Chapter 13: Problem 3

In the space \(L^{2}([0,1])\) which of the following sequences of functions (i) is a Cauchy sequence, (ii) converges to 0 , (iii) converges everywhere to 0, (iv) converges almost everywhere to 0 , and (v) converges almost nowhere to \(0 ?\). (a) \(f_{n}(x)=\sin ^{n}(x), n=1,2, \ldots\) (b) \(f_{n}(x)= \begin{cases}0 & \text { for } x<1-\frac{1}{n}, \\ n x+1-n & \text { for } 1-\frac{1}{n} \leq x \leq 1 .\end{cases}\) (c) \(f_{n}(x)=\sin ^{n}(n x)\) (d) \(f_{n}(x)=\chi_{L_{s}}(x)\), the characteristic function of the set \(U_{n}=\left[\frac{k}{2^{m}}, \frac{k+1}{2^{m}}\right]\) where \(n=2^{m}+k, m=0,1, \ldots\) and \(k=0 . \ldots .2^{m}-1\)

Short Answer

Expert verified

(a) Converges almost everywhere to 0 but it's not a Cauchy sequence. (b) Doesn't converge to 0. (c) Converges almost nowhere to 0. (d) Converges everywhere to 0.

Step by step solution

Analysis of Function (a)

Look at the function \(f_{n}(x)=\sin ^{n}(x)\). Since |sin(x)| ≤ 1 for all x, raising it to any power n gives us a function that converges pointwise to 0 for any x ∈ [0,1] ≠ k π, k ∈ Z (since sin(k π) = 0 for all k ∈ Z). So, the function converges almost everywhere to 0.

Analysis of Function (b)

Now consider the function \(f_{n}(x)= \begin{cases}0 & \text { for } x<1-\frac{1}{n}, \ n x+1-n &\text { for } 1-\frac{1}{n} \leq x \leq 1 .\end{cases}\). This function tends to 0 for x<1 and tends to 1 for x=1 as n tends to infinity. Therefore, it converges pointwise to the function f(x) = 0 on the interval [0, 1). However, at x = 1, it converges pointwise to 1. Therefore, the function \(f_{n}(x)\) doesn't converge to 0.

Analysis of Function (c)

Inspecting the function \(f_{n}(x)=\sin ^{n}(n x)\). This function oscillates more quickly as n increases and doesn't converge pointwise at any x ∈ [0,1]. Hence, it doesn't converge to 0 anywhere, making it a sequence that converges almost nowhere to 0.

Analysis of Function (d)

Lastly, explore the function \(f_{n}(x)=\chi_{L_{s}}(x)\), the characteristic function of the set \(U_{n}=\left[\frac{k}{2^{m}}, \frac{k+1}{2^{m}}\right]\). When k and m vary as specified in the exercise, each interval of the form \([j/2^{n}, (j+1)/2^{n}]\) appears for infinitely many n. Moreover, on each such interval, the function \(f_{n}(x)\) is eventually 0 for sufficiently large n. Hence, \(f_{n}(x)\) converges to 0 everywhere.