How Can I Prove Boundedness and Continuity in Isomorphism Problems?

[dirk_mec1]

Homework Statement

http://img297.imageshack.us/img297/1434/25931863lt2.png

Homework Equations

http://img297.imageshack.us/img297/4654/35953374xl9.png



I don't see how you can show that T and T^-1 are bounded. Furthermore I don't understand the notation Tf is that T*f or Tf as a whole?

 

[cellotim]

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.

 

[dirk_mec1]

Tf = f'. That is, the operator applied to f, in this case a differential operator.

Ok, that's clear.

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.

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

 

[cellotim]

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..

 

[dirk_mec1]

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

 

[cellotim]

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

 

[dirk_mec1]

http://img179.imageshack.us/img179/2341/64146669mv5.png

If I want to proof that it is a Banach space I need to show that the norm is complete in other words every cauchy sequence has a limit. Do I need to start with a sequence of sequences?

 

[cellotim]

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.

 

[dirk_mec1]

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].

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

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.

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?