# Thread: very important question

1. ## very important question

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

and thank you very much

2. ## Re: very important question

Hint: $\theta$ is not an isomorphism because it is not surjective.

3. ## Re: very important question

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.

4. ## Re: very important question

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

5. ## Re: very important question

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, $\theta$ is not surjective. Consider various functions on [0, 1] and see if they are images of $\theta$.

6. ## Re: very important question

Originally Posted by emakarov
As I said, $\theta$ is not surjective. Consider various functions on [0, 1] and see if they are images of $\theta$.
Thank you very much, I have found an example which is not continous

7. ## Re: very important question

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.

8. ## Re: very important question

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

9. ## Re: very important question

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.