Restriction mapping and banach algebra If I am looking at the Banach algebra $A(\mathbb{D})$ and let 
$$\Phi: A(\mathbb{D}) \to C(\mathbb{T})$$
be the restriction mapping. Let $\mathcal{B}= \Phi(A(\mathbb{D})) \subseteq C(\mathbb{T})$. 
Then I want to show that 
(i) $\Vert  \Phi(f) \Vert_\infty = \Vert f \Vert_\infty$ for all $f \in A(\mathbb{D})$ 
(ii) That $\mathcal{B}$ is a Banach algebra and that $\Phi: A(\mathbb{D}) \to \mathcal{B}$ is an isometric isomorphism. Conclude that $\sigma_\mathcal{A}(f) =\sigma_\mathcal{B}(\Phi(f)) $ for all $f \in A(\mathbb{D})$
(iii) Let $f_0 \in C(\mathbb{T})$ be given by $f_0(z)=z$. Show that $f \in \mathcal{B}$ and show next that $\sigma_\mathcal{B}(f_0)=\mathbb{\overline{D}}$ and $\sigma_{C(\mathbb{T})}(f_0)= \mathbb{T}$. 
However, I am not sure how to do this. In (i) it is hinted that I should use the maximum modulus principle but I am not sure how to as I have some trouble seeing what the restriction map really is. 
 A: $A(\mathbb D)$ consists of continuous functions on the closed unit disk which are analytic in the open unit disk. The map $\Phi$ simply takes any $f$ in  $A(\mathbb D)$ to its restriction to the boundary of the disk. 
(i) is immediate from Maximum Modulus Principle: The maximum of $|f|$ is attained on  the boundary. 
(ii) is a general fact about isometric isomorphisms; they always preserve spectra as you can see directly from the definition of spectrum.
(iii) The fist part is from (ii). For the spectrum of $f(z)=z$ in $C(\mathbb T)$ you have to observe that $f(z)-\lambda$ has an inverse in this space (which means there is a continuous function $g$ on the circle with $f(z)(z-\lambda)=1$ for all $z$ iff $|\lambda | \neq 1$. 
A: The restriction map is defined as follows:
If $f A(\mathbb{D})$, then $\Phi(f):= f_{| \mathbb T}$, where $f_{| \mathbb T}$ is the restriction of $f$ to $ \mathbb T.$
Now you should see that $(i)$ follows from the  maximum modulus principle.
Can you proceed ?
A: Assume $f$ is constant on the interior of the disk. Then clearly the two norms are equal to each other (by continuity, $f$ is also constant on the boundary of the disk with the same constant value). 
Next, assume $f$ is not constant on the interior of the disk. The maximum modulus theorem implies that $f$ can't attain a maximum on the interior of the disk. But a maximum must be attained somewhere, because the disk is compact. Thus, the maximum must be attained on the boundary of the disk and the two supremumnorms are evidently equal.
This shows that $\Phi$ is isometric and thus also injective. It is obvious to see that $\Phi$ is an algebra-morphism and $\Phi$ preserves the unit, which is in both cases the constant function $1$. 
For such a map, assuming that $\sigma(x)$ denotes the spectrum of the element $x$, we have
$$\sigma(x)=\sigma(\Phi(x))$$
(use that $\Phi$ is an isomorphism (of algebra's) onto its image preserving $1$, the norm structure isn't important here).
The reason you probably had to make this exercise was that it gives an example of a Banach algebra inclusion $B\subseteq A$ where the reverse inclusion between the spectra becomes strict.
