By Michel Herve

**Read Online or Download Analytic and plurisubharmonic functions in finite and infinite dimensional spaces. Course given at the University of Maryland, Spring 1970 PDF**

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**Additional resources for Analytic and plurisubharmonic functions in finite and infinite dimensional spaces. Course given at the University of Maryland, Spring 1970**

**Sample text**

Definition: Rational functions are functions of the form f (x) = • x3 + 3x2 + 1 1 and are both rational functions. x2 + 1 x Composition Once we have defined a few elementary functions, we can create new functions by combining them together using +, −, ·, ÷, or by introducing the composition operator ◦. Definition: If f : A → B and g : B → C then we define g ◦ f : A → C to be the function that takes x ∈ A to g(f (x)) ∈ C. • 2 f (x) = x√ +1 g(x) = 2 x√ g(f (x)) = 2 x2 + 1 f : R → [1, ∞), g : [1, ∞) → [2, ∞), g ◦ f : R → [2, ∞).

7 (Sequence Limit Ratio Test)): Let {an }∞ n=1 be a sequence such that an+1 lim = r, where r ∈ [0, 1). Let s = (1 + r)/2, so that 0 ≤ r < s < 1. n→∞ an (a) Show that there exists a number N such that n ≥ N ⇒ Consider ǫ = s − r > 0. an+1 an ≤ s. Hint: (b) Use part (a) and induction to show for n ≥ N that 0 ≤ |an | sN ≤ |aN | sn . (c) Prove that lim an = 0. n→∞ (d) Apply this result to prove for any x ∈ R that xn = 0, n→∞ n! lim (e) If r > 1 show that {an } is divergent. Hint: consider the sequence {1/an }.

Given ǫ > 0, ∃N ∋ n > N ⇒ |an − L| < ǫ. 4) Let {ank } be a subsequence of {an }. Then k > N ⇒ nk ≥ k > N. Considering only the indices n = nk in Eq. 4) ⇒ |ank − L| < ǫ. ∴ lim ank = L. k→∞ “⇐” Suppose each subsequence of {an } is convergent with limit L. Note that {an } is a subsequence of itself. Hence {an } is convergent with limit L. • {(−1)n } is not convergent since and −1 = 1. {(−1)2n } = {1, 1, 1, . } → 1 {(−1)2n+1 } = {−1, −1, −1, . } → −1 In the examples to follow, we will make use of a lemma.