/*! 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 10 Theorem. Given a Radon integral ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 10

Theorem. Given a Radon integral \(\int\) on a locally compact Hausdorff space \(X\), the class \(\mathscr{L}^{1}(X)\) of integrable functions is a vector space containing \(C_{e}(X)\), which is closed under the lattice operations \(\bigvee\) and \(\bigwedge\). Moreover, \(\int: \mathscr{L}^{1}(X) \rightarrow \mathbb{R}\) is a positive functional that extends the original integral on \(C_{c}(X)\). Proof. Simple manipulations with the definition of integrability, in connection with \(6.1 .6\) (and the corresponding result for lower integrals), show that if \(f\) and \(g\) both belong to \(\mathscr{L}^{1}(X)\) and \(t \geq 0\), then \(t f+g \in \mathscr{L}^{1}(X)\) with $$ \int(t f+g)=t \int f+\int g \text {. } $$ Since \(-C_{c}(X)_{m}=C_{c}(X)^{m}\), it is also easy to verify that if \(f \in \mathscr{L}^{1}(X)\), then \(-f \in \mathscr{L}^{1}(X)\) with \(\int-f=-\int f .\) Taken together this means that \(\mathscr{L}^{1}(X)\) is areal vector space, and that \(\int\) is a functional on it. Since \(\int^{*} f=\int_{*} f=\int f\) if \(f \in C_{c}(X)\), cf. 6.1.2, it follows that \(C_{c}(X) \subset \mathscr{L}^{1}(X)\) and that \(\int\) extends the original integral on \(C_{c}(X) .\) Moreover, if \(f \geq 0\), then \(\int * f \geq 0\), so that \(\int\) is a positive functional on \(\mathscr{L}^{1}(X)\). Take \(f_{1}\) and \(f_{2}\) in \(\mathscr{L}^{1}(X)\) and for \(\varepsilon>0\) choose \(g_{1}, g_{2}\) in \(C_{e}(X)^{m}, h_{1}, h_{2}\) in \(C_{c}(X)_{m}\), such that \(h_{k} \leq f_{k} \leq g_{k}\) and \(\int * g_{k}-\int_{*} h_{k}<\varepsilon\) for \(k=1,2 ;\) cf. (*) in 6.1.9. Note that $$ g_{1} \vee g_{2}-h_{1} \vee h_{2} \leq\left(g_{1}-h_{1}\right)+\left(g_{2}-h_{2}\right) $$ so that \(\int^{*} g_{1} \vee g_{2}-\int_{*} h_{1} \vee h_{2}<2 \varepsilon\). It follows that \(f_{1} \vee f_{2} \in \mathscr{L}^{1}(X)\). Similarly, or by exploiting the identity \(f_{1} \wedge f_{2}+f_{1} \vee f_{2}=f_{1}+f_{2}\), we see that \(f_{1} \wedge f_{2} \in \mathscr{L}^{1}(X)\).

Short Answer

Expert verified
\(\mathscr{L}^{1}(X)\) is a vector space allowing lattice operations, and \(\int\) is a positive functional extending the integral on \(C_c(X)\).

Step by step solution

01

Understanding the Problem

We are given a Radon integral on a locally compact Hausdorff space \(X\) and need to show certain properties about the class \(\mathscr{L}^{1}(X)\) of integrable functions. This class must satisfy properties of a vector space and support lattice operations while maintaining the functional properties of the integral \(\int\).
02

Prove Vector Space Properties

To show that \(\mathscr{L}^{1}(X)\) is a vector space, we identify that for \(f, g \in \mathscr{L}^{1}(X)\) and \(t \geq 0\), the function \(tf + g\) is also in \(\mathscr{L}^{1}(X)\) and \(\int(tf+g) = t\int f + \int g\). Also, for any \(f \in \mathscr{L}^{1}(X)\), its negative \(-f \in \mathscr{L}^{1}(X)\) with \(\int -f = -\int f\). This confirms closure under scalar multiplication and addition, and existence of additive inverses.
03

Extend the Integral

The integral \(\int\) on \(\mathscr{L}^{1}(X)\) extends the integral on \(C_{c}(X)\), the space of continuous functions with compact support. This shows that \(C_{c}(X) \subset \mathscr{L}^{1}(X)\), ensuring \(\int\) maintains its properties on the smaller space.
04

Prove Lattice Operations

For lattice operations, we need to show that for \(f_1, f_2 \in \mathscr{L}^{1}(X)\), both \(f_1 \vee f_2\) and \(f_1 \wedge f_2\) are in \(\mathscr{L}^{1}(X)\). Using functions \(g_1, g_2 \in C_e(X)^m\) and \(h_1, h_2 \in C_c(X)_m\) with \(h_k \leq f_k \leq g_k\) and ensuring their integrals approximate those of \(f_k\) within a small \(\varepsilon\), conclude that these operations maintain membership in \(\mathscr{L}^{1}(X)\).
05

Ensure Positivity

For \(f \geq 0\), \(\int f \geq 0\) confirms that \(\int\) is a positive functional. This aligns with the positivity defined for integrals of non-negative functions, ensuring adherence to properties needed in integration.

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Key Concepts

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

Vector Space
A vector space is a fundamental concept in linear algebra. It's a collection of objects, called vectors, which can be added together and multiplied by scalars (real numbers) while satisfying specific rules. When discussing integrable functions on a locally compact Hausdorff space, we treat these functions as vectors within the vector space
  • Closure under Addition: For any two functions, \(f\) and \(g\), in the class \(\mathscr{L}^{1}(X)\), their sum \(f + g\) is also in \(\mathscr{L}^{1}(X)\).
  • Closure under Scalar Multiplication: If you multiply any function \(f\) in \(\mathscr{L}^{1}(X)\) by a scalar \(t\), the product \(tf\) is still within the class.
  • Additive Inverse: For every function \(f\) in the space, its negative \(-f\) is also included in the space.
This elegant structure allows us to perform operations such as scaling and addition, crucial for analyzing complex interactions between functions.
Integrable Functions
The concept of integrable functions pertains to those functions which can be effectively 'summed up' in a meaningful way across a given space. In the context of a Radon integral on a locally compact Hausdorff space, the focus is on the space \(\mathscr{L}^{1}(X)\) of integrable functions. This space includes certain continuous functions and respects the properties of integration:
  • Extension of Classical Integration: The Radon integral \(\int\) on \(\mathscr{L}^{1}(X)\) extends traditional integration methods used on \(C_{c}(X)\), which is the space of continuous functions with compact support.
  • Manipulation and Calculation: For functions \(f\) and \(g\) both in \(\mathscr{L}^{1}(X)\) and a non-negative scalar \(t\), the integral of \(tf + g\) is equal to \(t \int f + \int g\). This reflects how integration respects vector space operations.
These properties ensure the class of integrable functions aligns with our intuitive understanding of summing or integrating functions.
Locally Compact Hausdorff Space
A locally compact Hausdorff space is a specific type of topological space that serves as the backdrop for many mathematical analyses, especially in integration theory.
  • Locally Compact: Every point has a neighborhood that is compact, meaning it is closed and bounded.
  • Hausdorff Property: Any two distinct points can be separated by neighborhoods, ensuring a form of 'closeness' that avoids ambiguity.
This framework is essential because it provides a controlled environment where integrable functions and vector spaces behave predictably. The structure of a locally compact Hausdorff space is what makes the analysis of integrable functions and their properties feasible in practical applications.
Lattice Operations
Lattice operations deal with the mechanisms of combining and comparing elements to form new structures. In our context, these are applied to functions within the class \(\mathscr{L}^{1}(X)\).
  • Join (\(\vee\)): Given two functions \(f_1\) and \(f_2\), the function \(f_1 \vee f_2\) represents their pointwise maximum at each point in the space.
  • Meet (\(\wedge\)): Similarly, \(f_1 \wedge f_2\) is their pointwise minimum.
These operations must yield results within \(\mathscr{L}^{1}(X)\), promoting a coherent structure that remains closed under such transformations. By ensuring the class of integrable functions supports lattice operations, we can effectively manipulate and analyze the relations between different function values within the space.

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Most popular questions from this chapter

Theorem. For a locally compact group \(G\) with Haar integral \(\int\), the space \(L^{1}(G)\) is a Banach algebra with an isometric involution. Product and involution in \(L^{1}(G)\) are given by the formulas $$ f \times g(x)=\int f(y) g\left(y^{-1} x\right) d y, \quad f^{*}(x)=\overline{f\left(x^{-1}\right)} \Delta(x)^{-1} $$ Proof. It follows from \(6.6 .20\) that \(f \times g \in L^{1}(G)\) with \(\|f \times g\|_{1} \leq\|f\|_{1}\|g\|_{1}\). It is elementary to check that the convolution product is distributive with respect to the sum, but it requires Fubini's theorem to show that the product is associative. Indeed, if \(f, g\), and \(h\) belong to \(\mathscr{L}^{1}(G)\), both functions \((f \times g) \times h\) and \(f \times(g \times h)\) belong to \(\mathscr{L}^{1}(G)\); and they are equal almost everywhere because the function $$ (x, y, z) \rightarrow f(z) g\left(z^{-1} y\right) h\left(y^{-1} z\right) $$ belongs to \(\mathscr{L}^{1}(G \times G \times G)\). The operation \(f \rightarrow f^{*}\) is con jugate linear and isometric with period two by 6.6.18. Moreover, we have $$ \begin{aligned} g^{*} \times f^{*}(x) &=\int g^{*}(y) f^{*}\left(y^{-1} x\right) d y \\ &=\int \overline{f\left(x^{-1} y\right)} \Delta\left(x^{-1} y\right) \overline{g\left(y^{-1}\right)} \Delta\left(y^{-1}\right) d y \\ &\left.=\int \overline{f(y)} \overline{g\left(y^{-1}\right.} x^{-1}\right) d y \Delta\left(x^{-1}\right)=(f \times g) *(x) \end{aligned} $$ which shows that \(*\) is antimultiplicative, and thus is an involution on \(L^{1}(G)\).

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. To each Radon charge \(\Phi\) on a locally compact, \(\sigma\)-compact Hausdorff space \(X\) there is a Radon integral \(\int\) and a Borel function u such that \(|u|=1\) almost everywhere \((\) with respect to \(f)\) and \(\Phi=\int \cdot u\).Proof. For each \(f\) in \(C_{c}(X)_{+}\)define $$ \int f=\sup \left\\{|\Phi(g)|\left|g \in C_{c}(X),\right| g \mid \leq f\right\\} . $$ Then \(0 \leq \int f<\infty\) and \(f \rightarrow \int f\) is a positive homogeneous function. Evidently, we may replace \(|\Phi(g)|\) with \(\operatorname{Re} \Phi(g)\) in the definition above, from which it follows that \(\int\) is superadditive \(\left[\int f_{1}+\int f_{2} \leq \int\left(f_{1}+f_{2}\right)\right]\), since \(\left|g_{1}\right| \leq f_{1}\) and \(\left|g_{2}\right| \leq f_{2}\) imply \(\left|g_{1}+g_{2}\right| \leq f_{1}+f_{2}\). To show that \(\int\) is additive it therefore suffices to show that it is subadditive. Given \(f_{1}\) and \(f_{2}\) and \(\varepsilon>0\), choose \(g\) in \(C_{c}(X)\) such that \(|g| \leq f_{1}+f_{2}\) and $$ \int f_{1}+f_{2} \leq \varepsilon+|\Phi(g)| $$ Put \(g_{j}=g f_{j}\left(f_{1}+f_{2}\right)^{-1}\) for \(j=1,2\), and note that these are continuous functions. Moreover, \(g_{1}+g_{2}=g\) and $$ \left|g_{j}\right|=|g| f_{j}\left(f_{1}+f_{2}\right)^{-1} \leq f_{j}, \quad j=1,2 $$ Thus, $$ |\Phi(g)| \leq\left|\Phi\left(g_{1}\right)\right|+\left|\Phi\left(g_{2}\right)\right| \leq \int f_{1}+\int f_{2} $$ In conjunction with the previous estimate this shows that \(\int\) is subadditive. Thus, \(\int\) is a positive homogeneous, additive function on \(C_{c}(X)_{+}\), and therefore extends uniquely to a positive functional on \(C_{c}(X)\), i.e. a Radon integral on \(X\). Assume first that \(\int 1<\infty\). Then for each \(f\) in \(C_{\mathrm{c}}(X)\) we have $$ |\Phi(f)|^{2} \leq\left(\int|f|\right)^{2} \leq\left(\int 1\right)\left(\int|f|^{2}\right) $$ As in the proof of \(6.5 .4\) this produces a Borel function \(u\) in \(\mathscr{L}^{2}(X)\) such that $$ \Phi(f)=(f \mid \bar{u})=\int f u, \quad f \in C_{c}(X) $$ Let \(N=\\{|u| \geq 1+\varepsilon\\}\) and use \(6.4 .11\) to find a sequence \(\left(u_{n}\right)\) in \(C_{c}(X)\) such that \(\left\|u_{n}-\bar{u}|u|^{-1}[N]\right\|_{1} \rightarrow 0\) and \(\left\|u_{n}\right\|_{\infty} \leq 1 .\) Since \(\left|u_{n} f\right| \leq f\) for every \(f\) in \(C_{c}(X)_{+}\), we have $$ \begin{aligned} \int f & \geq \lim \sup \left|\Phi\left(u_{n} f\right)\right|=\lim \sup \left|\int f u_{n} u\right| \\ &=\int f[N]|u| \geq(1+\varepsilon) \int f[N] \end{aligned} $$since \(C\) and \(\varepsilon\) are arbitrary, it follows that \(|u| \leq 1\) almost everywhere. On the other hand, for each \(f\) in \(C_{c}(X)_{+}\)and \(\varepsilon>0\) we can find \(g\) in \(C_{c}(X)\) such that \(|g| \leq f\) and $$ \int f \leq \varepsilon+|\Phi(g)|=\varepsilon+\left|\int g u\right| \leq \varepsilon+\int f|u| . $$ Therefore, \(\int f \leq \int f|u|\) for every \(f\) in \(C_{\mathrm{c}}(X)_{+}\), thus also for every \(f\) in \(\mathscr{B}(X)_{+}\), and since \(|u| \leq 1\) this means that \(\int f=\int f|u|\) for every \(f\) in \(\mathscr{B}(X)_{+}\), i.e. \(|u|=1\) almost everywhere. To prove the general case, use the \(\sigma\)-compactness of \(X\) to find a function \(h\) in \(C_{0}(X)_{+}\)such that \(\int h<\infty\) but \(h(x)>0\) for every \(x\) in \(X\). Then consider the Radon charge \(\Phi(\cdot h)\) and the finite Radon integral \(\int \cdot h\), and note that \(\int \cdot h\) is obtained from \(\Phi(\cdot h)\) as in the first part of the proof. There is therefore a Borel function \(u\) on \(X\), with \(|u|=1\) almost everywhere (with respect to \(\int \cdot h\) ) such that \(\Phi(\cdot h)=\int \cdot h u\). Since \(h>0\), the integral \(\int \cdot h\) and \(\int\) have the same null sets, so that \(|u|=1\) almost everywhere with respect to \(\int .\) Moreover, \(h C_{c}(X)=\) \(C_{\mathrm{c}}(X)\), so \(\Phi(f)=\int f u\) for every \(f\) in \(C_{\mathrm{c}}(X)\). But then \(\int f \geq(1+\varepsilon) \int f[N]\) for every Borel function \(f \geq 0\), in particular for \(f=[C]\), where \(C\) is a compact subset of \(N .\) We conclude that \(\int[C]=0 ;\) and

We wish to consider the complex linear span of Radon integrals on \(X\). This will be referred to as the space of Radon charges on \(X\), and consists of certain linear functionals on \(C_{c}(X)\). Equipped with the weak topology \(\tau\) induced by the seminorms \(f \rightarrow\left|\int f\right|\), where \(\int\) ranges over all Radon integrals on \(X\), it follows from \(2.4 .1\) and \(2.4 .4\) that the Radon charges constitute the dual space of the topological vector space \(\left(C_{c}(X), \tau\right) .\) Clearly, a more constructive description of these elements is desirable. We therefore define a Radon charge to be a functional \(\Phi\) on \(C_{c}(X)\), which for every \(f\) in \(C_{c}(X)_{+}\)satisfies $$ \sup \left\\{\left|\Phi(g) \| g \in C_{c}(X),\right| g \mid \leq f\right\\}<\infty . $$
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