Continuous endomorphism on Hilbert space Let $u$ and $v$ two continuous endomorphism over X (an Hilbert Space ) such that  $||u(x)|| \leq ||v(x)||$.
Show that there is exist an endomorphism $w$ over $X$ such that $||w|| \leq 1$ and $u=w(v) $
Any hint will be appreciate
 A: You have to see the Douglas majorization Theorem (R. G. Douglas, On majorization and range inclusion of operators in Hilbert
space, Proc. Amer. Math. Soc., 17 (1966), 413-416.)
$\textbf{Theorem:}$ Let $A$ and $B$ be (bounded) operators on the Hilbert space
$X$. The following statements are equivalent:
1- $Range(A)\subset Range(B)$.
2-$AA^{\star}\leq \lambda^{2}BB^{\star}$ for some real $\lambda$.
3- There exist a bounded operator $C$ such that $A=BC$.
In your case, if you just take  $v^{\star}=B$ and $u^{\star}=A$, you obtain 
$u^{\star}=v^{\star}C$. Hence, $u=C^{\star}v=wv$.
A: Some hints:
First define $w$ only on the set that is the image of $v$.
That is, we define a map $w:\text{Im}(v) \to X$. If $y = v(x) \in \text{Im}(v)$, how should we define $w(y)$? 
Now extend that map continuously to the closure of $\text{Im}(v)$.
So think about the following: if $x \in \text{Cl}({\text{Im}(v)})$ then there must be some sequence $(x_n)_n$ in $\text{Im}(v)$ such that $x_n$ converges to $x$. How should we define $w(x)$?
Then define $w$ on the entirety of $X$. Note that $X = Y \oplus Y^\perp$ where $Y = \text{Cl}(\text{Im}(v))$. How should we define $w(x)$ for $x \in Y^\perp$?
Finally prove that $w$ indeed satisfies your requirements, i.e. $\| w\| \leq 1$ and $u = wv$.
