Thread: very important question

Dears;
I need the solution of the following exercise as soon as possible



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.

Originally Posted by Prove It

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.

I have proved that it is a linear and preserve the inner product but I face a difficult in the second part

Originally Posted by raed

I have proved that it is a linear and preserve the inner product but I face a difficult in the second part

As I said, is not surjective. Consider various functions on [0, 1] and see if they are images of .

Originally Posted by emakarov

As I said, is not surjective. Consider various functions on [0, 1] and see if they are images of .

Thank you very much, I have found an example which is not continous

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.

Originally Posted by emakarov

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.

for example let g=x²+1 belongs to V₂ consider θ(f)=xf(x)=g(x) then f(x)=3x+1/x which is not continuous at x=0 so f is not belongs to v₁

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.