Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need help with an operator, I am not very good at functional analysis and need to find some properties of following operator: $X=C[(0,1)], A \in B(X); A[f(t)]=f(t^2)$

1. I need to show that an inverse operator exists
2. Find the norms $||A||, ||A^{-1}||$
3. show that the spectrum $\sigma(A) \subset \{y, ||y||=1\}$ Thanks a lot I am clueless.

share|improve this question
1  
Hint: what is the inverse of the homeomorphism $t\longmapsto t^2$ of $[0,1]$ onto itself? –  1015 Jun 23 '13 at 18:27
add comment

1 Answer

For each Hausdorff compact $X$ and continuous map $\phi:X\to X$ denote by $\mathcal{C}(\phi)$ the operator $\mathcal{C}(\phi):C(X)\to C(X):f\to f\circ\phi$. It is straightforward to check that $$ \mathcal{C}(\phi)\mathcal{C}(\psi)=\mathcal{C}(\psi\circ\phi)\qquad\qquad \mathcal{C}(1_X)=1_{C(X)} $$

Now assume that $\phi$ is a homeomorphism, id est invertible with continuous inverse. Then as an easy consequence

1) $\mathcal{C}(\phi)$ is invertible and $\mathcal{C}(\phi)^{-1}=\mathcal{C}(\phi^{-1})$

2) for all $f\in C(X)$ we have $ \Vert \mathcal{C}(\phi)(f)\Vert_\infty =\max_{y\in \phi(X)}|f(y)| =\max_{y\in X}|f(y)| =\Vert f\Vert_\infty $ i.e. $\mathcal{C}(\phi)$ is an isometry so $\Vert \mathcal{C}(\phi)\Vert=1$

3) spectrum of every isometry on a $C^*$-algebra (and in parrticlar on $C(X)$) is contained in the unit circle. See theorem $3.1$ here.

In your particular case you need to put $X=[0,1]$ and consider homeomorphism $\phi(t)=t^2$. Then $A=\mathcal{C}(\phi)$, and it is remains to apply results above.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.