I was reading the definition of m-extremal and m-geodesic and was confused on a basic result. Let $\mathbb D$ be the unit disk in $\mathbb C$ and let $O(D_1,D_2)$ be the holomorhpic function that map from $D_1$ to $D_2$ ($D_1,D_2$ are domains in $ \mathbb C^n$ and $\mathbb C^m$ respectivly).
$f\in O(\mathbb D,D)$ is m-extremal for the points $\{ \lambda_1,\dots,\lambda_m\}$ if there is no $g\in O(\mathbb D,D)$ such that $g(\lambda_j)=f(\lambda_j)$ and $g(\mathbb D)$ is a relatively compact subset of $D$. If $f$ is a weak m-extremal for any set of $k$ points, then it is called an weak m-extremal.
$f\in O(\mathbb D,D)$ is an m-geodesic if there is a function $F\in O(D,\mathbb D)$ such that $F\circ f$ is a finite Blasche factor of degree at most $m-1$.
The text I'm reading states that all m-geodesics are m-extremals. I'm not sure how to prove this. I get that $$ F\circ g(\lambda_j)=e^{i\theta}\prod_{k}\frac{\lambda_j-a_k}{1-\bar a_j \lambda_k}=F\circ f(\lambda_j).$$ Since blasche factors extend to the unit circle, then I can extend $f$ to the unit circle. I'm not sure how to how that there cannot be a function $g$ as in the definition of the m-extremal.
I was able to write a proof for my question as soon as I become aware of the following property of Blacshke products. Let $B$ be a finite Blaschke product of degree $m-1$. Let $g(z)$ be a function analytic on $\mathbb D$ and extends continuously to the boundary of $\mathbb D$. If $g$ and $B$ agree at $m$ points in the interior of $\mathbb D$ then $|g(z)|\geq |B(z)|$ at some point $|z|=1$. Indeed, suppose that $|g(z)|<|B(z)|$ for all $|z|=1$. Then $B(z)$ and $B(z)-g(z)$ have the same number of zeros in $\mathbb D$ by Rouche's Theorem. But $B-g$ will have $n$ zeros while $B$ will have $n-1$ zeros, a contradiction.
Back to my question. Let $f$ be an m-geodesic and let $F$ be such that $F\circ f= B(z)$ where $B$ is a Blaschke factor of degree $m-1$. Suppose there is a function $g:\mathbb D\rightarrow D$ such that $f(\lambda_j)=g(\lambda_j)$ for $j=1,\dots,m$ and $g(\mathbb D)$ is a relatively compact subset of $D$. We can extend $g$ to the boundary of the unit disk continuously because its image is relatively compact in $D$. Then by the property above we have a $z$ such that $|z|=1$ and $|F\circ g(z)|\geq |F\circ f(z)|=1$. But this is a contradiction because $F$ would have to map $g(\overline{\mathbb D})\subset \mathbb D$.