I need the solution of the following exercise as soon as possible
Attachment 23914Attachment 23914
and thank you very much
Hint: is not an isomorphism because it is not surjective.
Another hint for the OP - read the rules of this forum. You are expected to show some effort, and the helpers are not allowed to give full solutions as it is expected that students do their own work. We can, however, give you some hints and help clear misconceptions IF you show what you have tried and exactly where you are stuck.
Could you share your example? Continuity is not essential here. We could as well consider V to be the space of functions with at most countable number of points of discontinuity (in order for the integrals to make sense), and still the problem would make sense and have the same solution.
Yes, for all f ∈ V1 we have θ(f)(0) = 0, so if g(0) ≠ 0, then g is not in the image of θ.