$||\phi_z - \phi_w||_{A(\mathbb{D})^*}= 2$ if and only if $z$ or $w$ in $\mathbb{T}$ Let $\mathbb{D}$ be the open unit disc in the complex plane $\mathbb{C}$. Let $A(\mathbb{D})$ stand for the space of all continuous functions on $\overline{\mathbb{D}}$ which are holomorphic in $\mathbb{D}$. We give supremum norm on $A(\mathbb{D})$, so that it becomes a Banach algebra. 
Let $z,w\in \overline{\mathbb{D}}$. Consider the evaluation functionals $\phi_z , \phi_w$. I want to prove that $||\phi_z - \phi_w||_{A(\mathbb{D})^*}= 2$ if and only if at least one of $z$ and $w$ is in $\mathbb{T}$. 
By maximum modulus theorem , any $f\in A(\mathbb{D})$ with unit norm satisfies $|f(z)|<1$ whenever $z\in \mathbb{D}$. The problem I face is that I cannot claim that the norm $||\phi_z - \phi_w||_{A(\mathbb{D})^*}$ is attained. So, we can find a sequence $f_n\in A(\mathbb{D})$ such that $|f_n (z) - f_n (w)|$ converges to $2$. So the maximum modulus theorem is pretty useless here.
Thank you.
 A: Any sequence $f_n \in A(\mathbb{D}), |f_n(z)| \le 1$ is normal, so there is a subsequence that converges normally to $f$; now $f$ may not be in $A(\mathbb{D})$, though it is definitely holomorphic on the open disc and bounded by $1$
(as we can just take any bounded holomorphic function $g$ on the open unit disc, which form a huge Banach algebra with an even huger dual space so to speak, with $|g|<1$ that is not in $A(\mathbb{D})$, e.g. any infinite Blaschke product or any singular inner function, and then $g_n(z)=g(a_nz)$ for say your favorite sequence $0<a_n<1, a_n \to 1$ are in $A(\mathbb{D})$ and converge normally to $g$)
However if both $|z|, |w|<1$, we can take the limit of the subsequence pointwise at them (since we can include $z,w$ in a compact set contained in the open disc) and note first that $f$ cannot be constant since $|f(z)-f(w)|=2$ and then we get the contradiction $2=|f(z)-f(w)| \le |f(z)|+|f(w)| < 2$ so we are done
Edit - just realized that the question is an if and only if so let's do the other part: let $|z|=1$:
case 1: $|w|<1$; note that given $a_n$ as above, $0<a_n<1, a_n \to 1$, we can construct a disc automorphism $h_n$ sending $1-a_n \to 0, -1 \to z$ by first doing the usual $\frac{u-1+a_n}{1-u(1-a_n)}$ and then rotating by the angle needed to send $-1 \to z$ (multiplying by $e^{i\theta}$ for the right $\theta$) and similarly we can construct a disc automorphism $h$ which sends $w \to 0, z \to z$ and then $f_n(u)=h_n^{-1}(h(u))$ sends $w \to 1-a_n, z \to -1$, so it works as $f_n(w)-f_n(z)=2-a_n$ etc
case 2: $|w|=1, w \ne z$, we first rotate the argument to make $z=1$ and then assuming $w \ne -1$ (otherwise $f(u)=u$ works) we notice that for any $|w_1|=1, w_1 \ne \pm 1, w_1 \ne \bar{w}$ there is a unique real $t=\frac{w-w_1}{1-ww_1}, t \ne \pm 1$ s.t. $f_t(u)=\frac{u-t}{1-tu}$ sends $1 \to 1$ and $w \to w_1$; $t$ is a continuos function of $w_1$ on the half circle (upper/lower) containing $w$ and it is $0$ at $w$ so it follows $-1 < t < 1$ on the corresponding half circle, so  for $t_n \to 1, t_n < 1, w_{1,n} \to -1$ and the appropriate disc automorphism $f_n$ satisfies $f_n(1)=1, f_n(w) \to -1$ hence the result 
(note that this gives another example of $f_n$ in the disc algebra that has a subsequence that converges normally to a function that is holomorphic in the open disc, bounded by $1$ there, but not in the disc algebra as its boundary function is discontinuous!)
