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Denitions:

Chapter 8:

Denition of Analytic functions.Denition of Trig Polynomial (8.9)Denition of Orthogonal and orthonormal system (8.10)Denition of Dirichlet kernel (8.13)

Chapter 9

Denition of differentiability at a point and of total derivative (9.11)Denition of partial derivatives (9.16)Denition of contraction (9.22)

Theorems

Chapter 8:

Statement and Proof of Thms AC and UC from the notesStatement and Proof of Thm 8.11 (and the obvious Corollary, Thm 8.12)Statement and Proof of Thm 8.14

Statement of Thm 8.15Statement and proof of Thm 8.16

Chapter 9.

Statement of Thm 9.15Statement and Proof of Thm 9.21,9.23, 9.24

All HW assignments.

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Theorem AC. (Analytic continuation) Let f : I R R be an analytic func-tion. If there exists x0 I such that f ( k ) (x 0) = 0 for all k 0 then f 0.

Proof. Denote byF := {x I : f ( k ) (x ) = 0 k 0}.

By our assumptions x0 F , hence F = . We want o show that F is both openand closed in I . Since I is connected and F = , this implies that F = I whichgives the desired conclusion. To show that F is closed notice that,

F =

k =0

{(f k ) 1(0)}.

Since each f k is continuous, the inverse image of the closed set {0} via f k is alsoclosed and the intersection of any family of closed sets is closed. To show that F isopen, let x F and let us show that there exists a small interval around x which isfully contained in F . Since f is analytic, f can b expanded in power series around

x that isf (x ) =

k =0

ck (x x )k , for |x x | < .

Moreover,

ck = f k (x )

k! .

Since x F we conclude that ck = 0 for all ks and hence f 0 on the interval|x x | < . In particular this interval is contained in F , as desired.

Theorem UC. (Uniform convergence of Fourier Series.) Let f be dened on [ , ] and periodic of period 2. If f C 2 then the Fourier sum sN (f ) =

N n =1 cn e

inx converges uniformly to f as N .

Proof. Integrating by parts, we obtain the following:

2c n =

f (x )e inx dx =

1in

[e inx f (x )] + 1in

f (x )e inx dx,

thus (the rst term is zero by the periodicity of f )

2c n = 1in

f (x )e inx dx.

By a second integration by parts we then obtain,

2c n = 1n 2

f (x )e inx dx.

In conclusion, since f is C 2 the second derivative is bounded on [ , ] and we

have |cn | M 1n 2

.

Hence, by Theorem 7.10 in Rudin (Weierstrass test), s N converges uniformly. More-over under our assumption, the hypothesis (79) in Theorem 8.14 (point wise con-vergence theorem) is satised, thus s N must converge to f .

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