9. Let the domain of the function $f(x)$ be $\mathbf{R}$. If there exists a constant $\omega>0$ such that $|f(x)| \leqslant \omega|x|$ for all real numbers $x$, then $f(x)$ is called a "conditionally bounded function". Now, given the following functions:
(1) $f(x)=4 x$;
(2) $f(x)=x^{2}+2$;
(3) $f(x)=\frac{2 x}{x^{2}-2 x+5}$;
(4) $f(x)$ is an odd function defined on the real number set $\mathbf{R}$, and for all $x_{1}, x_{2}$, we have
$$
f\left(x_{1}\right)-f\left(x_{2}\right) \leqslant 4\left|x_{1}-x_{2}\right| .
$$

Among these, the functions that are "conditionally bounded functions" are $\qquad$ (write down all the sequence numbers that meet the condition).
For (1), it is obvious that $|f(x)| \leqslant 4|x|;$  
For (2), for any $M>0$, take $x=M,|f(M)|=M^{2}+2>M^{2}$, which does not meet the requirement;  
For (3), since $\frac{2}{x^{2}-2 x+5}=\frac{2}{(x-1)^{2}+4} \leqslant \frac{1}{2}$, taking $\omega=\frac{1}{2}$ is sufficient;  
For (4), taking $x_{1}=x, x_{2}=0$, we have $|f(x)| \leqslant 4|x|$.  
Conversely, if $|f(x)| \leqslant 4|x|$, then $f\left(x_{1}\right)-f\left(x_{2}\right) \leqslant 4\left|x_{1}\right|-4\left|x_{2}\right| \leqslant 4\left|x_{1}-x_{2}\right|$, which satisfies the requirement.  
In summary, the sequence numbers that are "condition-constrained functions" are (1)(3)(4).