/*! 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 9 Proposition. Given a Cauchy sequ... [FREE SOLUTION] | 91Ó°ÊÓ

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Proposition. Given a Cauchy sequence in \(\mathscr{L}^{p}(X)\), for \(1 \leq p<\infty\), there is for each \(\varepsilon>0\) a subsequence \(\left(f_{n}\right)\), an open set \(A\) with \(\int[A]<\varepsilon\) and a null set \(N\), such that the sequence \(\left(f_{n}\right)\) converges uniformly on \(X \backslash A\) and converges pointwise on \(X \backslash N\). PROOF. Choose the subsequence such that \(\left\|f_{n+1}-f_{n}\right\|_{p} \leq\left(\varepsilon 2^{-n(p+1)}\right)^{1 / p}\) for every \(n\). With $$ B_{n}=\left\\{\left|f_{n+1}-f_{n}\right| \geq 2^{-n}\right\\} $$ we therefore have $$ \int\left[B_{n}\right] \leq 2^{n p} \int\left|f_{n+1}-f_{n}\right|^{p}<2^{n p} \varepsilon 2^{-n(p+1)}=\varepsilon 2^{-n} $$ Taking \(A_{n}=\bigcup_{m>n} B_{m}\) we obtain the estimate \(\int\left[A_{n}\right]<\varepsilon 2^{-n}\). In particular, \(\int\left[A_{0}\right]<\varepsilon\) and \(\int[N]=0\), where \(N=\bigcap A_{n} .\) By \(6.2 .15\) we can choose an open set \(A \supset A_{0}\) such that \(\int[A]<\varepsilon\). If \(x \notin A_{n}\), we have \(\left|f_{m+1}(x)-f_{m}(x)\right|<2^{-m}\) as soon as \(m>n\). In particular, \(\left(f_{n}\right)\) is uniformly convergent on \(X \backslash A_{0}\), hence on \(X \backslash A\). Moreover, since \(X \backslash N=\bigcup\left(X \backslash A_{n}\right)\), it follows that \(\left(f_{n}\right)\) converges pointwise on \(X \backslash N\).

Short Answer

Expert verified
A Cauchy sequence in \( \mathscr{L}^p(X) \) has a subsequence that converges uniformly outside a small open set and pointwise outside a null set.

Step by step solution

01

Understand the Problem

We are given a Cauchy sequence in the space \( \mathscr{L}^{p}(X) \) for \( 1 \leq p < \infty \). Our goal is to find a subsequence of this sequence along with an open set and a null set to show uniform and pointwise convergence as stated in the proposition.
02

Choose a Suitable Subsequence

Select a subsequence \( (f_{n_k}) \) based on the inequality \( \|f_{n+1}-f_n\|_p \leq (\varepsilon 2^{-n(p+1)})^{1/p} \), ensuring it diminishes quickly enough to ensure convergence properties.
03

Define Transition Sets

Define sets \( B_n = \{|f_{n+1} - f_n| \geq 2^{-n}\} \). These sets help control where large differences occur. Use the Cauchy property to bound the measure of these sets: \( \int[B_n] < \varepsilon 2^{-n} \).
04

Construct the Open Set Approximations

Define \( A_n = \bigcup_{m>n} B_m \) and hence \( \int[A_n] < \varepsilon 2^{-n} \). This ensures that beyond each \( A_n \), the sequence elements are close. Choose an open set \( A \supset A_0 \) with \( \int[A] < \varepsilon \).
05

Establish Uniform Convergence

For \( x otin A_n \), since \( |f_{m+1}(x) - f_m(x)| < 2^{-m} \) for \( m > n \), the sequence is uniformly convergent on \( X \backslash A \). This follows from the inequality bounding \(|f_{m+1} - f_m|\) on these spaces.
06

Confirm Pointwise Convergence

Set \( N = \bigcap A_n \) where \( \int[N] = 0 \). The sequence converges pointwise on \( X \backslash N \) since \( X \backslash N = \bigcup (X \backslash A_n) \), demonstrating pointwise convergence in those regions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

L^p space
The concept of an \(L^p\) space revolves around a collection of measurable functions for which the \(p\)-th power of the absolute value is integrable. In simpler terms, these spaces serve essentially as a framework within which functions have a bound on their "size" measured through integration.

Key points about \(L^p\) spaces include:
  • Considered over a measure space \((X, \mathscr{F}, \mu)\), where \(X\) is the set, \(\mathscr{F}\) is the \(\sigma\)-algebra, and \(\mu\) is the measure.
  • A function \(f\) belongs to the \(L^p(X)\) space if \(\int_X |f|^p \, d\mu < \infty\).
  • The values of \(p\) range from 1 to infinity. For \(p = 2\), \(L^2\) specifically represents the space of square-integrable functions.
For students beginning with \(L^p\) spaces, it's crucial to grasp that these spaces provide a generalized setting for examining convergence and function behavior in mathematical analysis.
subsequence
A subsequence is derived by selecting elements from an original sequence while preserving order. It plays a crucial role in understanding convergence, particularly in proving properties of sequences like the one in \(L^p\) space.

Some important aspects of subsequences include:
  • They are not necessarily contiguous in the parent sequence, but ordered identically.
  • Subsequences can assist in demonstrating convergence, even when the parent sequence's behavior is complex.
  • For Cauchy sequences in \(L^p\), selecting a suitable subsequence can help show uniform and pointwise convergence.
In this context, subsequences help identify stable sequences, thereby making it easier to work with and analyze functions' behavior.
uniform convergence
Uniform convergence is a strong form of convergence for sequences of functions. It signifies that the functions not only converge pointwise, but do so at a uniform rate across the entire domain (or a specified subset).

A few characteristics of uniform convergence to understand include:
  • For a sequence \((f_n)\) to converge uniformly to a function \(f\) on a set, the maximum difference \(|f_n(x) - f(x)|\) needs to become arbitrarily small, uniformly independent of \(x\).
  • This type of convergence allows one to interchange limits with integration and differentiation operations, which is not generally the case for pointwise convergence.
  • In the given problem, uniform convergence outside a particular set \(A\) shows that within the remaining domain, the sequence stabilizes homogeneously.
Uniform convergence is integral in ensuring that function sequences effectively approximate desired behaviors across entire intervals.
pointwise convergence
Pointwise convergence refers to the basic form of convergence for a sequence of functions. This form implies that at each point in the domain, the functions in the sequence individually approach a target function as the sequence progresses.

The essentials of pointwise convergence:
  • A sequence \((f_n)\) pointwise converges to a function \(f\) if for every point \(x\), \(\lim_{n \to \infty} f_n(x) = f(x)\).
  • Unlike uniform convergence, the rate of convergence can vary across different points in the domain.
  • In the proposition example, pointwise convergence occurs outside a null set \(N\), indicating convergence is achieved at almost all points.
Pointwise convergence is a valuable concept, particularly in real analysis, for understanding how functions grow closer to one another across specific regions.

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Most popular questions from this chapter

For a topological space \(X\) the system \(\mathscr{B}\) of Borel sets is defined as the smallest \(\sigma\)-algebra of subsets of \(X\) that contains all open sets (equivalently, contains all closed sets). The theory of Borel sets is due to Baire and Lebesgue. Taking \(X=\mathbb{R}^{n}\) we know that every open set is the countable union of open boxes, and each such is the intersection of \(2 n\) open half-spaces. Thus, the system of Borel subsets of \(\mathbb{R}^{n}\) is the \(\sigma\)-ring generated by all half-spaces of the form \(\left\\{x \in \mathbb{R}^{n} \mid x_{k}>t\right\\}\), where \(1 \leq k \leq n\) and \(t \in \mathbb{R}\).

Lemma. For a real function \(f\) on \(X \times Y\) and \(y\) in \(Y\), consider the function \(f(\cdot, y)\) on \(X .\) If \(f \in C_{c}(X \times Y)\), then \(f(\cdot, y) \in C_{c}(X) .\) If \(f \in C_{c}(X \times Y)^{m}\), then \(f(\cdot, y) \in C_{c}(X)^{m} .\) If \(f \in \mathscr{B}(X \times Y)\), then \(f(\cdot, y) \in \mathscr{B}(X)\). Similar results hold for each function \(f\left(x,^{\circ}\right), x \in X\). Proof. Let \(t_{y}: X \rightarrow X \times Y\) be the continuous function given by \(l_{y}(x)=(x, y)\). Then \(f(\cdot, y)=f \circ t_{y}\). From this we see immediately that if \(f \in C_{c}(X \times Y)\), then \(f(\cdot, y)\) is continuous. Furthermore, it is clear that the support of \(f(\cdot, y)\) is contained in the projection from \(X \times Y\) onto \(X\) of the support of \(f\), so that \(f(\cdot, y) \in C_{c}(X)\). It follows from the definitions that \(f(\cdot, y) \in C_{c}(X)^{m}\) if \(f \in\) \(C_{c}(X \times Y)^{m}\). Finally, since composition of Borel functions again produce Borel functions, we see from the identity \(f(\cdot, y)=f \circ t_{y}\) that \(f(\cdot, y) \in \mathscr{B}(X)\) if \(f \in \mathscr{B}(X \times Y)\).

Theorem. If a function \(f\) on \(X\) is the pointwise limit of an increasing sequence \(\left(f_{n}\right)\) in \(\mathscr{L}^{1}(X)\) such that \(\sup \int f_{n}<\infty\), then \(f \in \mathscr{L}^{1}(X)\) and \(\int f=\lim \int f_{n}\).

Riesz' representation theorem \((6.3 .4)\) (F. Riesz 1909) shows indirectly that Daniell's extension theorem is not best possible, if the goal is to make the class of integrable functions as large as possible. (It isn't.) If in the class \(\mathscr{L}(X)\) of measurable functions on \(X\) we define $$ \mathscr{L}_{e s s}^{1}(X)=\left\\{f \in \mathscr{L}(X)\left|\int_{*}\right| f \mid<\infty\right\\} $$ then the essential integral \(\int_{e s s}\), defined on \(\mathscr{L}_{e s s}^{1}(X)\) by $$ \int_{e_{s s}} f=\int_{*} f \vee 0+\int^{*} f \wedge 0, $$ is a positive linear functional on \(\mathscr{L}_{e s s}^{1}(X)\) extending \(\int\) and satisfying Lebesgue's convergence theorems. We have a strict inclusion \(\mathscr{L}^{1}(X) \subset \mathscr{L}_{\text {ess }}^{1}(X)\), whenever the two measures \(\mu_{*}\) and \(\mu^{*}\) in \(6.3 .2\) are different. However, under the innocent assumption (with regards to the applications) that the space \(X\) is a countable union of compact subsets-a \(\sigma\)-compact space-the dichotomy between \(\mu_{*}\) and \(\mu^{*}\) disappears.

Even though the integration theory is at heart concerned only with real-valued functions, there is no difficulty in extending it to complex functions. We say that a function \(f: X \rightarrow \mathbb{C}\) is Borel/measurable provided that the same holds for the two functions \(\operatorname{Re} f\) and \(\operatorname{Im} f\). In fact, this is equivalent with the demand that \(f^{-1}(B) \in \mathscr{B}\) (respectively \(\mathscr{M}\) ) for every Borel subset \(B\) in \(\mathbb{C}\) (because it suffices to consider sets \(B\) of the form \(\\{\alpha \in \mathbb{C} \mid \operatorname{Re} \alpha>t\\}\) and \(\\{\alpha \in \mathbb{C} \mid \operatorname{Im} \alpha>t\\}, \mathrm{cf} .6 .2 .5\), and the counterimages of such sets depend only on \(\operatorname{Re} f\) or on \(\operatorname{Im} f)\). In the same manner we define the complex \(\mathscr{L}^{p}\)-spaces and the complex null functions as the complexifications of the corresponding real spaces. This means that the integral \(\int\) becomes a complex linear functional on the vector space of complex, integrable functions [which we still, with an absolute disregard for consistency, shall denote by \(\left.\mathscr{L}^{1}(X)\right]\). This extension process is painless, but the following complex version of \(6.1 .11\) and \(6.2 .16\) requires a special proof.

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