If an irreducible representation is unitarizable, it has at most one unitarization (up to equivalence). Going back from $S_2$ to $S_1$, you only have to consider the invariant subspace of smooth vectors, which has a naturally topology associated to it. The smooth vectors are always a dense subset, so can not be zero.

For the second question, I am not quite sure what bounded equivalence means, but it seems that you do not get surjectivity from $S_2$ to $S_3$. I assume it means isomorphism, but not necessarily unitary. There are smooth, admissible representations, which are not unitarizable. A neccessary condition for unitarizability is that the central character is unitary. This is also a sufficient criteria for the Steinberg and the supercuspidal representations. These are square integrable. For the principal series, it is not. Only the continuous series and complementeray series representation are unitarizable. So $S_2$ is a proper subset of $S_3$.

If an irreducible representation is unitarizable, it has at most one unitarization (up to equivalence). Going back from $S_2$ to $S_1$, you only have to consider the invariant subspace of smooth vectors, which has a naturally topology associated to it. The smooth vectors are always a dense subset, so cannot be zero.

For the second question, I am not quite sure what bounded equivalence means, but it seems that you do not get surjectivity from $S_2$ to $S_3$. I assume it means isomorphism, but not necessarily unitary. There are smooth, admissible representations, which are not unitarizable. A necessary condition for unitarizability is that the central character is unitary. This is also a sufficient criteria for the Steinberg and the supercuspidal representations. These are square integrable. For the principal series, it is not. Only the continuous series and complementary series representation are unitarizable. So $S_2$ is a proper subset of $S_3$.

If an irreducible representation is unitarizable, it has at most one unitarization (up to equivalence). Going back from $S_2$ to $S_1$, you only have to consider the invariant subspace of smooth vectors, which has a naturally topology associated to it.

Certainly, there are smooth, admissible representations, which are not unitarizable. A neccessary condition for unitarizability is that the central character is unitary. This is also a sufficient criteria for the Steinberg and the supercuspidal representations. These are square integrable. For the principal series, it is not. Only the continuous series and complementeray series representation are unitarizable. So $S_2$ is a proper subset of $S_3$.

If an irreducible representation is unitarizable, it has at most one unitarization (up to equivalence). Going back from $S_2$ to $S_1$, you only have to consider the invariant subspace of smooth vectors, which has a naturally topology associated to it. The smooth vectors are always a dense subset, so can not be zero.

For the second question, I am not quite sure what bounded equivalence means, but it seems that you do not get surjectivity from $S_2$ to $S_3$. I assume it means isomorphism, but not necessarily unitary. There are smooth, admissible representations, which are not unitarizable. A neccessary condition for unitarizability is that the central character is unitary. This is also a sufficient criteria for the Steinberg and the supercuspidal representations. These are square integrable. For the principal series, it is not. Only the continuous series and complementeray series representation are unitarizable. So $S_2$ is a proper subset of $S_3$.