Posts

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 3, Problem 13

Problem Prove that the Cauchy product of two absolutely convergent series converges absolutely. Answer Let \sum a_n, \sum b_n be two absolutely convergent series. There Cauchy product is defined as c_n:=\sum_{k=0}^n a_kb_{n-k}. \sum_{n=0}^N |c_n| \leq \sum_{n=0}^N \Bigg|\sum_{k=0}^n a_kb_{n-k}\Bigg| \leq \sum_{n=0}^N \sum_{k=0}^n |a_k||b_{n-k}| = \sum_{n=0}^N |a_n|\sum_{k=0}^{N-n} |b_k| \leq \sum_{n=0}^\infty |a_n| \sum_{k=0}^\infty |b_k| < \infty. Since C_N:= \sum_{n=0}^N |c_n| is bounded and increasing, C_N convergence, and therefore, \sum c_n converges absolutely.

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 23

Problem A real-valued function f defined in (a,b) is said to be convex if f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda) f(y) whenever a<x<b,a<y<b,0<\lambda<1. Prove that every convex function is continuous. Prove that every increasing convex function of a convex function is convex. (For example, if f is convex, so is e^f.) If f is convex in (a,b) and if a<s<t<u<b, show that \frac{f(t) - f(s)}{t-s} \leq \frac{f(u) - f(s)}{u-s} \leq \frac{f(u) - f(t)}{u-t}. Answer (1) Let f be a convex function in (a,b) and a<s<t<u<b. Then note that t = \frac{t-s}{u-s}u + (1-\frac{t-s}{u-s})s. Then f(t) \leq \frac{t-s}{u-s}f(u) + (1-\frac{t-s}{u-s})f(s). From the above inequality, we can deduce the following two inequalities. \frac{f(t) - f(s)}{t-s} \leq \frac{f(u) - f(s)}{u-s}, \frac{f(u) - f(s)}{u-s} \leq \frac{f(u) - f(t)}{u-t}. (2) Fix x \in (a,b) and \epsilon >0. For y \in (a,b)...

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 3, Problem 8

Problem If \sum a_n converges, and if \{b_n\} is monotonic and bounded, prove that \sum a_nb_n converges. Answer Since b_n is monotonic and bounded, b_n \rightarrow b. If b_n is increasing, then \sum a_n (b-b_n) converges by Theorem 3.42. Then \sum a_nb_n = \sum a_n b - \sum a_n (b - b_n) also converges. If b_n is decreasing, then \sum a_n (b_n-b) converges by Theorem 3.42. Then \sum a_nb_n = \sum a_n b + \sum a_n (b_n - b) also converges. 

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 5, Problem 6

Problem Suppose (a) f is continuous for x \geq 0, (b) f'(x) exists for x > 0, (c) f(0)=0, (d) f' is monotonically increasing. Put g(x) = \frac{f(x)}{x}\;\;(x>0) and prove that g is monotonically increasing. Answer For x>0, g'(x) = \frac{xf'(x) - f(x)}{x^2}. Note that there exists c \in (0,x) such that g(x) = \frac{f(x)}{x} = \frac{f(x) - f(0)}{x- 0} = f'(c)  \Rightarrow  xf'(x) > xf'(c) = f(x)  \Rightarrow  g'(x)>0.

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 5, Problem 3

Problem Suppose g is a real function on \mathbb{R}^1, with bounded derivative (say |g'| \leq M). Fix \epsilon >0, and define f(x) = x + \epsilon g(x). Prove that f is one-to-one if \epsilon is small enough. (A set of admissible values of \epsilon can be determined which depends only on M.) Answer It is enough to show that f'>0 for a small \epsilon>0. Note that f'(x) = 1+ \epsilon g'(x) \geq 1 - \epsilon M. Hence, if we choose \epsilon < 1/M, f is one-to-one.

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 4

Problem Let f and g be continuous mappings of a metric space X into a metric space Y, and let E be a dense subset of X. Prove that f(E) is dense in f(X). If g(p)=f(p) for all p \in E, prove that g(p)=f(p) for all p \in X. (In other words, a continuous mapping is determined by its values on a dense subset of its domain.) Answer For y \in f(X), there exists x \in X such that y=f(x). For \epsilon > 0, there exists \delta > 0 such that d_X(x,x') < \delta  \Rightarrow  d_Y(f(x),f(x'))<\epsilon. Since E is dense in X, there exists z \in E such that z \in N_\delta(x), which implies f(x) \in N_\epsilon(f(z)). This means f(E) is dense in f(X). For p \in X and \epsilon >0, there exists \delta > 0 such that d_X(x,x') < \delta  \Rightarrow  d_Y(f(x),f(x'))<\epsilon/2, d_Y(g(x),g(x'))<\epsilon/2. For q \in E satisfying q \in N_\delta(p), $d_Y(g(p),f(p)) \leq d_Y(g(p),g(q)) + d_Y(g(q)...

Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 1

Problem If f is a continuous mapping of a metric space X into a metric space Y, prove that f(\overline{E}) \subset \overline{f(E)} for every set E \subset X. (\overline{E} denotes the closure of E.) Show, by an example, that f(\overline{E}) can be a proper subset of \overline{f(E)}. Answer For y \in f(\overline{E}), there exists x \in \overline{E} such that y=f(x). For \epsilon > 0, there exists \delta > 0 such that d_X(x,x')<\delta  \Rightarrow  d_Y(f(x),f(x'))<\epsilon. Since x \in \overline{E}, there exists z \in E such that d_X(x,z) < \delta, which implies f(z) \in N_\epsilon(f(x)). This means y=f(x) \in \overline{f(E)}.