.