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If \(f\) is a measurable function on \(X\), define the essential range \(R_{f}\) of \(f\) to be the set of all \(z \in \mathbb{C}\) such that \(\\{x:|f(x)-z|<\epsilon\\}\) has positive measure for all \(\epsilon>0\). a. \(R_{f}\) is closed. b. If \(f \in L^{\infty}\), then \(R_{f}\) is compact and \(\|f\|_{\infty}=\max \left\\{|z|: z \in R_{f}\right\\}\).

Short Answer

Expert verified
(a) \(R_f\) is closed; (b) If \(f \in L^{\infty}\), \(R_f\) is compact and \(\|f\|_{\infty} = \max \{|z| : z \in R_f\}\).

Step by step solution

01

Understand Essential Range

The essential range, \(R_f\), of a measurable function \(f\) consists of all complex numbers \(z\) such that for any \(\epsilon > 0\), the set \(\{x : |f(x) - z| < \epsilon\}\) has positive measure. Essentially, \(z\) is close to \(f(x)\) for a non-negligible set of \(x\).
02

Prove that \(R_f\) is Closed

To show that \(R_f\) is closed, take any sequence \((z_n) \subset R_f\) converging to \(z\). For each \(n\), the set \(\{x : |f(x) - z_n| < \epsilon\}\) has positive measure. Given \(\epsilon > 0\), there exists \(N\) such that if \(n \geq N\), then \(|z_n - z| < \epsilon/2\), so \(|f(x)-z| < |f(x)-z_n| + |z_n-z| < \epsilon\), ensuring \(\{x : |f(x)-z| < \epsilon\}\) also has positive measure. Thus, \(z \in R_f\), showing \(R_f\) is closed.
03

Show that \(R_f\) is Bounded if \(f \in L^{ ext{∞}}\)

If \(f \in L^{\infty}\), it means that there exists an \(M\) such that \(|f(x)| \leq M\) almost everywhere. Therefore, for \(z \in R_f\), \(|z|\) must also be \(\leq M\), implying that \(R_f\) is bounded.
04

Prove \(R_f\) is Compact

Since \(R_f\) is a subset of a closed ball in \(\mathbb{C}\), it is bounded. We've shown it is closed. In a metric space, a set that is closed and bounded is compact according to the Heine-Borel theorem. Thus, \(R_f\) is compact.
05

Prove \(\|f\|_{\infty} = \max \{|z| : z \in R_f\}\)

Since \(R_f\) is compact, it attains its maximum value. Let \(z_0\) be the point where this maximum is achieved. Any \(|f(x)|\) almost everywhere is smaller than or equal to \(z_0\), implying \(\|f\|_{\infty} = \max \{|z| : z \in R_f\}\) since we already know \(R_f\) contains all essential values in the range of \(f\).

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Key Concepts

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

Measurable Function
A measurable function is a fundamental concept in real analysis and measure theory. It refers to a function that is compatible with the structure of the space it is defined on. This means the function must be measurable concerning the measure on that space, ensuring that the preimage of any measurable set is also measurable. This concept is crucial when working with functions that may not necessarily be continuous or differentiable.
  • To be measurable, a function, say \(f: X \to \mathbb{R}\), must satisfy that for every real number \(a\), the set \(\{x \in X : f(x) > a\}\) is measurable.
  • Measurable functions allow for the integration and analysis of functions that could be quite irregular.
  • The essential range relates to measurable functions by focusing on values that the function takes across sets of significant size.
Understanding measurable functions is key to grasping more advanced topics like integration, probability, and essential range. They help in dealing with functions in more abstract settings where calculus alone may not suffice.
Compactness
Compactness is a powerful concept in mathematics, particularly in the context of analysis and topology. It gives the notion of being "small" or "finite in extent," but in a very specific, formal way.
  • A set is compact if every open cover of the set has a finite subcover. Informally, you can think of compactness as having no "edges" or being "closed and bounded" in finite-dimensional spaces.
  • In the context of metric spaces, like \(\mathbb{R}^n\), compact sets are those that are closed and bounded, thanks to the Heine-Borel theorem.
  • For the essential range \(R_f\) of a function \(f\), if \(f\) is bounded (as in \(L^{\infty}\)), then we can show that \(R_f\) is compact since it is both closed and bounded.
The importance of compactness arises from its many properties such as guaranteeing the existence of maximum and minimum values, making many sophisticated proofs and results accessible.
L-infinity Space
The \(L^{\infty}\) space, or essentially bounded function space, plays a crucial role when dealing with bounded functions. Functions in \(L^{\infty}\) are characterized by being bounded almost everywhere.
  • If a function \(f\) belongs to \(L^{\infty}\), it means there's a real constant \(M\) such that \(|f(x)| \leq M\) for almost all \(x\).
  • This space is equipped with the "essential supremum" norm, given by \(\|f\|_{\infty} = \inf \{ M : |f(x)| \leq M\ \text{almost everywhere}\}\).
  • The essential range \(R_f\) of a function in \(L^{\infty}\) is particularly interesting because it captures the behavior of the function about these bounds, ensuring \(R_f\) not only is bounded but also compact.
  • Moreover, in \(L^{\infty}\) spaces, \(\|f\|_{\infty}\) equals the maximum modulus of the essential values of \(f\), fulfilling the condition where the essential range holds as a meaningful representation of the function's range.
This concept is vital when studying functional spaces, helping mathematicians deal with spaces of functions in a structured and manageable way.

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Most popular questions from this chapter

Suppose \(1 \leq p<\infty\). If \(\left\|f_{n}-f\right\|_{p} \rightarrow 0\), then \(f_{n} \rightarrow f\) in measure, and hence some subsequence converges to \(f\) a.e. On the other hand, if \(f_{n} \rightarrow f\) in measure and \(\left|f_{n}\right| \leq g \in L^{P}\) for all \(n\), then \(\left\|f_{n}-f\right\|_{p} \rightarrow 0\).

(The Vitali Convergence Theorem) Suppose \(1 \leq p<\infty\) and \(\left\\{f_{n}\right\\}_{1}^{\infty} \subset L^{p}\). In order for \(\left\\{f_{n}\right\\}\) to be Cauchy in the \(L^{p}\) norm it is necessary and sufficient for the following three conditions to hold: (i) \(\left\\{f_{n}\right\\}\) is Cauchy in measure; (ii) the sequence \(\left\\{\left|f_{n}\right|^{P}\right\\}\) is uniformly integrable (see Exercise 11 in \(\S 3.2\) ); and (iii) for every \(\in>0\) there exists \(E \subset X\) such that \(\mu(E)<\infty\) and \(\int_{E^{c}}\left|f_{n}\right|^{p}<\epsilon\) for all \(n\). (To prove the sufficiency: Given \(\epsilon>0\), let \(E\) be as in (iii), and let \(A_{\operatorname{mn}}=\\{x \in E\) : \(\left.\left|f_{m}(x)-f_{n}(x)\right| \geq \epsilon\right\\}\). Then the integrals of \(\left|f_{n}-f_{m}\right|^{p}\) over \(E \backslash A_{m n}, A_{m n}\), and \(E^{c}\) are small when \(m\) and \(n\) are large - for three different reasons.)

Suppose \(0

If \(0<\alpha

Let \((X, \mathcal{M}, \mu)\) be a measure space. A set \(E \in \mathcal{M}\) is called locally null if \(\mu(E \cap F)=0\) for every \(F \in \mathcal{M}\) such that \(\mu(F)<\infty\). If \(f: X \rightarrow \mathbb{C}\) is a measurable function, define $$ \|f\|_{*}=\inf \\{a:\\{x:|f(x)|>a\\} \text { is locally null }\\}, $$ and let \(\mathcal{L}^{\infty}=\mathcal{L}^{\infty}(X, \mathcal{M}, \mu)\) be the space of all measurable \(f\) such that \(\|f\|_{0}<\infty\). We consider \(f, g \in \mathcal{L}^{\infty}\) to be identical if \(\\{x: f(x) \neq g(x)\\}\) is locally null. a. If \(E\) is locally null, then \(\mu(E)\) is cither 0 or \(\infty\). If \(\mu\) is semifinite, then every locally null set is null. b. \(\|\cdot\|_{*}\) is a norm on \(\mathcal{L}^{\infty}\) that makes \(\mathcal{L}^{\infty}\) into a Banach space. If \(\mu\) is semifinite, then \(\mathcal{L}^{\infty}=L^{\infty}\).

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