My question is about the derivation of Ward identities. I will sketch it here in the case of an O(N) symmetric model and point out what it bothering me when I am done. I am being very sloppy with the notation. Please ask in the comments if you don't understand it. I assume that you know about the generating functional of connected correlation functions $W[j] = \ln[Z[j]]$ and its Legendre transform $\Gamma[\phi]$.
Consider an action that depends on an $N$ component field, $\phi_a$, and is invariant under rotations in this space, $$ \phi_a \rightarrow \phi_a = U(\theta_\alpha)_{ab} \phi_b' = \left[\text{e}^{i \theta_\alpha J^\alpha}\right]_{ab} \phi_b' \, .$$ $J^\alpha$ are the generators of the rotations and $\theta_\alpha$ the corresponding "angles".
The ward identities are derived by making a change of variables in the generating functional $$ Z[j^a] = \int D\phi \, \text{e}^{S\left[ \phi_a \right]+\int j^a \phi_a} = \int D \phi' \, \text{e}^{i S[U(\theta)\phi']+\int j^a U(\theta)^{ab} \phi_b'} \, .$$ After renaming $\phi'\rightarrow \phi$ and exploiting the symmetry of the action $S[U(\theta)\phi']=S[\phi']$ we write $$ Z[j^a] = \int D \phi \, \text{e}^{i S[\phi]+\int j^a U(\theta)^{ab} \phi_b} \, .$$ Finally we expand to linear order in $\theta$, $$ Z[j^a] = \int D \phi \, \text{e}^{i S[\phi]+\int j^a \phi_a} \left(1 + i \theta _\alpha \int j^a \left[J^{\alpha}\right]_{ab}\phi^b\right) \, .$$ We conclude that $$\int j^a \left[J^{\alpha}\right]_{ab} \langle \phi^b \rangle = 0 \, . \tag{1}$$ $\langle \phi^b \rangle$ is the expectation value of $\phi^b$ with the source $j$, $$\langle \phi^b \rangle = \frac{\delta Z}{\delta j^b} \, .$$
Next this is expressed in terms of the effective action, $\Gamma[\phi] = -\log[Z[j]] + \int j^a \phi_a \, ,$
$$\int \frac{\delta \Gamma}{\delta \phi_a} \left[J^{\alpha}\right]_{ab} \phi^b = 0 \, . \tag{2}$$
Taking one field derivative of the last equation and evaluation at physical solutions, $j=\delta\Gamma/\delta \phi = 0$ leads to $$\int \frac{\delta^2 \Gamma}{\delta \phi_c \delta \phi_a} \left[J^{\alpha}\right]_{ab} \phi^b = 0 \, .$$ When the fourier transform of this equation is taken we find Goldstones' theorem $$\frac{\delta^2 \Gamma}{\delta \phi_c \delta \phi_a}(p=0) \left[J^{\alpha}\right]_{ab} \phi^b = 0\, . $$ I.e. for each generator of the symmetry there is one mode with zero mass.
My question is the following: As I understand it, Eq. (1) is valid for any choice of the source $j$. However if I use it to constrain correlation functions, I get (after one derivative with respect to $j^c$ and evaluating at $j=0$, $$\int \left[J^{\alpha}\right]_{cb} \langle \phi^b \rangle = 0 \, .\tag{3}$$ I find that the correlation functions are symmetric, $\langle \phi^b \rangle = 0$. This simply tells me that there is no symmetry breaking. What went wrong here? Why can I use the ward identity in terms of $\Gamma$, Eq. (2) and not the one which is written in terms of $Z$, Eq. (1)?