For the complex scalar field theory
$$L = -\partial_{\mu}\phi^{*}\partial_{\mu}\phi - m^{2}\phi^{*}\phi + J\phi^{*}+J^{*}\phi,$$
Why is there no factor of 1/2 in the lagrangian like in the real scalar field?
Can we say $ Y = 0$ (renormalization) because we know the two-point function $<0|T\phi(x)\phi(x')|0> = 0$ and so $<0|\phi(x)|0> = 0$ is satisfied?
Without the factor $1/2$ for a complex field all observables constructed out of the Lagrangian in the standard way vie Noether theorem, like the energy $H:= \int T_{00} dx$ or the momentum $P_i = \int T_{i0} dx$, turn out automatically to be the ones of a system of identical particles of two types, {\em proper particles} and {\em anti particles}. E.g., $$H = \int d^3k\: k^0 ( a^*_ka_k + b^*_kb_k)$$ This is the standard interpretation of the quanta associated to a complex, also known as charged, field. The presence of the factor $1/2$ would instead produce $$H = \frac{1}{2}\int d^3k\: k^0 ( a^*_ka_k + b^*_kb_k)\:.$$ (I cannot answer the second question as I do not know what you mean by $Y$.)
A real scalar field has one degree of freedom...Here we have two degrees of freedom (two real scalar field) and we treat $\phi$ and $\phi^{*}$ as independent field. When we put the lagrangian in Euler-Lagrange's equation we generate a factor of 2 and 1/2 cancel it,e.g. $m^{2}\partial_{\phi}(\phi^{2})=2m^{2}\phi$..and similar for the derivative term... But here it won't happen since $\phi $ and $\phi^{*}$ are independent field and you get only one term... Yes you can write the generating functional $Z[J,J^{*}]=Exp[i\int J^{*}[x] k^{-1}(x-y)J(y)\,dx dy$ And we can see that $<{\phi(x)}>=(1/i)\frac{\partial Z}{\partial J^{*}}$ is zero and similarly for $\phi^{*}(x)$. So i don't need to add any linear term in $\phi$ or $\phi^{*}$. I think that's the reason. You check with others.
There is no $1/2$ because such factors arise whenever there are products of the same field. For example, you probably have seen the interaction term in $\phi^4$ theory written as $\frac{\lambda}{4!}\phi^4$. This is because when taking functional derivatives to get the Feynman rules, these combinatoric factors will arise from the derivative hitting any of the four fields.
Say we want to compute the two point correlation function $\langle 0|T \phi(x)\phi(y)|0\rangle$, i.e. the propagator, in a real scalar field theory. To do this we calculate the generating functional $Z[J] = \int \mathcal{D}\phi \exp(iS)$, take two functional derivatives with respect to $J$, and then set $J=0$. Well, in free field theory it turns out that $Z = \exp(-i/2 \int\ dx\ dy J(x)D(x-y)J(y))$, with $D$ the Feynman propagator. The $1/2$ comes from the very same $1/2$ in the Lagrangian, and would not be there in your complex scalar Lagrangian. When taking derivatives with respect to $J$, these derivatives can hit the $J$'s in two different ways, and both give the same result, giving a factor of $2$: the $1/2$ cancels it. When you have $J$ and $J^*$, there is only one possible combination so there is no factor of $2$, and hence we do not put the $1/2$ to cancel it.
Of course, we can put whatever numerical factors we want in the Lagrangian and we will get the same results since it amounts to a change of variables, but there are a couple of advantages to the way it's usually done. One is that the Feynman rules are similar for any kind of theory; you don't need to remember if you should put factors of $2$ or not. Another advantage is that with the usual convention, the parameter called $m$ is the actual mass (at least in free field theory); otherwise these combinatorial factors would creep in.
The two point function is definitely not zero; it is the propagator. As outlined in Srednicki's book (and probably others), we need $\langle 0 | \phi(x) | 0\rangle =0$ to use the LSZ formula; a $Y\phi$ term would make this vacuum expectation value non zero, and so we leave it out. In an interacting theory, this term is needed, but it turns out that it plus the requirement that $\langle 0 | \phi(x) | 0\rangle =0$ is equivalent to leaving out diagrams with tadpoles and simply ignoring $Y$.
