What is the definition of $f(\infty)$ in $\mathbb{C}^\infty$? Supposing that I have a function like $f(z) = \frac{az+b}{cz+d}$, how I can evaluate the function at $z=\infty$? Because $f(\infty) = \frac{\infty}{\infty}$ which is undefined.
My questions:
- Can be $f(\infty)$ determined by evaluating f($\frac{1}{z}$) in $z=0$, so $f(\frac{1}{z}) = \frac{a+bz}{c+dz}=g(z)$ and $g(0)=\frac{a}{c} \implies f(\infty) = \frac{a}{c}$ ?
- If $f$ is continuous at infinity ($\iff f(\frac{1}{z})$ is continuous at $0$) we have that $f(\infty) = f(\lim_{z \to \infty} z) = \lim_{z \to \infty} f(z) ?$
- If $f(\frac{1}{z})$ is not defined (so 1. don't work) and $f$ is not continuous at $\infty$ (so 2 don't work) how I calculate $f(\infty)$. What's the general definition? I searched many textbooks and don't appear a definition for this. They just use $f(\infty)$.
In general, there is no way to take an arbitrary function defined on $\mathbb{C}$ and determine a unique extension to $\mathbb{C}^{\infty}$. There are infinitely many functions which are identical on $\mathbb{C}$ but different at $\infty$, because with no further restrictions the value may be arbitrary.
For your statement (2), the sentence "$f$ is continuous at $\infty$" is meaningless until $f$ has been given a value at $\infty$. However, the formula "$f(z) \to w$ as $z \to \infty$" can be stated and evaluated whether or not $f$ is defined at $\infty$. The usual definition is something like:
\begin{aligned} \forall \varepsilon > 0\, \exists \rho\,\, \forall z\,\, |z| > \rho &\implies |f(z)-w| < \varepsilon \quad &(w \in \mathbb{C})&\\ \forall R > 0\,\, \exists \rho \,\,\forall z \,\,|z| > \rho &\implies |f(z)| > R \quad &(w = \infty)& \end{aligned}
It is a good exercise to show that (1) is equivalent to the formula "$f(1/z) \to w$ as $z \to 0$" and (2) is equivalent to "$1/f(1/z) \to 0$ as $z \to 0$", where those limits are defined completely internally to $\mathbb{C}$ (so no mention of $\infty$ at all). In fact, by putting absolute value signs in the right places, these can be reduced to just limits over $\mathbb{R}$, which you must surely be familiar with.
If the limit $L = \lim_{z \to \infty}f(z)$ exists in $\mathbb{C}^{\infty}$, then the function $\hat{f}$ which agrees with $f$ on $\mathbb{C}$ and is defined to $\hat{f}(\infty) = L$, will be continuous on $\mathbb{C}^{\infty}$, and $\hat{f}$ is the unique such extension of $f$. This can be proved if you define the right topology on $\mathbb{C}^\infty$ (but for the "right" topology it is a bit tautological).
Moreover, if $f$ is given by a "nice" formula, so that it is obviously holomorphic on a neighbourhood of 0, then the function $g(z) = f(1/z)$ is automatically holomorphic on a punctured neighbourhood of 0. If $g$ is additionally bounded on a neighbourhood of 0, then it is a theorem (Riemann's theorem) that $g$ has a unique holomorphic extension to a neighbourhood of 0 (and $f(\infty)$ should be defined as $\lim_{z \to 0} g(z)$).
In fact, if a function $g$ is holomorphic on a punctured neighbourhood of $a \in \mathbb{C}$, then $a$ is called a singularity of $g$ and there are only three possibilities for the behaviour:
In case 2, we can prove that taking $g(0) = \infty$ produces a continuous function on $\mathbb{C}^\infty$ - in fact, if you define what it means to be differentiable on $\mathbb{C}^{\infty}$, then this extension of $g$ has that property (such a function is called meromorphic).
In case 3, we can actually prove that the limit never exists. There is a famous theorem (Picard's Great Theorem) which states that when $a$ is an essential singularity of $g$, then it maps every punctured neighbourhood to a dense set in $\mathbb{C}$. In fact, it says much more - but the point is that in this case the limit cannot exist (because the limit definition says that $g$ avoids all large balls if the neighbourhood is sufficiently small).
Functions that are merely continuous on $\mathbb{C}$ and bounded need not have any of these nice classifications. Being holomorphic is really an extremely strong property.