I am going to present the most basic theorems in complex analysis for the case where the functions have values in a Banach space. See Chapter 13 above for a short discussion of the main properties of these spaces. There are good reasons for allowing functions to have values in a complex Banach space. In particular, when X is a Banach space, so is ℒ
I assume the reader knows about vector spaces. In this book, all vector spaces are with respect to the field of complex numbers. A norm
 (15.1) 
 (15.2) 
 (15.3) 
The last inequality above is called the triangle inequality. Another version of this is
 (15.4) 
Note this shows that x →
Here is a very interesting result which is usually obtained in calculus from a use of the mean value theorem which of course does not apply to the case of a function having values in a Banach space.
Lemma 15.1.1 Let Y be a normed vector space and suppose h :

Then

Proof: Let ε > 0 be given and let

Then a ∈ S. Let t = supS. Then by continuity of h it follows
 (15.5) 
Suppose t < b. If strict inequality holds, then by continuity of h strict inequality will persist for s near t and t≠supS after all. Thus it must be the case that

Since t < b, there exist positive numbers, h_{k} decreasing to 0 such that

Then

Now divide by h_{k} and take a limit. This shows

which does not happen. Hence t = b and so

since ε > 0 is arbitrary, it follows that h
Also there is a fundamental theorem about intersections. As before, a set F is closed if its complement is open or equivalently if it contains all of its limit points. The proof is exactly as done earlier.
Theorem 15.1.2 Let F_{n} ⊇ F_{n+1}
Proof: Obviously there can be no more than one point in the intersection because if x,y are two points, then