ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

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$Math 33B/1 S00 K. Solutions to Midterm Exam G1, Version B If you ...$

$Math 3C Homework 3 Solutions$

34 <strong>ANALYSIS</strong> QUALS Proof. First, since f is continuous on a compact set, it is uniformly continuous, and the image is compact. Let M be the maximum value of f. Let ε > 0. Choose δ > 0 such that |f(x − t) − f(x)| < ε whenever |t| < δ for every x ∈ R/Z. Then for any x ∈ R/Z, we have that |f ∗ Kn(x) − f(x)| = [f(x − t) − f(x)]Kn(t)dt R/Z = [f(x − t) − f(x)]Kn(t)dt + |t|
so y ∈ B(x, ε), so y ∈ B(x (n1) i , 1 n1 B(x (2N) k , 1 2N ) ⊂ B(x(n1) i <strong>ANALYSIS</strong> QUALS 35 , 1 n1 ) ∪ B(x(n2) j ) ∪ B(x(n2) j , 1 ). Thus, we have that n2 , 1 n2 ). Hence, B is a base for the topology, so X is second countable. (c) Let {x1, x2, . . .} be the dense subset of X found in (a). By Urysohn’s Lemma, for n, m ≥ 1, there exists ψn,m ∈ C(X) such that ψn,m = 1 on B(xn, 1 m ) and ψn,m is supported on B(xn, 2 Then {ψn,m} separates points, as the balls are a base for the topology by (b). By the Stone-Weierstrass theorem, the algebra of rational polynomial combinations of ψn,m is dense, so C(X) is separable. Problem 4: Let f, g ∈ L2 (R) be two square-integrable functions on R (with the usual Lebesgue measure). Show that the convolution f ∗ g(x) := f(y)g(x − y)dy of f and g is a bounded continuous function on R. Solution. First, we show it is bounded: |f ∗ g(x)| = f(y)g(x − y)dy R 1/2 ≤ R |f(y)| R 2 dy = f2 g2 . m ). |g(x − y)| R 2 1/2 dy Hence, f ∗ g is bounded. Now we show that f ∗ g is continuous. |f ∗ g(x1) − f ∗ g(x2)| = f(y)g(x1 − y)dy − f(y)g(x2 − y)dy R R = f(y) [g(x1 − y) − g(x2 − y)] dy R = f(x1 + y) [g(y) − g(x2 − x1 − y)] dy R ≤ f 2 g − g(x2−x1) , 2 where g (x2−x1) is the translation. But g − g(x2−x1) → 0 as |x2 − x1| → 2 0. Hence f ∗ g is continuous. Problem 5: Let H be a Hilbert space, and let T : H → H be a bounded linear operator on H. (a) Show that if the operator norm T of T is strictly less than 1, then the operator 1 − T is invertible. (b) Let σ(T ) denote the set of all complex numbers z such that T − zI is not invertible. (This set is known as the spectrum of T .) Show that σ(T ) is a compact subset of C.
Page 1 and 2: ANALYSIS QUALIFYING EXAM PROBLEMS B
Page 3 and 4: ANALYSIS QUALS 3 Proof. (a) First,
Page 5 and 6: ∂∆n ANALYSIS QUALS 5 0, 1, 2, .
Page 7 and 8: ANALYSIS QUALS 7 Thus, as |h ′ (0
Page 9 and 10: ANALYSIS QUALS 9 But by the geometr
Page 11 and 12: Then Γ (j) k ANALYSIS QUALS 11 f(
Page 13 and 14: ANALYSIS QUALS 13 = 1 √ |f(w)|
Page 15 and 16: ANALYSIS QUALS 15 Then B has the s
Page 17 and 18: ∂Q ANALYSIS QUALS 17 Problem 8: L
Page 19 and 20: ANALYSIS QUALS 19 Consider ˆxn ∈
Page 21 and 22: ANALYSIS QUALS 21 b) Suppose S −
Page 23 and 24: ANALYSIS QUALS 23 Problem 12: Let f
Page 25 and 26: ANALYSIS QUALS 25 b) I present the
Page 27 and 28: ANALYSIS QUALS 27 Solution. Picard
Page 29 and 30: ANALYSIS QUALS 29 Problem 2: Is eve
Page 31 and 32: δ 0 Then f(ξ) δf(w) = δ dξ
Page 33: ANALYSIS QUALS 33 But as D is compa
Page 37 and 38: Problem 10: Let the power series f(
Page 39 and 40: ANALYSIS QUALS 39 where the inequal
Page 41 and 42: ANALYSIS QUALS 41 Problem 8: Let f
Page 43 and 44: ANALYSIS QUALS 43 Winter 2007 Probl
Page 45 and 46: Then for N sufficiently large, we h
Page 47 and 48: ANALYSIS QUALS 47 see that on ( √
Page 49: ANALYSIS QUALS 49 If z0 ∈ K, choo

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