# Isomorphism homework help

1. Sep 25, 2008

### dirk_mec1

Last edited by a moderator: May 3, 2017
2. Sep 25, 2008

### cellotim

Re: isomorphism

Tf = f'. That is, the operator applied to f, in this case a differential operator. T is bounded because ||Tf|| = ||f||. $$T^{-1}$$ is defined to be $$T^{-1}f' = f$$ and so is also bounded by the same reason. This comes from the definition of boundedness of linear operators.

3. Sep 25, 2008

### dirk_mec1

Re: isomorphism

Ok, that's clear.

I don't see how you prove that it is bounded. Why is||Tf|| = ||f||. $$T^{-1}$$ ?

4. Sep 25, 2008

### cellotim

Re: isomorphism

The definition of boundedness of linear operators on normed spaces:

$$||Tf||_\infty \leq M||f||_E$$ for some M. If we let M=1, then we have the proof for T and its inverse..

5. Sep 25, 2008

### dirk_mec1

Re: isomorphism

OK clear. But how can I compute its inverse? I know it has to go from $$||f'||_{\infty}$$ to f,right?

6. Sep 25, 2008

### cellotim

Re: isomorphism

You don't need to compute anything. You know that $$T^{-1}f' = f$$, that's all you need.

7. Sep 25, 2008

### dirk_mec1

Last edited by a moderator: May 3, 2017
8. Sep 25, 2008

### cellotim

Re: isomorphism

You need to show that given a sequence $$f_n\in E$$, where $$||f_m - f_n||_E < 1/k$$ for any m, n>N, some N, and any k, that its limit is in E, i.e. that $$lim_{n\rightarrow\infty} f_n = f$$ such that f(0) = 0 and f is continuously differentiable on the interval [0,1]. The first part is easy. To prove C1, the key is in the norm. The norm is the sup of the derivative meaning that sup of the derivative of the difference of two members of the sequence becomes smaller. This keeps the derivative of f from exploding.

9. Sep 25, 2008

### dirk_mec1

Re: isomorphism

What I want to show is that $$||f - f_n ||_E < \epsilon$$ for n>N is that the same?

I don't understand: the difference of two members is just in the infinity-norm. How can you see that it's decreasing? (the only useful thing you can do is the triangle inequality, right?

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