**Subject :**Math -3

**Unit :**Function of Complex variable

## Sequences of Holomorphic Functions and their Derivatives

**Sequences of Holomorphic Functions and their Derivatives:**

A sequence of functions gj defined on a common domain E is said to converge uniformly to a limit function g if, for each ε > 0, there is a number N > 0 such that for all j > N it holds that | gj (x)− g (x) | < for every x E . The key point is that the degree of closeness of gj (x) to g (x) is independent of x E.

Let fj : U → C, j = 1, 2, 3 . . . , be a sequence of holomorphic functions on an open set U in C. Suppose that there is a function f : U → C such that, for each compact subset E (a compact set is one that is closed and bounded of U, the restricted sequence f _{j|E} converges uniformly to f |E. Then f is holomorphic on U. [In particular, f C^{}(U).] If fj , f, U are as in the preceding paragraph, then, for any k {0, 1, 2, . . . }, we have,

uniformly on compact sets.^{1 }The proof is immediate from, which we derived from the Cauchy integral formula, for the derivative of a holomorphic function.

**Zero Set of a Holomorphic Function:**

Let f be a holomorphic function. If f is not identically zero, then it turns out that f cannot vanish at too many points. This once again bears out the dictum that holomorphic functions are a lot like polynomials. To give this concept a precise formulation, we need to recall the topological notion of connectedness.

**Discreteness of the Zeros of a Holomorphic Function:**

Let U C be a connected open set and let f : U → C be holomorphic. Let the zero set of f be Z = {z U : f(z) = 0}. If there are a z0 Z and { zj }_{j =1} Z \ {z0} such that zj → z0, then f ≡ 0.

Let us formulate the result in topological terms. We recall that a point z0 is said to be an accumulation point of a set if there is a sequence { zj } \ {z0} with lim _{j→} zj = z0. Then the theorem is equivalent to the statement: If f : U → C is a holomorphic function on a connected open set U and if Z = {z U : f(z) = 0} has an accumulation point in U, then f ≡ 0. For the proof, suppose that the point 0 is an interior accumulation point of zeros {zj} of the holomorphic function f. Thus f (0) = 0. We may write f (z) = z · f *(z). But f vanishes at { zj } and 0 is still an accumulation point of { zj }. It follows that f (0) = 0. Hence f itself has a zero of order 2 at 0. Continuing in this fashion, we see that f has a zero of infinite order at 0. So the power series expansion of f about 0 is identically 0. It then follows from an easy connectedness argument (more on this below) that f ≡ 0.

**Discrete Sets and Zero Sets:**

There is still more terminology concerning the zero set of a holomorphic function in the above equations. A set S is said to be discrete if for each s S there is an ε > 0 such that D(s,ε ) S = { s }. People also say, in a slight abuse of language, that a discrete set has points that are “isolated” or that S contains only “isolated points.” The result in equations thus asserts that if f is a non-constant holomorphic function on a connected open set, then its zero set is discrete or, less formally, the zeros of f are isolated. It is important to realize that the result in equations does not rule out the possibility that the zero set of f can have accumulation points in C\U; in particular, a non-constant holomorphic function on an open set U can indeed have zeros accumulating at a point of ∂U. Consider, for instance, the function f (z) = sin(1/[1−z]) on the unit disc. The zeros of this f include {1−1/[ j π]}, and these accumulate at the boundary point 1.

**Figure of Zeros accumulating at a boundary point:
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