# Bounded operator inverse, norm and spectrum

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.

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

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.

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