/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Suppose \(\mu\) is a measure, \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 17

Suppose \(\mu\) is a measure, \(0

0,\( there exists a simple function \)g \in \mathcal{L}^{p}(\mu)\( such that \)\|f-g\|_{p}<\varepsilon .$ [This exercise extends 3.44.]

Short Answer

Expert verified
Given a real-valued measurable function \(f \in \mathcal{L}^p(\mu)\) and any \(\varepsilon > 0\), we applied Egorov's theorem to find a set \(B\) such that \(\mu(B)< \varepsilon^{p/(p-1)}\) and f converges uniformly on \(B^c\). We constructed a simple function \(g\) by setting \(g(x) = 0\) for \(x \in B\) and approximating f on \(B^c\) with a finite partition. By analyzing the L^p norm of the difference between f and g, we showed that \(\|f-g\|_p < \varepsilon\), proving the existence of a simple function g in \(\mathcal{L}^p(\mu)\) such that the L^p norm of the difference is less than 蔚.

Step by step solution

01

Recall the definition of L^p space and simple functions

The L^p space is defined as the set of all measurable functions whose absolute value is integrable to the p-th power with respect to measure 渭: \[\mathcal{L}^p(\mu) = \{f: \int |f|^p d\mu < \infty\}\] A simple function is a function that takes only finitely many values. Every simple function can be written in the form: \[g = \sum_{i=1}^n a_i \chi_{A_i}(x)\] where \(a_i\) are distinct real numbers, \(A_i\) are measurable sets, and \(\chi_{A_i}\) is the characteristic function of the set A_i.
02

Apply Egorov's theorem

By Egorov's theorem, since f is measurable and has finite L^p norm, for any 蔚 > 0, there exists a measurable set B such that 渭(B) < 蔚^(p/(p-1)) and f converges uniformly on B^c.
03

Construct a simple function on B^c

Since f converges uniformly on B^c, there exists a finite partition \(\{A_1, \dots, A_n\}\) of B^c such that: \[\sup_{x \in A_i}|f(x) - a_i| < \frac{\varepsilon}{2(1 + \int_B |f|^p d\mu)^{1/p}n}\] for some constants \(a_i\). Define the simple function g on B^c as: \[g(x) = \sum_{i=1}^n a_i \chi_{A_i}(x)\]
04

Extend the simple function on the whole space

Now extend the simple function g on the whole space by setting: \[g(x) = \begin{cases} \sum_{i=1}^n a_i \chi_{A_i}(x), & \text{ if } x \in B^c \\ 0, & \text{ if } x \in B \end{cases}\] It is clear that g is a simple function and g 鈭� L^p(渭), as all its values are finite and it only differs from f on a set of measure less than 蔚^(p/(p-1)).
05

Estimate the L^p norm of the difference

Now, we will estimate the L^p norm of the difference between f and g: \[\begin{aligned} \|f-g\|_p^p &= \int |f-g|^p d\mu \\ &= \int_{B^c} |f-g|^p d\mu + \int_{B} |f-g|^p d\mu \\ &\leq \int_{B^c} \left(\frac{\varepsilon}{2(1 + \int_B |f|^p d\mu)^{1/p}n}\right)^p d\mu + \int_{B} |f|^p d\mu + \int_B |g|^p d\mu \\ &< \frac{\varepsilon^p}{2^p n^p} \mu(B^c) + \int_B |f|^p d\mu \\ &\leq \frac{\varepsilon^p}{2^p n^p}(\mu(B)+\mu(B^c)) + \int_B |f|^p d\mu \\ &< \varepsilon^p \end{aligned}\] Therefore, we have: \[\|f-g\|_p < \varepsilon\] This completes the proof.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Suppose \(1 \leq p<\infty\). Prove that there is a countable subset of \(\ell^{p}\) whose closure equals \(\ell^{p}\) (b) Prove that there does not exist a countable subset of \(\ell^{\infty}\) whose closure equals \(\ell^{\infty}\).

Suppose \(1 \leq p \leq 2\) (a) Prove that if \(w, z \in \mathbf{C},\) then $$ \frac{|w+z|^{p}+|w-z|^{p}}{2} \leq|w|^{p}+|z|^{p} \leq \frac{|w+z|^{p}+|w-z|^{p}}{2^{p-1}} $$ (b) Prove that if \(\mu\) is a measure and \(f, g \in \mathcal{L}^{p}(\mu),\) then $$ \frac{\|f+g\|_{p}^{p}+\|f-g\|_{p}^{p}}{2} \leq\|f\|_{p}^{p}+\|g\|_{p}^{p} \leq \frac{\|f+g\|_{p}^{p}+\|f-g\|_{p}^{p}}{2^{p-1}} . $$

Suppose \(p, q \in(0, \infty]\), with \(p \neq q\). Prove that neither of the sets \(\mathcal{L}^{p}(\mathbf{R})\) and \(\mathcal{L}^{q}(\mathbf{R})\) is a subset of the other.

Suppose \(1 \leq p<\infty\) and \(f \in \mathcal{L}^{p}(\mathbf{R})\). Prove that $$ \lim _{t \downarrow 0} \frac{1}{2 t} \int_{b-t}^{b+t}|f-f(b)|^{p}=0 $$ for almost every \(b \in \mathbf{R}\).

Suppose \(1 \leq p<\infty\) and \(f \in \mathcal{L}^{p}(\mathbf{R})\) (a) For \(t \in \mathbf{R}\), define \(f_{t}: \mathbf{R} \rightarrow \mathbf{R}\) by \(f_{t}(x)=f(x-t)\). Prove that the function \(t \mapsto\left\|f-f_{t}\right\|_{p}\) is bounded and uniformly continuous on \(\mathbf{R}\). (b) For \(t>0\), define \(f_{t}: \mathbf{R} \rightarrow \mathbf{R}\) by \(f_{t}(x)=f(t x)\). Prove that $$ \lim _{t \rightarrow 1}\left\|f f_{t}\right\|_{p}=0 $$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.