# very important question

• May 20th 2012, 07:43 AM
raed
very important question
Dears;
I need the solution of the following exercise as soon as possible

Attachment 23914Attachment 23914

and thank you very much
• May 20th 2012, 08:23 AM
emakarov
Re: very important question
Hint: $\theta$ is not an isomorphism because it is not surjective.
• May 20th 2012, 08:40 AM
Prove It
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.
• May 21st 2012, 12:59 AM
raed
Re: very important question
Quote:

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
• May 21st 2012, 01:56 AM
emakarov
Re: very important question
Quote:

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 $\theta$.
• May 21st 2012, 06:21 AM
raed
Re: very important question
Quote:

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
• May 21st 2012, 06:27 AM
emakarov
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.
• May 21st 2012, 01:54 PM
raed
Re: very important question
Quote:

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
• May 21st 2012, 02:25 PM
emakarov
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.