Let's assume $M$ is a symplectic manifold with the group action $G$. If $Lie(G)$ is semi simple then why the Hamiltonian condition, which requires the existence of linear map $Lie(G)\to C^{\infty}(M,R)$ is always satisfied? 



If Lie(G) is semi simple then the moment map exists!
Let $\mathfrak{g}=Lie(G)$. The action of $G$ on $M$ gives a morphism of Lie algebras $a:\mathfrak{g}\rightarrow Vect\_{symp}(M)$.


Since $\mathfrak{g}$ has trivial abelianization, $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, i.e. any element can be decomposed into commutators. An easy computation shows that a commutator of two symplectic vector fields is a Hamiltonian vector field, so you can define a linear map $b:\mathfrak{g}\rightarrow C^\infty(M)$, which is *not* a morphism of Lie algebras in general.


Pick any two elements $x,y\in\mathfrak{g}$ and observe, that $b([x,y])-\{b(x),b(y)\}$ is a constant, since $\{b(x),b(y)\}$ is a Hamiltonian function for $[x,y]$. Call it $c(x,y)$: it defines a two-cocycle on $\mathfrak{g}$ which is furthermore trivial (by semisimplicity $H^2(\mathfrak{g})=0$). Therefore, there is an element $f\in\mathfrak{g}^\*$, such that $b([x,y])-\{b(x),b(y)\}=c(x,y)=f([x,y])$.


Finally, define the map $\mathfrak{g}\rightarrow C^\infty(M)$ by $x\mapsto b(x)-f(x)$, you can easily check that it is a morphism of Lie algebras.


