# Allowed values for the "differentiability limit" in complex analysis

Tags:
1. Nov 16, 2014

### V0ODO0CH1LD

In complex analysis differentiability for a function $f$ at a point $z_0$ in the interior of the domain of $f$ is defined as the existence of the limit
$$\lim_{h\rightarrow{}0}\frac{f(z_0+h)-f(z_0)}{h}.$$
But why are the possible $z_0$'s in the closure of the domain of the original function? And what are the possible "candidates" for this limit?

I know that for a function $f:A\rightarrow{}B$, where the domain $A$ is a subset of some topological space $X$ and the image $B$ is a subset of some other topological space $Y$, the limit as $x$ approaches $x_0$ of $f(x)$ equals some $L$, i.e.
$$\lim_{x\rightarrow{}x_0}f(x)=L,$$
if and only if for all neighborhoods $V$ of $L$ there exists a neighborhood $U$ of $x_0$ such that $f(U\cap{}A-\{x_0\})\subseteq{}V\cap{}B$.

In this case $x_0$ is required to be a limit point of $A$ and $L$ in the closure of $B$.

So are the allowed $z_0$'s in the expression above limit points of something (i.e. is the interior of the domain of the original function equal to the set of limit points of something)? And are the allowed values of the "differentiability limit" in the closure of something else?

I know the definition of limits for a function only requires that both the domain and the image be subsets of some topological spaces, which means that I could define limits for complex-valued functions of a complex variable just from the topological structure of $\mathbb{C}$. However, differentiability requires more structure, right? From the reverse perspective, differentiability in complex analysis involves some operations, which are not part of the topological structure of $\mathbb{C}$. So does my question about the allowed values for the "differentiability limit" even make sense?

2. Nov 17, 2014

### lavinia

Complex derivatives are defined in the same way as real derivatives using Newton quotients. The only difference is that one uses real arithmetic while the other uses complex arithmetic.

3. Nov 17, 2014

### HallsofIvy

Staff Emeritus
This makes no sense. You can take a limit as you approach a point in the closure of the domain but in order to find the derivative, $$f(z_0)$$ itself must exist so $$z_0$$ must be in the domain of f, not just in its closure.

4. Nov 18, 2014

### Stephen Tashi

As HallsofIvy said, in the definition of a derivative, the limit is special kind of limit because the function involved (which is $\frac{ f(z0 + h) - f(z_0)}{h}$ ) contains the term $f(z_0)$ explicitly.

In contrast to taking a limit of $f(z)$ , the function used in the definition of the derivative requires that $f(z_0)$ exists. This is by the convention that the existence of a limit of a function that's defined in terms of several parts implies the parts themselves exist.

For example the existence of $lim_{h \rightarrow a}\ (g(b + h) + g(c) + k)$ implies that $a,b,g(c), k$ exist. It does not imply that $g(b+h)$ exists for all values of $h$.

5. Nov 27, 2014

### WWGD

But notice that division in the Real case, or in the $\mathbb R^n$ case is not part of the topological structure of $\mathbb R^n$ either. (though for n>1 we use the Euclidean n-norm), and this does not create any problem.