I'm interested in the basic amplification procedure for QMA: the prover sends $O(r)$ copies of his witness to the verifier, which decreases the error probability to $2^{-O(r)}$ (Chernoff bound). The only difficulty is to show when $x \notin \mathcal{L}$ that the prover will not gain anything if he entangles the copies. There are complicated proofs of this result (see [KW00],[KSV02]), but I have also seen one-line proofs that use convexity. See these lectures notes from Aaronson from instance:
Now what if Merlin sends witness states that are entangled with each other? By convexity, some pure state $|\psi\rangle$ maximizes the probability that Arthur accepts. Even after Arthur has performed an experiment on a register, Merlin can hope for nothing better than for the next register Arthur acts on to contain $|\psi\rangle$. But then, why not just put the unentangled state $|\psi\rangle$ there? So, entangling registers doesn't win Merlin anything.
I don't understand the convexity argument here. Assume that the acceptance probability is given by $||\Pi |\psi\rangle ||^2$ for some projector $\Pi$. I know that $||\cdot||^2$ is convex but I don't see how to use it here. For instance, if $|\psi\rangle = \sum_i \alpha_i |\psi^i_1\rangle \otimes \cdots \otimes |\psi^i_r\rangle$ for orthogonal unit vectors $|\psi^i_1\rangle \otimes \cdots \otimes |\psi^i_r\rangle$ (thus $\sum_i |\alpha_i|^2 = 1$) we would like to have $||\Pi |\psi\rangle ||^2 \leq \sum_{i,j} |\alpha_i|^2 \cdot ||\Pi |\psi^i_j\rangle ||^2$ but this is not what convexity means...
