/*! 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. Each space \(L^{p}(X), ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 10

Theorem. Each space \(L^{p}(X), 1 \leq p \leq \infty\), is a Banach space. Proof. Let \(Q: \mathscr{L}^{p}(X) \rightarrow L^{p}(X)\) denote the quotient map. If \(p<\infty\) and \(\left(Q f_{n}\right)\) is a Cauchy sequence in \(L^{P}(X)\), then \(\left(f_{n}\right)\) is a Cauchy sequence in \(\mathscr{L}^{P}(X)\) (forthe seminorm \(\|\cdot\|_{p}\) ), and we can find a subsequence as described in \(6.4 .9\), which we shall continue to call \(\left(f_{n}\right)\). Define \(f(x)=\lim f_{n}(x)\) if \(x \notin N\) and \(f(x)=0\) if \(x \in N\). Taking \(g=\sum\left|f_{n+1}-f_{n}\right|\), we have \(\|g\|_{p} \leq \sum\left\|f_{n+1}-f_{n}\right\|_{p}<\infty\) so \(g^{p}\) is an integrable function that majorizes \(\left|f-f_{n}\right|^{p}\) on \(X \backslash N\). By Lebesgue's dominated convergence theorem \((6.1 .15)\) it follows that \(\left(\left\|f-f_{n}\right\|_{p}\right)^{p} \rightarrow 0\); whence \(f \in \mathscr{L}^{p}(X)\), and \(Q f_{n} \rightarrow Q f\) in \(L^{p}\). If \(\left(Q f_{n}\right)\) is a Cauchy sequence in \(L^{\infty}(X)\) we set $$ N_{n_{m}}=\left\\{\left|f_{n}-f_{m}\right|>\left\|f_{n}-f_{m}\right\|_{\infty}\right\\} \quad \text { and } \quad N_{n}^{\prime}=\left\\{\left|f_{n}\right|>\left\|f_{n}\right\|_{\infty}\right\\} . $$ All these sets are null sets and since \(\mathscr{N}\) is a \(\sigma\)-ring, also their union \(N=\left(\bigcup N_{n m}\right) \cup\left(\bigcup N_{n}^{\prime}\right)\) is a null set. On \(X \backslash N\) the sequence \(\left(f_{n}\right)\) converges uniformly to a bounded function \(f .\) Defining \(f=0\) on \(N\) we have that \(f \in \mathscr{L}^{\infty}(X)\) and that \(Q f_{n} \rightarrow Q f\) in \(L^{\infty}(X)\).

Short Answer

Expert verified
Each space \(L^p(X), 1 \leq p \leq \infty\), is a Banach space because every Cauchy sequence in \(L^p(X)\) converges.

Step by step solution

01

Understand the Problem Statement

We need to prove that each space \(L^p(X), 1 \leq p \leq \infty\), is a Banach space. This means showing that every Cauchy sequence in \(L^p(X)\) converges to a limit in \(L^p(X)\).
02

Define the Quotient Map

Denote \(Q: \mathscr{L}^p(X) \rightarrow L^p(X)\) as the quotient map from the space of equivalence classes of measurable functions (\(\mathscr{L}^p(X)\)) to the \(L^p(X)\) space.
03

Handle the Case \(1 \leq p < \infty\)

Assume \( (Q f_n) \) is a Cauchy sequence in \( L^p(X) \). Then \( (f_n) \) is a Cauchy sequence in \( \mathscr{L}^p(X) \) in the seminorm \( \|\cdot\|_p \). Using a subsequence, define \( f(x) = \lim f_n(x) \) for \( x otin N \) (a null set), and \( f(x) = 0 \) for \( x \in N \).
04

Apply Dominated Convergence Theorem

Define \( g = \sum |f_{n+1} - f_n| \). Since \( \|g\|_p < \infty \), \( g^p \) majorizes \( |f - f_n|^p \) outside \( N \). By the Lebesgue dominated convergence theorem, \( \left(\|f - f_n\|_p\right)^p \rightarrow 0 \). Hence, \( f \in \mathscr{L}^p(X) \) and \( Q f_n \rightarrow Q f \) in \( L^p(X) \).
05

Handle the Case \(p = \infty\)

For a Cauchy sequence \((Q f_n)\) in \( L^\infty(X) \), define sets \( N_{nm} \) and \( N_{n}' \) for \( |f_n - f_m| > \|f_n - f_m\|_\infty \) and \( |f_n| > \|f_n\|_\infty \), respectively, which are null sets.
06

Show Bounded Convergence

The union \( N = (\cup N_{nm}) \cup (\cup N_{n}') \) is a null set. On \( X \backslash N \), \((f_n)\) converges uniformly to a bounded function \( f \). Defining \( f = 0 \) on \( N \), then \( f \in \mathscr{L}^\infty(X) \) and \( Q f_n \rightarrow Q f \) in \( L^\infty(X) \).
07

Conclude Each \(L^p(X)\) is a Banach Space

Both cases show that Cauchy sequences in \(L^p(X)\) converge within \(L^p(X)\), thus proving each \(L^p(X)\) is a Banach space.

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Key Concepts

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

Lp spaces
The concept of Lp spaces is a fundamental topic in functional analysis and measure theory. These spaces are built around measurable functions and their integrability in terms of the p-norm. Here, the variable "p" refers to the power to which function values are raised before integration.
  • In mathematical terms, for a measurable function \( f \) over a measure space \( X \), the Lp space consists of equivalence classes of functions for which the integral \( \int_X |f(x)|^p \, dx \) is finite.
  • The value of "p" can range between 1 and infinity, denoted as \( 1 \leq p \leq \infty \).
**Why are Lp Spaces Important?**
Lp spaces allow us to work with functions that might not be well-behaved as we would like. They are crucial for the study of partial differential equations, probability, and other fields that require control over function behavior through integration.
Cauchy sequence
Understanding a Cauchy sequence is crucial to really grasp what it means for a space to be complete, like a Banach space. Intuitively, a sequence \( (a_n) \) is Cauchy if, as you progress through the sequence, the terms eventually get arbitrarily close to each other.
**Mathematically Speaking**
  • A sequence \( (a_n) \) in a metric space \( M \) is called a Cauchy sequence if for every \( \epsilon > 0 \), there exists an integer \( N \) such that for all \( m, n > N \), the distance \( d(a_m, a_n) < \epsilon \).
**Importance in Lp Spaces**
In the context of Lp spaces, demonstrating that every Cauchy sequence converges within the space itself is key to showing that the space is complete, hence a Banach space. This property assures that Lp spaces are resilient and predictable, making them incredibly useful in mathematical analysis.
Dominated convergence theorem
The Dominated Convergence Theorem is a significant result in measure theory that allows for the interchange of limits and integrals under certain conditions. It is particularly useful in proving that Lp spaces are Banach spaces.
**How It Works**
  • If we have a sequence of measurable functions \( f_n \) that converge almost everywhere to a function \( f \), and if there exists a function \( g \) integrable over \( X \) that dominates \( |f_n| \) (i.e., \( |f_n(x)| \le g(x) \) for all \( x \)), then \( \lim_{n \to \infty} \int f_n \) is equal to \( \int f \).
**Application**
Using this theorem, we can show that the limit of a sequence of functions, under certain conditions, is within Lp(X) when the sequence itself is Cauchy in Lp(X), hence proving the completeness of the space.
Measurable functions
Measurable functions form the foundational building blocks, not only of Lp spaces but also of many concepts in real analysis and probability. These functions are fundamentally connected to the concept of measure, which generalizes the idea of "size" or "volume".
**What Makes a Function Measurable?**
  • A function \( f: X \to \mathbb{R} \) is measurable if, for any real number \( a \), the set \( \{ x \in X | f(x) \leq a \} \) is a measurable set.
**Relevance to Lp Spaces**
In Lp spaces, we primarily deal with equivalence classes of these measurable functions. Their measurability ensures that we can apply integral calculus, including Lebesgue integration, which is more suitable for analysis in Lp spaces compared to standard Riemann integration. Being measurable is essentially the first step towards dealing with functions and their properties in a rigorous, analytic manner.

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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 \(L^{1}(G)\).

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

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 \(f_{1} \wedge f_{2} \in \mathscr{L}^{1}(X)\).

Proposition. If \(f\left(f_{n}\right)\) is a sequence in \(\mathscr{L}^{1}(X)\) such that \(f_{n}(x) \leq f_{n+1}(x)\) for almost all \(x\) and every \(n\), and if \(\lim \int f_{n}<\infty\), there is an element \(f\) in \(\mathscr{L}^{1}(X)\) such that \(\int f=\lim \int f_{n}\) and \(f(x)=\lim f_{n}(x)\) almost everywhere. Proof. For each \(n\) there is a null set \(N_{n}\) such that \(f_{n}(x) \leq f_{n+1}(x)\) for \(x \notin N_{n}\). With \(N=\bigcup N_{n}\) we have a null set \(N\) and an extended-valued function \(f\) such that \(f_{n}(x) \nearrow f(x)\) for \(x \notin N\). Since6.4. \(L^{p}\)-Spaces 241 $$ \int\left[\left\\{f_{n}>m\right\\}\right] \leq \int m^{-1} f_{n} $$ for every \(n\) and \(m\), we see from \(6.1 .13\) that $$ \int[\\{f>m\\}] \leq m^{-1} \lim \int f_{n^{*}} $$ It follows that \(N_{\infty}=\\{f=\infty\\}\) is a null set, so that \(6.1 .13\) can be applied to the restriction of \(\left(f_{n}\right)\) and \(f\) to \(X \backslash\left(N_{\infty} \cup N\right)\).

The convolution product can be defined for other classes of functions. Thus \(f \times g\) exists as an element in \(C_{0}(G)\) whenever \(f \in \mathscr{L}^{p}(G)\) and \(g{g} \in \mathscr{L}^{q}(G)\) with \(p^{-1}+q^{-1}=1\) [because \(f \times g(x)=\int f(y)_{x} \check{g}(y) d y\) ]. In the case \(p=1\), \(q=\infty\), however, we only have \(f \times g\) as a uniformly continuous, bounded function on \(G\). We wish to define the convolution product of finite Radon charges (cf. 6.5.8). If \(\Phi, \Psi \in M(G)\), we define \(\Phi \otimes \Psi\) as a finite Radon charge on \(G \times G\), either by mimicking the proof of \(6.6 .3\) or by taking polar decompositions \(\Phi=|\Phi|\left(u^{*}\right)\) and \(\Psi=|\Psi|\left(v^{\circ}\right)\) as in \(6.5 .6\) and \(6.5 .8\) and then setting \((\Phi \otimes \Psi) h=(|\Phi| \otimes|\Psi|)((u \otimes v) h), \quad h \in C_{c}(G \times G) .\) \((\Phi \times \Psi) f=(\Phi \otimes \Psi)(f \circ \pi), \quad f \in C_{c}(G)\), \((\Phi \times \Psi) f=(\Phi \otimes \Psi)(f \circ \pi), \quad f \in C_{c}(G)\), \(\rightarrow G\) is the product map \(\pi(x, y)=x y .\) Note that although where \(\pi: G \times G \rightarrow G\) is the product map \(\pi(x, y)=x y .\) Note that although \(f \circ \pi\) is a bounded continuous function on \(G \times G\) it does not belong to \(C_{c}(G \times G)\) (if \(f \neq 0\) ), so that we need the assumption that \(\Phi\) and \(\Psi\) are finite Having done this, we define the product in \(M\) $$ (\Phi \times \Psi) f=(\Phi \otimes \Psi)(f \circ \pi) \text {, } $$ where \(\pi: G \times G \rightarrow G\) is the product map \(\pi(x, y)\) \(f \circ \pi\) is a bounded continuous function on \(G \times\) \(C_{c}(G \times G)(\) if \(f \neq 0)\), so that we need the assumption that charges. It follows from \((*)\) and (**) that \(\Phi \times \Psi \in M(G)\), with $$ \|\Phi \times \Psi\| \leq\|\Phi \otimes \Psi\|=\|\Phi\|\|\Psi\| \text {. } $$ $$ ((\Phi \times \Psi) \times \Omega) f=(\Phi \otimes \Psi \otimes \Omega)(f \circ \tau)=(\Phi \times(\Psi \times \Omega)) f $$ where \(\tau(x, y, z)=x y z\), so that the product is associative. In conjunction with \(6.5 .9\) this shows that \(M(G)\) is a unital Banach algebra (the point measure \(\delta_{1}\) at 1 being the unit). Defining $$ \Phi^{*} f=\overline{\Phi f}, \quad f \in C_{c}(G), $$ where \(f(x)=\overline{f\left(x^{-1}\right)}\), we see that $$ \left(\Psi * \otimes \Phi^{*}\right)(f \otimes g)=\overline{\Psi(f) \Phi(\tilde{g})}=\overline{(\Phi \otimes \Psi)(\tilde{g} \otimes f)}=\overline{(\Phi \otimes \Psi)(f \otimes g)^{s}} $$ where \(h^{s}(x, y)=h\left(y^{-1}, x^{-1}\right)\) for every function on \(G \times G\). It follows from (**) that for each \(f\) in \(C_{c}(G)\) we have $$ \begin{aligned} \left(\Psi^{*} \times \Phi^{*}\right) f &=\left(\Psi^{\otimes} \otimes^{*}\right)(f \circ \pi)=\overline{(\Phi \otimes \Psi)(f \circ \pi)^{s}} \\ &=\overline{(\Phi \otimes \Psi)(f \circ \pi)}=(\Phi \times \Psi)^{*} f \end{aligned} $$ so that \(*\) is an isometric involution on \(M(G)\).
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