What you need is not that "T uniquely determines T*" but that "T* uniquely determines T". Otherwise it is possible that two different Ts determine the same T* and the mapping would not be one to one.
Let be a Hilbert Space and be an operator in .
Prove that the mapping is an one to one and onto mapping of into itself,where is the adjoint of and is the set of bounded operator on .
Since we know that uniquely determines the adjoint and by the fact that is an operator in ,hence,we can claim that the mapping from is one to one and onto.
Is this the right way to prove?Can anyone help?
Use the definition of "adjoint"! If T* is the adjoint of T, the <Tu, v>= <u, T*v> for all u and v in H. < , > is the inner product in H. Suppose that, for a given T, there were two adjoints, and . Then we must have [tex]<Tu, v>= <u, T_1*v>[tex] and so that for all u and v in H. Can you show that from that?