Recent content by nyisles131

I finally get it...:smile: Thanks for your help!

Does anyone know if what I have stated 2 posts ago is the correct answer?

Is there anything else I need to do for this problem other than what I stated in the previous post?

Ok, so let me start over again... (FORGET ALL THE VARIABLES I USED BEFORE) I have: T: X --> U T': U' --> X' T": X" --> U" Then, take f in X" Therefore, T"(f) is in U". Then, T"(f)(L) = f(T'(L)) for L in U' Then T"(f)(L) = f(LT) because T'(L) = LT Thus, T"(f)(L) = LT(x) by duality relation...

I'm very confused right now. By definition, isn't (T"(f))(L) = f(T'(l)?...Can you give me any suggestions as to how to approach this problem?

Let me check with you if this is correct... Let T(x) = u Then, we have to show that T"(f) = u where f is in X"? We know the following: For f in X", L in X', T"(f)(L) = fT'(L) This implies: T"(f)(L) = T' (Lx) (by duality relation) Then, T"(f)(L) = LT(x) Then, T"(f)(L) = Lu But, this does not...

The question that I am stuck on is: Show that if X" (double dual of X) is identified with X and U" (double dual of U) with U via the duality relation, then T" (double transpose) = T. (Duality relation is f(L) = L (x) where f is in X", L is in X', and x is in X) So far, here is my work...