Question

Define the characteristic function \(\chi_E\) of a subset E in \(\mathbb{R}\) by
\[ \chi_E(x) = \begin{cases} 1, & \text{if } x \in E \\ 0, & \text{if } x \notin E \end{cases} \]
For \(1 \le p < 2\), let \(L^p[0,1] = \{f: [0,1] \to \mathbb{R} : f \text{ is Lebesgue measurable and } \int_0^1 |f(x)|^p dx < \infty\}\).
Let \(f: [0,1] \to \mathbb{R}\) be defined by
\[ f(x) = \sum_{n=1}^\infty \frac{2^n}{n^3} \chi_{[\frac{1}{2^{n+1}}, \frac{1}{2^n}]}(x). \]
Consider the following two statements:
P: \(f \in L^p[0,1]\) for every \(p \in (1, 2)\).
Q: \(f \in L^1[0,1]\).
Then

• A. P is TRUE
• B. Q is TRUE
• C. Q is FALSE
• D. P is FALSE

Hint:

For a function defined as a sum of characteristic functions on disjoint sets, \(\int |\sum c_n \chi_{E_n}|^p = \sum |c_n|^p m(E_n)\). The problem of checking if \(f \in L^p\) reduces to checking the convergence of a series. The Ratio Test is often a quick way to check convergence of such series.

Answer 1

The Correct Option is B, D

Step 1: Understanding the Concept:
We need to determine if a given function, defined as an infinite series of characteristic functions, belongs to the Lebesgue spaces \(L^p[0,1]\) for different values of p. This involves computing the \(L^p\) norm (or its p-th power) and checking for convergence.
Step 3: Detailed Explanation:
The function \(f(x)\) is a step function defined on a partition of \((0, 1/2]\). Let \(E_n = [1/2^{n+1}, 1/2^n]\). These intervals are disjoint. On each interval \(E_n\), the function has the constant value \(c_n = 2^n/n^3\). The measure of the interval \(E_n\) is \(m(E_n) = \frac{1}{2^n} - \frac{1}{2^{n+1}} = \frac{1}{2^{n+1}}\).
Analysis of Statement Q: \(f \in L^1[0,1]\) We check if the \(L^1\) norm is finite by computing its integral: \[ \int_0^1 |f(x)| dx = \sum_{n=1}^\infty \int_{E_n} |c_n| dx = \sum_{n=1}^\infty |c_n| m(E_n) \] \[ \int_0^1 |f(x)| dx = \sum_{n=1}^\infty \left(\frac{2^n}{n^3}\right) \left(\frac{1}{2^{n+1}}\right) = \sum_{n=1}^\infty \frac{1}{2n^3} = \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n^3} \] This is a p-series with \(p=3>1\), which converges. Therefore, \(f \in L^1[0,1]\). Q is TRUE. This means (B) is correct and (C) is incorrect.
Analysis of Statement P: \(f \in L^p[0,1]\) for every \(p \in (1, 2)\) We check if the p-th power of the \(L^p\) norm is finite: \[ \int_0^1 |f(x)|^p dx = \sum_{n=1}^\infty \int_{E_n} |c_n|^p dx = \sum_{n=1}^\infty |c_n|^p m(E_n) \] \[ \int_0^1 |f(x)|^p dx = \sum_{n=1}^\infty \left(\frac{2^n}{n^3}\right)^p \left(\frac{1}{2^{n+1}}\right) = \sum_{n=1}^\infty \frac{2^{np}}{n^{3p}} \frac{1}{2^{n+1}} = \frac{1}{2} \sum_{n=1}^\infty \frac{2^{n(p-1)}}{n^{3p}} \] We need to determine if this series converges for \(p \in (1,2)\). Let \(a_n = \frac{2^{n(p-1)}}{n^{3p}}\).
We can use the Ratio Test: \[ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{2^{(n+1)(p-1)}}{(n+1)^{3p}} . \frac{n^{3p}}{2^{n(p-1)}} = \lim_{n \to \infty} 2^{p-1} \left(\frac{n}{n+1}\right)^{3p} = 2^{p-1} \] For the series to converge, this limit must be less than 1. \[ 2^{p-1}<1 \implies p-1<0 \implies p<1 \] However, the statement P is for \(p \in (1, 2)\). In this range, \(p-1>0\), so \(2^{p-1}>1\). The ratio is greater than 1, which means the series diverges. Therefore, \(f \notin L^p[0,1]\) for \(p \in (1,2)\). P is FALSE. This means (D) is correct and (A) is incorrect.
Step 4: Final Answer:
The correct statements are (B) and (D).