Chapter 8

The LpL^{p} Spaces

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Section 8.1Lp\boldsymbol{L}^{\boldsymbol{p}} Space: The Case 1⩽p<+∞\mathbf{1} \leqslant \boldsymbol{p}<+\infty[10/90%]
Problem 8.1.1

Pick A,B∈Lp\mathfrak {A}, \mathfrak {B} \in \mathfrak {L}^{p}. Then A+B∈Lp\mathfrak {A}+\mathfrak {B} \in \mathfrak {L}^{p} and ∥A+B∥p⩽∥A∥p+∥B∥p\left\| \mathfrak {A}+\mathfrak {B}\right\|_{p} \leqslant \left\| \mathfrak {A}\right\|_{p}+\left\| \mathfrak {B}\right\|_{p}.

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Problem 8.1.2

Prove the Cauchy-Schwarz inequality

∣∑k=1nakbk∣⩽∑k=1nak2∑k=1nbk2. \left|\sum _{k=1}^{n} a_{k} b_{k}\right| \leqslant \sqrt{\sum _{k=1}^{n} a_{k}^{2}} \sqrt{\sum _{k=1}^{n} b_{k}^{2}} .
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Problem 8.1.3

Let 1<p1,…,pn<+∞1<p_{1}, \ldots , p_{n}<+\infty be such that 1/p1+⋯+1/pn=1 / p_{1}+\cdots +1 / p_{n}= 1 , and pick functions f1∈Lp1,…,fn∈Lpnf_{1} \in L^{p_{1}}, \ldots , f_{n} \in L^{p_{n}}. We wish to generalize Hölder's Inequality by showing that ∏i=1nfi∈L1\prod_{i=1}^{n} f_{i} \in L^{1} and ∣∫∏i=1nfi∣⩽∏i=1n∥fi∥pi\left|\int \prod_{i=1}^{n} f_{i}\right| \leqslant \prod_{i=1}^{n}\left\| f_{i}\right\|_{p_{i}}.

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(a)

Show first that a1⋯an⩽a1p1/p1+⋯+anpn/pna_{1} \cdots a_{n} \leqslant a_{1}^{p_{1}} / p_{1}+\cdots +a_{n}^{p_{n}} / p_{n} by generalizing the calculus result given in the section. [Here 0⩽a1,…,an<+∞0 \leqslant a_{1}, \ldots , a_{n}<+\infty.]

(b)

If ∥f1∥p1=0\left\| f_{1}\right\|_{p_{1}}=0 or …\ldots or ∥fn∥pn=0\left\| f_{n}\right\|_{p_{n}}=0, the claim is trivial.

(c)

Use (a) to prove the claim when ∥f1∥p1=⋯=∥fn∥pn=1\left\| f_{1}\right\|_{p_{1}}=\cdots =\left\| f_{n}\right\|_{p_{n}}=1.

(d)

Prove the claim when ∥f1∥p1,…,∥fn∥pn\left\| f_{1}\right\|_{p_{1}}, \ldots ,\left\| f_{n}\right\|_{p_{n}} are positive.

Problem 8.1.4

We have equality in Hölder's Inequality iff there are nonnegative numbers AA and BB, not both zero, with A∣f∣p=B∣g∣p′μA\left|f\right|^{p}=B\left|g\right|^{p^{\prime }} \mu-a.e. on Ω.

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Problem 8.1.5

Given f∈Lp[1<p<+∞]f \in L^{p}\left[1<p<+\infty \right], there is g∈Lp′g \in L^{p^{\prime }} with ∥g∥p′=1\left\| g\right\|_{p^{\prime }}=1 and ∫fg=∥f∥p\int f g=\left\| f\right\|_{p}.

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Problem 8.1.6

We now explore conditions for equality in Minkowski's Inequality. Let f,g∈Lpf, g \in L^{p}.

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(a)

When p=1,∥f+g∥p=∥f∥p+∥g∥pp=1,\left\| f+g\right\|_{p}=\left\| f\right\|_{p}+\left\| g\right\|_{p} iff there exists positive F/B∗\mathcal{F} / \mathscr {B}^{*} measurable h>0h>0 defined on Ω\Omega with fh=gμf h=g \mu-a. e. on {ω∈Ω:f(ω)g(ω)≠0}\left\{ \omega \in \Omega : f(\omega ) g(\omega ) \neq 0\right\}.

(b)

For 1<p<+∞1<p<+\infty, equality obtains iff there are nonnegative real numbers AA and BB, not both zero, such that Af=BgμA f=B g \mu-a. e. on Ω\Omega.

Problem 8.1.7

Let 1⩽p,q,r<+∞1 \leqslant p, q, r<+\infty be such that 1/r=1/p+1/q1 / r=1 / p+1 / q. Let f∈Lpf \in L^{p} and g∈Lqg \in L^{q}. Then fg∈Lrf g \in L^{r} and ∥fg∥r⩽∥f∥p∥g∥q\left\| f g\right\|_{r} \leqslant \left\| f\right\|_{p}\left\| g\right\|_{q}.

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Problem 8.1.8

If 1⩽p<+∞1 \leqslant p<+\infty, Minkowski's Inequality gives ∣∥f∥p−∥g∥p∣⩽∥f−g∥p\left|\left\| f\right\|_{p}-\left\| g\right\|_{p}\right| \leqslant \left\| f-g\right\|_{p} for every f,g∈Lpf, g \in L^{p}.

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Problem 8.1.9

Let ϕ\phi be a continuous and strictly increasing function defined on [0,+∞)\left[0,+\infty \right) with ϕ(0)=0\phi (0)=0 and lim⁡x→+∞ϕ(x)=+∞\lim_{x \rightarrow +\infty } \phi (x)=+\infty. Then for each 0⩽a,b<+∞0 \leqslant a, b<+\infty we have

ab⩽∫0aϕ(u) du+∫0bϕ−1(v) dv, a b \leqslant \int _{0}^{a} \phi (u) \, d u+\int _{0}^{b} \phi ^{-1}(v) \, d v,

with equality iff ϕ(a)=b\phi (a)=b.

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Remark.
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Suggestion: first show that ∫0cϕ(u) du+∫0ϕ(c)ϕ−1(v) dv=cϕ(c)\int_{0}^{c} \phi (u) \, d u+\int_{0}^{\phi (c)} \phi^{-1}(v) \, d v=c \phi (c) for every 0⩽c<+∞0 \leqslant c<+\infty. A picture might be helpful. The calculus result in the section may be obtained by setting ϕ(u)=up−1\phi (u)=u^{p-1} and ϕ−1(v)=v1p−1\phi^{-1}(v)=v^{\frac{1}{p-1}}.

Problem 8.1.10

Prove the following extension of part (c) from Claim 2 of Section 7.5, where 1⩽p<+∞1 \leqslant p<+\infty and y∈Rk\boldsymbol {y} \in \mathbb {R}^{k}: If f∈Lp(Rk,Lk,λˉk)f \in L^{p}\left(\mathbb {R}^{k}, \mathcal{L}^{k}, \bar\lambda_{k}\right), then τyf∈Lp(Rk,Lk,λˉk)\tau_{\boldsymbol {y}} f \in L^{p}\left(\mathbb {R}^{k}, \mathcal{L}^{k}, \bar\lambda_{k}\right) and ∥τyf∥p=∥f∥p\left\| \tau_{\boldsymbol {y}} f\right\|_{p}=\left\| f\right\|_{p}.

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Remark.
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Claim 2 of Section 7.5: Let f:Rk→R‾f: \mathbb {R}^{k} \rightarrow \overline{\mathbb {R}} be Lebesgue measurable, and let τy\tau_{\boldsymbol {y}} denote translation by some y∈Rk\boldsymbol {y} \in \mathbb {R}^{k}: τyf(x)=f(x+y)\tau_{\boldsymbol {y}} f(\boldsymbol {x})=f(\boldsymbol {x}+\boldsymbol {y}) for every x∈Rk\boldsymbol {x} \in \mathbb {R}^{k}. Then: (a) τyf\tau_{\boldsymbol {y}} f is Lebesgue measurable; (b) if ff is nonnegative, then ∫f dλˉk=∫τyf dλˉk\int f \, d\bar\lambda_{k}=\int \tau_{\boldsymbol {y}} f \, d\bar\lambda_{k}; (c) if f∈L1(Rk,Lk,λˉk)f \in L^{1}\left(\mathbb {R}^{k}, \mathcal{L}^{k}, \bar\lambda_{k}\right), then τyf∈L1(Rk,Lk,λˉk)\tau_{\boldsymbol {y}} f \in L^{1}\left(\mathbb {R}^{k}, \mathcal{L}^{k}, \bar\lambda_{k}\right), and again ∫f dλˉk=∫τyf dλˉk\int f \, d\bar\lambda_{k}=\int \tau_{\boldsymbol {y}} f \, d\bar\lambda_{k}.

Section 8.2The Riesz-Fischer Theorem[5/100%]
Problem 8.2.1

Return to the formal definition of LpL^{p}.

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(a)

Write out the formal definition of convergence in LpL^{p}.

(b)

State and prove the formal version of the Riesz-Fischer Theorem.

Problem 8.2.2

Let (X,ρ)(X, \rho ) denote a generic metric space. Let Cb(X)\mathscr {C}_{b}(X) denote the collection of continuous real-valued bounded functions on XX. For f∈Cb(X)f \in \mathscr {C}_{b}(X), write ∥f∥=sup⁡x∈X∣f(x)∣\left\| f\right\| =\sup_{x \in X}\left|f(x)\right|, the usual supremum norm. Then Cb(X)\mathscr {C}_{b}(X) is a Banach space.

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Problem 8.2.3

A function ff on Rk\mathbb {R}^{k} is said to vanish at infinity iff f(x)→0f(\boldsymbol {x}) \rightarrow 0 as ∥x∥→∞\left\| \boldsymbol {x}\right\| \rightarrow \infty. Show that the collection of continuous functions on Rk\mathbb {R}^{k} that vanish at infinity is a Banach space relative to the supremum norm given in the previous exercise.

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Problem 8.2.4

Let Cc(Rk)\mathscr {C}_{c}\left(\mathbb {R}^{k}\right) denote the collection of continuous functions on Rk\mathbb {R}^{k} with compact support, and again consider the supremum norm. This collection is dense in the collection in the previous exercise, but it fails to be a Banach space.

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Problem 8.2.5

A sequence {fn}\left\{ f_{n}\right\} in LpL^{p} may converge in pp th mean to some f∈Lpf \in L^{p} but at the same time fail to converge pointwise to ff at any point in Ω\Omega. Therefore, convergence in LpL^{p} does not in general imply convergence μ−\mu - a. e.

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Section 8.3LpL^{p} Space: The Case 0<p<10<p<1[8/50%]
Problem 8.3.1

Let f,g∈Lpf, g \in L^{p}, where 0<p<10<p<1. We know that f+g∈Lpf+g \in L^{p} by the Minkowski-like Inequality result given earlier.

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(a)

We have (a+b)p⩽ap+bp(a+b)^{p} \leqslant a^{p}+b^{p} for every 0<a,b<∞0<a, b<\infty.

(b)

From (a), we have ∫∣f+g∣p⩽∫∣f∣p+∫∣g∣p\int \left|f+g\right|^{p} \leqslant \int \left|f\right|^{p}+\int \left|g\right|^{p}.

(c)

If we write ∥f−g∥pp\left\| f-g\right\|_{p}^{p} for the distance between ff and gg, then this distance function is truly a metric, if we identify functions equal μ−\mu - a. e. on Ω\Omega.

(d)

Writing ∥f−g∥p\left\| f-g\right\|_{p} for the distance between ff and gg does not define a metric on LpL^{p}.

Problem 8.3.2

Consider the space Ω=(0,1)\Omega =(0,1) and let 0<p<10<p<1. Write B for the Borel subsets of (0,1)(0,1), and write λ\lambda for Lebesgue measure restricted to B. We will show that there exists no norm ∥⋅∥\left\| \cdot \right\| on LpL^{p} such that lim⁡k∥fk∥p=0\lim_{k}\left\| f_{k}\right\|_{p}=0 forces lim⁡k∥fk∥=0\lim_{k}\left\| f_{k}\right\| =0.

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(a)

Suppose that such a norm ∥⋅∥\left\| \cdot \right\| exists. Then there is C⩾0C \geqslant 0 such that ∥f∥⩽C∥f∥p\left\| f \right\| \leqslant C\left\| f\right\|_{p} for each f∈Lpf \in L^{p}.

Pick the minimal such CC from (a).

(b)

There is 0<c<10<c<1 with ∫0c∣f∣p=∫c1∣f∣p=12∫01∣f∣p\int_{0}^{c}\left|f\right|^{p}=\int_{c}^{1}\left|f\right|^{p}=\frac{1}{2} \int_{0}^{1}\left|f\right|^{p}.

(c)

Let g=f  1(0,c]g=f \; \mathbb {1}_{(0, c]} and h=f  1(c,1]h=f \; \mathbb {1}_{(c, 1]}, so that f=g+hf=g+h. Then ∥g∥p=∥h∥p=2−1/p∥f∥p\left\| g\right\|_{p}=\left\| h\right\|_{p}= 2^{-1 / p}\left\| f\right\|_{p} and ∥f∥⩽∥g∥+∥h∥⩽C∥g∥p+C∥h∥p=C×21−1/p∥f∥p\left\| f\right\| \leqslant \left\| g\right\| +\left\| h\right\| \leqslant C\left\| g\right\|_{p}+C\left\| h\right\|_{p}=C \times 2^{1-1 / p}\left\| f\right\|_{p}. Use the minimality of CC to obtain C⩽C21−1/pC \leqslant C 2^{1-1 / p}, and deduce C=0C=0.

(d)

Conclude that ∥f∥=0\left\| f\right\| =0 for every f∈Lpf \in L^{p}, and hence the assumption in (a) entails a contradiction.

Problem 8.3.3

Let 0<p0<∞0<p_{0}<\infty and let f∈Lp0f \in L^{p_{0}} be nonnegative. Let E1={ω∈Ω:f(ω)=0},E2={ω∈Ω:0<f(ω)⩽1}E_{1}=\left\{ \omega \in \Omega : f(\omega )=0\right\} , E_{2}=\left\{ \omega \in \Omega : 0<f(\omega ) \leqslant 1\right\}, and E3={ω∈Ω:f(ω)>1}E_{3}=\left\{ \omega \in \Omega : f(\omega )>1\right\}.

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(a)

lim⁡p→0+∫E2∣f∣p=μ(E2)\lim_{p \rightarrow 0^{+}} \int_{E_{2}}\left|f\right|^{p}=\mu \left(E_{2}\right).

(b)

lim⁡p→0+∫E3∣f∣p=μ(E3)\lim_{p \rightarrow 0^{+}} \int_{E_{3}}\left|f\right|^{p}=\mu \left(E_{3}\right).

(c)

lim⁡p→0+∫∣f∣p=μ({ω∈Ω:f(ω)≠0})\lim_{p \rightarrow 0^{+}} \int \left|f\right|^{p}=\mu (\left\{ \omega \in \Omega : f(\omega ) \neq 0\right\} ).

Problem 8.3.4

Say μ(Ω)=1\mu (\Omega )=1, and let f∈L1f \in L^{1} be nonnegative. Write log⁡0=−∞\log 0=-\infty.

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(a)

∫log⁡f⩽log⁡∫f\int \log f \leqslant \log \int f if log⁡f∈L1\log f \in L^{1}.

(b)

If log⁡f∉L1\log f \notin L^{1}, then ∫(log⁡f)+<∞\int (\log f)^{+}<\infty under the assumption f∈L1f \in L^{1}, so it must be the case that ∫(log⁡f)−=∞\int (\log f)^{-}=\infty. Conclude that even if log⁡f∉L1,∫log⁡f\log f \notin L^{1}, \int \log f still exists and equals −∞-\infty, giving the inequality in (a).

(c)

(fr−1)/r\left(f^{r}-1\right) / r decreases to log⁡f\log f as r→0+r \rightarrow 0^{+}, hence lim⁡r→0+(∫fr−1)/r=∫log⁡f\lim_{r \rightarrow 0^{+}}\left(\int f^{r}-1\right) / r=\int \log f.

(d)

Verify the inequalities

1r[∫fr−1]⩾1rlog⁡∫fr⩾1r∫log⁡fr=∫log⁡f \frac{1}{r}\left[\int f^{r}-1\right] \geqslant \frac{1}{r} \log \int f^{r} \geqslant \frac{1}{r} \int \log f^{r}=\int \log f
(e)

Conclude that lim⁡r→0+∥f∥r\lim_{r \rightarrow 0^{+}}\left\| f\right\|_{r} exists and equals exp⁡(∫log⁡f)\exp \left(\int \log f\right). If log⁡f∉L1\log f \notin L^{1}, this is interpreted as lim⁡r→0+∥f∥r=0\lim_{r \rightarrow 0^{+}}\left\| f\right\|_{r}=0.

Problem 8.3.5

Find conditions for equality in the Hölder-like Inequality.

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Remark.
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Hölder-like Inequality: Let 0<p<10<p<1. Let f∈Lpf \in L^{p} be nonnegative, and let gg denote a nonnegative F/B∗\mathcal{F} / \mathscr {B}^{*}-measurable function such that ∫gp′\int g^{p^{\prime }} exists and 0<∫gp′<+∞0<\int g^{p^{\prime }}<+\infty. Then ∫fg⩾∥f∥p(∫∣g∣p′)1/p′\int f g \geqslant \left\| f\right\|_{p}\left(\int \left|g\right|^{p^{\prime }}\right)^{1 / p^{\prime }}.

Problem 8.3.6

Find conditions for equality in the Minkowski-like Inequality.

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Remark.
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Minkowski-like Inequality: Let 0<p<10<p<1, and let f,g∈Lpf, g \in L^{p} be nonnegative. Then f+g∈Lpf+g \in L^{p} and ∥f+g∥p⩾∥f∥p+∥g∥p\left\| f+g\right\|_{p} \geqslant \left\| f\right\|_{p}+\left\| g\right\|_{p}.

Problem 8.3.7

For every a,b>0a, b>0 and 0<p<10<p<1, we have ab⩾app+bp′p′a b \geqslant \frac{a^{p}}{p}+\frac{b^{p^{\prime }}}{p^{\prime }}. Use this to give a calculus-based proof of the Hölder-like Inequality that is similar to the proof of the original Hölder Inequality.

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Remark.
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Let f(a,b)=ab−appf(a, b)=a b-\frac{a^{p}}{p} for a,b>0a, b>0, where 0<p<10<p<1 is fixed. For each b⩾0b \geqslant 0, we have that ∂∂af(a,b)=b−ap−1⩾0\frac{\partial }{\partial a} f(a, b)=b-a^{p-1} \geqslant 0 iff a⩾b1p−1a \geqslant b^{\frac{1}{p-1}} [here we use the fact that p−1<0p-1<0]. Therefore, we have that f(a,b)⩾f(b1p−1,b)f(a, b) \geqslant f\left(b^{\frac{1}{p-1}}, b\right) for every b>0b>0, which gives ab−app⩾p−1pbpp−1=1p′bp′a b-\frac{a^{p}}{p} \geqslant \frac{p-1}{p} b^{\frac{p}{p-1}}=\frac{1}{p^{\prime }} b^{p^{\prime }} for every b>0b>0.

Problem 8.3.8

Does the Riesz-Fischer Theorem necessarily hold in the case 0<p<10<p<1?

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Remark.
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Hint: note the relevance of the original Minkowski inequality in the proof of the Riesz-Fischer Theorem. When 0<p<10<p<1, the direction of the inequality is reversed.

Section 8.4LpL^{p} Space: The Case p=+∞p=+\infty[7/43%]
Problem 8.4.1

In this exercise, we deal with the formal definition of L∞L^{\infty }. We define L∞L^{\infty } to consist of those equivalence classes A∈E/∼\mathfrak {A} \in \mathcal{E} / \sim such that ess⁡sup⁡f<+∞\operatorname {ess} \sup f<+\infty for some f∈Af \in \mathfrak {A}.

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(a)

If f=g μf=g \: \mu-a.e. on Ω\Omega, then ess⁡sup⁡f=ess⁡sup⁡g\operatorname {ess} \sup f=\operatorname {ess} \sup g. Therefore, we may equivalently say that A∈L∞\mathfrak {A} \in L^{\infty } iff ess⁡sup⁡f<+∞\operatorname {ess} \sup f<+\infty for every f∈Af \in \mathfrak {A}.

(b)

If A,B∈L∞\mathfrak {A}, \mathfrak {B} \in L^{\infty } and c∈Rc \in \mathbb {R}, then cA∈L∞c \mathfrak {A} \in L^{\infty } and A+B∈L∞\mathfrak {A}+\mathfrak {B} \in L^{\infty }. Therefore, L∞L^{\infty } is a vector space.

We define ∥⋅∥∞\left\| \cdot \right\|_{\infty } on L∞L^{\infty } by writing ∥A∥∞=ess⁡sup⁡f\left\| \mathfrak {A}\right\|_{\infty }=\operatorname {ess} \sup f, where f∈Af \in \mathfrak {A}.

(c)

The definition of ∥⋅∥∞\left\| \cdot \right\|_{\infty } is well defined.

(d)

For A,B,C∈L∞\mathfrak {A}, \mathfrak {B}, \mathfrak {C} \in L^{\infty }, define ρ(A,B)=∥A−B∥∞\rho (\mathfrak {A}, \mathfrak {B})=\left\| \mathfrak {A}-\mathfrak {B}\right\|_{\infty }. Then

  1. 0⩽ρ(A,B)<+∞0 \leqslant \rho (\mathfrak {A}, \mathfrak {B})<+\infty.

  2. ρ(A,A)=0\rho (\mathfrak {A}, \mathfrak {A})=0, and ρ(A,B)=0\rho (\mathfrak {A}, \mathfrak {B})=0 iff A=B\mathfrak {A}=\mathfrak {B}.

  3. ρ(A,B)=ρ(B,A)\rho (\mathfrak {A}, \mathfrak {B})=\rho (\mathfrak {B}, \mathfrak {A}).

  4. ρ(A,C)⩽ρ(A,B)+ρ(B,C)\rho (\mathfrak {A}, \mathfrak {C}) \leqslant \rho (\mathfrak {A}, \mathfrak {B})+\rho (\mathfrak {B}, \mathfrak {C}). Therefore, ρ\rho is a metric on L∞L^{\infty }.

(e)

L∞L^{\infty } is complete in the metric ρ\rho. Therefore L∞L^{\infty } is a Banach space.

(f)
Remark.
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All parts of this exercise can be handled quickly by using the results of this section.

Problem 8.4.2

Consider the σ\sigma-finite measure space (Ω,F,μ)(\Omega , \mathcal{F}, \mu ).

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(a)

f∈L∞f \in L^{\infty } iff there is a bounded F/B∗\mathcal{F} / \mathfrak {B}^{*}-measurable function gg on Ω\Omega such that f=gμ−f=g \mu - a. e. on Ω\Omega.

(b)

If f∈L∞f \in L^{\infty }, then ∥f∥∞=inf⁡{sup⁡ω∈Ω∣g(ω)∣:g is as in (a)}\left\| f\right\|_{\infty }=\inf \left\{ \sup_{\omega \in \Omega }\left|g(\omega )\right|: g \text{ is as in (a)}\right\}.

Problem 8.4.3

Consider the measure space (Rk,Bk,λk)\left(\mathbb {R}^{k}, \mathcal{B}^{k}, \lambda_{k}\right). We have ess⁡sup⁡f=sup⁡x∈Rk∣f(x)∣\operatorname {ess} \sup f=\sup_{\boldsymbol {x} \in \mathbb {R}^{k}}\left|f(\boldsymbol {x})\right| if ff is continuous.

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Problem 8.4.4

Quickly prove the p=∞p=\infty version of Hölder's Inequality.

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Problem 8.4.5

Let ( Ω,F,μ\Omega , \mathcal{F}, \mu ) be such that μ(Ω)<∞\mu (\Omega )<\infty, and let ff denote a bounded F/B∗\mathcal{F} / \mathcal{B}^{*}-measurable function on Ω\Omega.

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(a)

For every 1⩽p⩽∞1 \leqslant p \leqslant \infty we have f∈Lpf \in L^{p}, hence ∥f∥p\left\| f\right\|_{p} exists.

(b)

lim⁡p→∞∥f∥p=∥f∥∞\lim_{p \rightarrow \infty }\left\| f\right\|_{p}=\left\| f\right\|_{\infty }.

Problem 8.4.6

For each k∈Nk \in \mathbb {N}, L∞(Rk,Bk,λk)L^{\infty }\left(\mathbb {R}^{k}, \mathcal{B}^{k}, \lambda_{k}\right) fails to contain an unbounded continuous function. Also, for each 1⩽p<+∞1 \leqslant p<+\infty, Lp(R,B,λ)L^{p}(\mathbb {R}, \mathcal{B}, \lambda ) contains an unbounded continuous function.

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Remark.
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See Exercise 8.4.3.

Problem 8.4.7

Let Ω=(0,1]\Omega =(0,1], let F\mathcal{F} denote the Borel subsets of (0,1](0,1], and let μ\mu denote Lebesgue measure restricted to the Borel subsets of (0,1](0,1]. Show in this case that L∞L^{\infty } is not separable, that is, there does not exist a countable dense subset of L∞L^{\infty }.

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Section 8.5Containment Relations for Lp\boldsymbol{L}^{\boldsymbol{p}} Spaces[8/63%]
Problem 8.5.1

Consider the measure space (R,B,λ)(\mathbb {R}, \mathfrak {B}, \lambda ). Let 1⩽p<q<∞1 \leqslant p<q< \infty, and let rr be such that 1/q<r<1/p1 / q<r<1 / p.

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(a)

Define ff on R\mathbb {R} by writing f(x)=x−r  1(0,1)(x)f(x)=x^{-r} \; \mathbb {1}_{(0,1)}(x) for each x∈Rx \in \mathbb {R}. Then f∈Lpf \in L^{p} but f∉Lqf \notin L^{q}. Therefore, we do not in general have Lp⊆LqL^{p} \subseteq L^{q} when p<qp<q.

(b)

Let g(x)=x−r  1(1,∞)(x)g(x)=x^{-r} \; \mathbb {1}_{(1, \infty )}(x) for every x∈Rx \in \mathbb {R}. Then g∈Lqg \in L^{q} but g∉Lpg \notin L^{p}. Therefore, we do not in general have Lq⊆LpL^{q} \subseteq L^{p} when p<qp<q.

Problem 8.5.2

Let 0<p<r<∞0<p<r<\infty. Then Lp∩L∞⊆LrL^{p} \cap L^{\infty } \subseteq L^{r}, and for any f∈Lp∩L∞f \in L^{p} \cap L^{\infty } we have ∥f∥r⩽∥f∥pp/r∥f∥∞1−p/r\left\| f\right\|_{r} \leqslant \left\| f\right\|_{p}^{p / r}\left\| f\right\|_{\infty }^{1-p / r}.

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Problem 8.5.3

For any 0<p<∞0<p<\infty and 0<M<∞0<M<\infty we have

(∫∣f∣p)1/p⩾Mμ({ω∈Ω:∣f(ω)∣⩾M})1/p. \left(\int \left|f\right|^{p}\right)^{1 / p} \geqslant M \mu (\left\{ \omega \in \Omega :\left|f(\omega )\right| \geqslant M\right\} )^{1 / p} .
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Problem 8.5.4

Let 0<r<∞0<r<\infty and assume that f∈Lr∩L∞f \in L^{r} \cap L^{\infty }, so that f∈Lpf \in L^{p} for every r<p<∞r<p<\infty by Exercise 8.5.2. We wish to show that lim⁡p→∞∥f∥p=∥f∥∞\lim_{p \rightarrow \infty }\left\| f\right\|_{p}=\left\| f\right\|_{\infty }. Follow this outline:

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(a)

Ignoring the trivial case where f=0μf=0 \mu-a.e. on Ω\Omega, let f≠0f \neq 0 on a set of positive measure, so that ∥f∥∞>0\left\| f\right\|_{\infty }>0. Show that lim inf⁡p→∞∥f∥p⩾∥f∥∞\liminf_{p \rightarrow \infty }\left\| f\right\|_{p} \geqslant \left\| f\right\|_{\infty }.

(b)

Show that lim sup⁡p→∞∥f∥p⩽∥f∥∞\limsup_{p \rightarrow \infty }\left\| f\right\|_{p} \leqslant \left\| f\right\|_{\infty }.

(c)

Put (a) and (b) together to prove the claim.

Problem 8.5.5

Let μ(Ω)=1\mu (\Omega )=1 and 1⩽p⩽q⩽∞1 \leqslant p \leqslant q \leqslant \infty. Show for arbitrary ff that

∫∣f∣⩽(∫∣f∣p)1/p=(∫∣f∣q)1/q⩽ess⁡sup⁡f \int \left|f\right| \leqslant \left(\int \left|f\right|^{p}\right)^{1 / p}=\left(\int \left|f\right|^{q}\right)^{1 / q} \leqslant \operatorname {ess} \sup f

so that ∥f∥1⩽∥f∥p⩽∥f∥q⩽∥f∥∞\left\| f\right\|_{1} \leqslant \left\| f\right\|_{p} \leqslant \left\| f\right\|_{q} \leqslant \left\| f\right\|_{\infty }.

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Problem 8.5.6

We know that lp⊆lq⊆l∞l^{p} \subseteq l^{q} \subseteq l^{\infty } when 0<p<q<+∞0<p<q<+\infty by Claim 5. Show further that

ess⁡sup⁡f⩽(∑ω∈Ω∣f(ω)∣q)1/q⩽(∑ω∈Ω∣f(ω)∣p)1/p \operatorname {ess} \sup f \leqslant \left(\sum _{\omega \in \Omega }\left|f(\omega )\right|^{q}\right)^{1 / q} \leqslant \left(\sum _{\omega \in \Omega }\left|f(\omega )\right|^{p}\right)^{1 / p}

for arbitrary f∈lpf \in l^{p}, so that ∥f∥∞⩽∥f∥q⩽∥f∥p\left\| f\right\|_{\infty } \leqslant \left\| f\right\|_{q} \leqslant \left\| f\right\|_{p} for every f∈lpf \in l^{p}.

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Remark.
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Claim 5: For every 0<p<q<+∞0<p<q<+\infty we have lp⊆lq⊆l∞l^{p} \subseteq l^{q} \subseteq l^{\infty }.

Problem 8.5.7

Consider (R,B,λ)\left(\mathbb {R}, \mathfrak {B}, \lambda \right).

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(a)

Find ff such that f∈Lpf \in L^{p} but f∉Lqf \notin L^{q} for every p<q<+∞p<q<+\infty.

(b)

Find ff such that f∈Lqf \in L^{q} for every 0<q<p0<q<p, but f∉Lpf \notin L^{p}.

(c)

Find a nonnegative finite-valued function ff with f∉Lpf \notin L^{p} for every 0<p<+∞0<p<+\infty.

(d)

Find ff such that f∈Lpf \in L^{p} but f∉Lqf \notin L^{q} for every q∈(0,∞)−{p}q \in (0, \infty )-\left\{ p\right\}.

Problem 8.5.8

Let (Ω,F,μ)(\Omega , \mathcal{F}, \mu ) be such that μ(A)>0\mu (A)>0 forces μ(A)⩾1\mu (A) \geqslant 1 for each A∈FA \in \mathcal{F}. Let 1⩽p⩽q<+∞1 \leqslant p \leqslant q<+\infty. Then Lp⊆Lq⊆L∞L^{p} \subseteq L^{q} \subseteq L^{\infty } and ∥f∥∞⩽∥f∥q⩽∥f∥p\left\| f\right\|_{\infty } \leqslant \left\| f\right\|_{q} \leqslant \left\| f\right\|_{p}.

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Section 8.6Approximation[10/0%]
Problem 8.6.1

The following is an outline of an alternative proof that Cc\mathfrak {C}_{c} is a dense subset of Lp(Rk,Bk,λk)L^{p}\left(\mathbb {R}^{k}, \mathscr {B}^{k}, \lambda_{k}\right) (where 1⩽p<+∞1 \leqslant p<+\infty) that doesn't use Lusin's Theorem [as was used in Claim 3]. Here Cc\mathfrak {C}_{c} denotes the collection of real-valued functions on Rk\mathbb {R}^{k} that are continuous and have compact support, and S0\mathfrak {S}_{0} denotes the collection of F/B∗\mathcal{F}/\mathfrak {B}^{*}-measurable finite-valued simple functions ss on Ω\Omega such that μ({ω∈Ω:s(ω)≠0})<∞\mu \left(\left\{ \omega \in \Omega : s(\omega ) \neq 0\right\} \right)<\infty.

Claim 3: Given any bounded f∈Lp(Rk,Bk,λk)f \in L^{p}\left(\mathbb {R}^{k}, \mathscr {B}^{k}, \lambda_{k}\right) and ε>0\varepsilon >0, we can find g∈Ccg \in \mathfrak {C}_{c} such that ∥f−g∥p<ε\left\| f-g\right\|_{p}<\varepsilon and sup⁡ω∈Rk∣g(ω)∣⩽sup⁡ω∈Rk∣f(ω)∣\sup_{\omega \in \mathbb {R}^{k}}\left|g(\omega )\right| \leqslant \sup_{\omega \in \mathbb {R}^{k}}\left|f(\omega )\right|. In particular, Cc\mathfrak {C}_{c} is a dense subset of Lp(Rk,Bk,λk)L^{p}\left(\mathbb {R}^{k}, \mathscr {B}^{k}, \lambda_{k}\right).

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(a)

Show that Cc\mathfrak {C}_{c} is dense in {  1A:A∈Bk is bounded}\left\{ \; \mathbb {1}_{A}: A \in \mathscr {B}^{k} \text{ is bounded}\right\}. [Let A∈BkA \in \mathscr {B}^{k} be bounded. Find compact KK and open GG with K⊆A⊆GK \subseteq A \subseteq G and λk(G−K)<εp\lambda_{k}(G-K)<\varepsilon^{p}. Use Urysohn's Lemma to find a g:Rk→Rg: \mathbb {R}^{k} \rightarrow \mathbb {R} that is continuous with 0⩽g⩽10 \leqslant g \leqslant 1, g=1g=1 on KK, and g=0g=0 on GcG^{c}. We have g∈Ccg \in \mathfrak {C}_{c} and g(x)−  1A(x)=0g(\boldsymbol {x})-\; \mathbb {1}_{A}(\boldsymbol {x})=0 for every x∈K∪Gc\boldsymbol {x} \in K \cup G^{c}. Then

∥g−  1A∥p=(∫∣g−  1A∣p)1/p=(∫G−K∣g−  1A∣p)1/p⩽(∫G−K1)1/p<ε. \left\| g-\; \mathbb {1}_{A}\right\| _{p}=\left(\int \left|g-\; \mathbb {1}_{A}\right|^{p}\right)^{1 / p}=\left(\int _{G-K}\left|g-\; \mathbb {1}_{A}\right|^{p}\right)^{1 / p} \leqslant \left(\int _{G-K} 1\right)^{1 / p}<\varepsilon .

]

(b)

Show that Cc\mathfrak {C}_{c} is dense in the collection of nonnegative finite-valued F/B∗\mathcal{F}/\mathfrak {B}^{*}-measurable simple functions ∑i=1nci  1Ai\sum_{i=1}^{n} c_{i} \; \mathbb {1}_{A_{i}} on Ω\Omega such that each AiA_{i} is bounded if ci>0c_{i}>0. [Without loss of generality, assume that each ci>0c_{i}>0, so that A1,…,AnA_{1}, \ldots , A_{n} are bounded. For each i∈{1,…,n}i \in \left\{ 1, \ldots , n\right\}, find gi∈Ccg_{i} \in \mathfrak {C}_{c} such that ∥gi−  1Ai∥p<ε1+∑i=1n∣ci∣\left\| g_{i}-\; \mathbb {1}_{A_{i}}\right\|_{p}<\frac{\varepsilon }{1+\sum_{i=1}^{n}\left|c_{i}\right|}; this requires the result of (a). Let g=∑i=1ncigig=\sum_{i=1}^{n} c_{i} g_{i}. Then g∈Ccg \in \mathfrak {C}_{c} and it is easy to show that ∥g−∑i=1nci  1Ai∥p<ε\left\| g-\sum_{i=1}^{n} c_{i} \; \mathbb {1}_{A_{i}}\right\|_{p}<\varepsilon.]

(c)

Show that Cc\mathfrak {C}_{c} is dense in the class of nonnegative functions in LpL^{p} that are zero outside a bounded set. [Show that the functions in (b) are dense in the collection here in (c) as follows: let f⩾0f \geqslant 0 be in LpL^{p} with f=0f=0 outside some bounded A∈BkA \in \mathscr {B}^{k}. Find a nondecreasing sequence {sn}n=1∞\left\{ s_{n}\right\}_{n=1}^{\infty } of functions in (b) with sn→fs_{n} \rightarrow f. Use DCT to show that lim⁡n∥sn−f∥p=0\lim_{n}\left\| s_{n}-f\right\|_{p}=0 by using the fact that ∣f−sn∣p⩽fp∈L1\left|f-s_{n}\right|^{p} \leqslant f^{p} \in L^{1}. Since Cc\mathfrak {C}_{c} is dense in the collection of functions in (b), and since the collection of functions in (b) has been shown by this outline here in (c) to be dense in the collection of functions given here in (c), the claim of (c) follows.]

(d)

Show that Cc\mathfrak {C}_{c} is dense in the class of functions in LpL^{p} that are zero outside a bounded set. [Let f∈Lpf \in L^{p} be zero outside a bounded set, and pick ε>0\varepsilon >0. Write f=f+−f−f=f^{+}-f^{-}. Both f+f^{+}and f−f^{-}are nonnegative functions in LpL^{p} that are zero outside a bounded set. By (c), there are functions g1,g2∈Ccg_{1}, g_{2} \in \mathfrak {C}_{c} with ∥f+−g1∥p\left\| f^{+}-g_{1}\right\|_{p} and ∥f−−g2∥p\left\| f^{-}-g_{2}\right\|_{p} both less than ε2\frac{\varepsilon }{2}. Let g=g1−g2g=g_{1}-g_{2}. Then g∈Ccg \in \mathfrak {C}_{c} and ∥f−g∥p<ε\left\| f-g\right\|_{p}<\varepsilon.]

(e)

Show that Cc\mathfrak {C}_{c} is dense in LpL^{p}. [Let f∈Lpf \in L^{p}. Since ∣f  1B(0,n)−f∣p⩽∣f∣p∈L1\left|f \; \mathbb {1}_{B(\mathbf{0}, n)}-f\right|^{p} \leqslant \left|f\right|^{p} \in L^{1}, DCT gives lim⁡n∥f  1B(0,n)−f∥p=0\lim_{n}\left\| f \; \mathbb {1}_{B(\mathbf{0}, n)}-f\right\|_{p}=0. Thus ff can be approximated by a function in LpL^{p} that vanishes outside a bounded set. Since (d) says that Cc\mathfrak {C}_{c} is dense in the class of functions in LpL^{p} that vanish outside a bounded set, it follows that ff can be approximated by a function in Cc\mathfrak {C}_{c}.]

Problem 8.6.2

Claim 2: S0\mathfrak {S}_{0} is a dense subset of LpL^{p}, where S0\mathfrak {S}_{0} denotes the collection of F/B∗\mathcal{F}/\mathfrak {B}^{*}-measurable finite-valued simple functions ss on Ω\Omega such that μ({ω∈Ω:s(ω)≠0})<∞\mu \left(\left\{ \omega \in \Omega : s(\omega ) \neq 0\right\} \right)<\infty.

We have shown for a general (Ω,F,μ)(\Omega , \mathcal{F}, \mu ) that S0\mathfrak {S}_{0} is a dense subset of LpL^{p} for 1⩽p<+∞1 \leqslant p<+\infty (Claim 2 above).

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(a)

It is also the case that S0\mathfrak {S}_{0} is a dense subset of L∞L^{\infty }. Therefore, S0\mathfrak {S}_{0} is a dense subset of LpL^{p} for 1⩽p⩽+∞1 \leqslant p \leqslant +\infty. [This should pose no problem.]

(b)

What about the case where 0<p<10<p<1? [Do Claims 1 and 2 in any way depend on whether or not p⩾1p \geqslant 1?]

Problem 8.6.3

Claim 3: Given any bounded f∈Lp(Rk,Bk,λk)f \in L^{p}\left(\mathbb {R}^{k}, \mathscr {B}^{k}, \lambda_{k}\right) and ε>0\varepsilon >0, we can find g∈Ccg \in \mathfrak {C}_{c} such that ∥f−g∥p<ε\left\| f-g\right\|_{p}<\varepsilon and sup⁡ω∈Rk∣g(ω)∣⩽sup⁡ω∈Rk∣f(ω)∣\sup_{\omega \in \mathbb {R}^{k}}\left|g(\omega )\right| \leqslant \sup_{\omega \in \mathbb {R}^{k}}\left|f(\omega )\right|. In particular, Cc\mathfrak {C}_{c} is a dense subset of Lp(Rk,Bk,λk)L^{p}\left(\mathbb {R}^{k}, \mathscr {B}^{k}, \lambda_{k}\right). Here Cc\mathfrak {C}_{c} denotes the collection of real-valued functions on Rk\mathbb {R}^{k} that are continuous and have compact support.

Claim 4: S00(R,B,λ)\mathfrak {S}_{00}\left(\mathbb {R}, \mathscr {B}, \lambda \right) is a dense subset of Lp(R,B,λ)L^{p}\left(\mathbb {R}, \mathscr {B}, \lambda \right), where S00(R,B,λ)\mathfrak {S}_{00}\left(\mathbb {R}, \mathscr {B}, \lambda \right) denotes the collection of functions on R\mathbb {R} of the form ∑i=1nci  1Ai\sum_{i=1}^{n} c_{i} \; \mathbb {1}_{A_{i}}, where c1,…,cn∈Rc_{1}, \ldots , c_{n} \in \mathbb {R} and each AiA_{i} is a bounded interval of any form.

Claim 5: Let f∈Lp(Rk,Bk,λk)f \in L^{p}\left(\mathbb {R}^{k}, \mathscr {B}^{k}, \lambda_{k}\right), where 1⩽p<+∞1 \leqslant p<+\infty. Then for every y∈Rk\boldsymbol {y} \in \mathbb {R}^{k} it follows that τyf∈Lp(Rk,Bk,λk)\tau_{\boldsymbol {y}} f \in L^{p}\left(\mathbb {R}^{k}, \mathscr {B}^{k}, \lambda_{k}\right), where τyf(x)=f(x+y)\tau_{\boldsymbol {y}} f(\boldsymbol {x})=f(\boldsymbol {x}+\boldsymbol {y}) for every x∈Rk\boldsymbol {x} \in \mathbb {R}^{k}. Moreover, lim⁡y→0∥τyf−f∥p=0\lim_{\boldsymbol {y} \rightarrow \mathbf{0}}\left\| \tau_{\boldsymbol {y}} f-f\right\|_{p}=0.

Each of Claims 3, 4, and 5 is stated (and was proved) for 1⩽p<+∞1 \leqslant p<+\infty.

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(a)

Analyze the cases 0<p<10<p<1 and p=+∞p=+\infty for Claim 3.

(b)

Analyze the cases 0<p<10<p<1 and p=+∞p=+\infty for Claim 4.

(c)

Analyze the cases 0<p<10<p<1 and p=+∞p=+\infty for Claim 5.

Problem 8.6.4

Let (Ω,F,μ)(\Omega , \mathcal{F}, \mu ) be such that μ\mu is σ\sigma-finite and F\mathcal{F} is countably generated. We will show that LpL^{p} is separable for 1⩽p<+∞1 \leqslant p<+\infty.

Claim 1: S0⊆Lp\mathfrak {S}_{0} \subseteq L^{p}, where S0\mathfrak {S}_{0} denotes the collection of F/B∗\mathcal{F}/\mathfrak {B}^{*}-measurable finite-valued simple functions ss on Ω\Omega such that μ({ω∈Ω:s(ω)≠0})<∞\mu \left(\left\{ \omega \in \Omega : s(\omega ) \neq 0\right\} \right)<\infty.

Claim 2: S0\mathfrak {S}_{0} is a dense subset of LpL^{p}; moreover, given ε>0\varepsilon >0 and f∈Lpf \in L^{p}, the approximating s∈S0s \in \mathfrak {S}_{0} with ∥s−f∥p<ε\left\| s-f\right\|_{p}<\varepsilon may be chosen with ∣s∣⩽∣f∣\left|s\right| \leqslant \left|f\right|.

Assume that F\mathcal{F} is generated by the countable collection C\mathcal{C} and Ω\Omega is covered by a countable collection D\mathcal{D} of F\mathcal{F}-sets of finite measure. Enumerate the sets in the countable collection C∪D\mathcal{C} \cup \mathcal{D} as E1,E2,…E_{1}, E_{2}, \ldots. For each n∈Nn \in \mathbb {N}, write

Pn={F1∩⋯∩Fn:Fi=Ei or Fi=Eic,i=1,…,n}, \mathcal{P}_{n}=\left\{ F_{1} \cap \cdots \cap F_{n}: F_{i}=E_{i} \text{ or } F_{i}=E_{i}^{c}, i=1, \ldots , n\right\} ,

and let Fn\mathcal{F}_{n} denote the collection of finite unions of Pn\mathcal{P}_{n}-sets. Write F0=⋃n=1∞Fn\mathcal{F}_{0}=\bigcup_{n=1}^{\infty } \mathcal{F}_{n}.

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(a)

Each Fn\mathcal{F}_{n} is a field on Ω\Omega.

(b)

F0\mathcal{F}_{0} is a countable field that generates F\mathcal{F}: σ(F0)=F\sigma \left(\mathcal{F}_{0}\right)=\mathcal{F}.

(c)

μ\mu is σ\sigma-finite on F0\mathcal{F}_{0}.

Let VV denote the collection of functions ∑i=1mαi  1Ai\sum_{i=1}^{m} \alpha_{i} \; \mathbb {1}_{A_{i}} defined on Ω\Omega [the function equals 00 on (A1∪⋯∪An)c\left(A_{1} \cup \cdots \cup A_{n}\right)^{c}], where the αi\alpha_{i}'s are rational and the AiA_{i}'s are F0\mathcal{F}_{0}-sets of finite measure.

(d) VV is a countable collection.

(e) Let f∈Lpf \in L^{p} and ε>0\varepsilon >0. There is g=∑i=1mαi  1Big=\sum_{i=1}^{m} \alpha_{i} \; \mathbb {1}_{B_{i}} with the αi\alpha_{i}’s and μ(Bi)\mu \left(B_{i}\right)’s finite and such that ∥f−g∥p<ε\left\| f-g\right\|_{p}<\varepsilon; we may assume without loss of generality that each αi\alpha_{i} is nonzero and rational. [Use Claim 2 above.]

(f) For i=1,…,mi=1, \ldots , m, there is Ai∈F0A_{i} \in \mathcal{F}_{0} with μ(Ai△Bi)<(εm∣αi∣)p\mu \left(A_{i} \triangle B_{i}\right)<\left(\frac{\varepsilon }{m\left|\alpha_{i}\right|}\right)^{p}. [Use an approximation result from Chapter 3.]

(g) Let h=∑i=1mαi  1Aih=\sum_{i=1}^{m} \alpha_{i} \; \mathbb {1}_{A_{i}}. Then h∈Vh \in V and ∥g−h∥p<ε\left\| g-h\right\|_{p}<\varepsilon, hence ∥f−h∥p<2ε\left\| f-h\right\|_{p}<2\varepsilon. Conclude that LpL^{p} is indeed separable.

Remark.
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The remaining exercises in this section are connected. Some parts are straightforward, while for other parts ample hints are provided. We consider the measure space (Rk,Bk,λk)\left(\mathbb {R}^{k}, \mathscr {B}^{k}, \lambda_{k}\right). We define C∞\mathfrak {C}^{\infty } as the collection of real-valued functions on Rk\mathbb {R}^{k} having continuous partial derivatives of all orders; Cc∞\mathfrak {C}_{c}^{\infty } will denote those elements of C∞\mathfrak {C}^{\infty } with compact support.

Problem 8.6.5

For every t∈Rt \in \mathbb {R} write h(t)=e−1t  1(0,+∞)(t)h(t)=e^{-\frac{1}{t}} \; \mathbb {1}_{(0,+\infty )}(t). The goal here is to show that hh has continuous derivatives of all orders.

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(a)

Verify for each n∈Nn \in \mathbb {N} that h(n)(t)=Pn(1t)e−1th^{(n)}(t)=P_{n}\left(\frac{1}{t}\right) e^{-\frac{1}{t}} for t>0t>0, where PnP_{n} is a polynomial of degree 2n2n with no degree zero terms. [Straightforward; use induction.]

(b)

We know that h(n)(t)=0h^{(n)}(t)=0 for every n∈Nn \in \mathbb {N} and t<0t<0, so all that remains is to discuss the case t=0t=0. Show that h(n)(0)=0h^{(n)}(0)=0 for every n∈Nn \in \mathbb {N}. [Use the definition of the derivative. For n=1n=1, we have lim⁡δ→0−δ−1[h(δ)−h(0)]=0\lim_{\delta \rightarrow 0^{-}} \delta^{-1}\left[h(\delta )-h(0)\right]=0 and lim⁡δ→0+δ−1[h(δ)−h(0)]=lim⁡x→+∞xe−x=0\lim_{\delta \rightarrow 0^{+}} \delta^{-1}\left[h(\delta )-h(0)\right]=\lim_{x \rightarrow +\infty } x e^{-x}=0, so that h′(0)h^{\prime }(0) exists and equals zero. Now assume that h(n)(0)=0h^{(n)}(0)=0. Show that lim⁡δ→0−δ−1[h(n)(δ)−h(n)(0)]=0\lim_{\delta \rightarrow 0^{-}} \delta^{-1}\left[h^{(n)}(\delta )-h^{(n)}(0)\right]=0 and

lim⁡δ→0+δ−1[h(n)(δ)−h(n)(0)]=lim⁡δ→0+δ−1Pn(δ−1)e−1δ=lim⁡x→+∞xPn(x)e−x=0, \lim _{\delta \rightarrow 0^{+}} \delta ^{-1}\left[h^{(n)}(\delta )-h^{(n)}(0)\right]=\lim _{\delta \rightarrow 0^{+}} \delta ^{-1} P_{n}\left(\delta ^{-1}\right) e^{-\frac{1}{\delta }}=\lim _{x \rightarrow +\infty } x P_{n}(x) e^{-x}=0,

so that h(n+1)(0)h^{(n+1)}(0) exists and equals 00.]

(c)

Put (a) and (b) together to conclude that hh indeed possesses continuous derivatives of all orders.

Problem 8.6.6

Let g:Rk→Rg: \mathbb {R}^{k} \rightarrow \mathbb {R} be defined by g(x)=h(1−∥x∥2)g(\boldsymbol {x})=h\left(1-\left\| \boldsymbol {x}\right\|^{2}\right), where hh is the function of Exercise 8.6.5. Prove the following easy properties of gg:

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(a)

g∈C∞g \in \mathfrak {C}^{\infty }.

(b)

∥x∥⩾1\left\| \boldsymbol {x}\right\| \geqslant 1 forces g(x)=0g(\boldsymbol {x})=0, hence g∈Cc∞g \in \mathfrak {C}_{c}^{\infty }.

(c)

0⩽∥x∥<10 \leqslant \left\| \boldsymbol {x}\right\| <1 forces g(x)>0g(\boldsymbol {x})>0, hence ∫g=∫B(0,1)g>0\int g=\int_{B(\mathbf{0}, 1)} g>0.

(d)

gg is bounded.

Problem 8.6.7

Let ϕ:Rk→R\phi : \mathbb {R}^{k} \rightarrow \mathbb {R} be defined by ϕ(x)=1cg(x)\phi (\boldsymbol {x})=\frac{1}{c} g(\boldsymbol {x}), where c=∫g>0c=\int g>0 [from part (c) of Exercise 8.6.6, gg being the function defined there]. We have the following very easy properties of ϕ\phi.

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(a)

ϕ∈Cc∞\phi \in \mathfrak {C}_{c}^{\infty }.

(b)

ϕ⩾0\phi \geqslant 0 and ϕ\phi is bounded.

(c)

ϕ(x)>0\phi (\boldsymbol {x})>0 iff 0⩽∥x∥<10 \leqslant \left\| \boldsymbol {x}\right\| <1.

(d)

∫ϕ=1\int \phi =1.

Problem 8.6.8

For each 0<a<+∞0<a<+\infty and x∈Rk\boldsymbol {x} \in \mathbb {R}^{k}, write ϕa(x)=a−nϕ(xa)\phi_{a}(\boldsymbol {x})=a^{-n} \phi \left(\frac{\boldsymbol {x}}{a}\right), where ϕ\phi is the function of Exercise 8.6.7. Prove the following properties of ϕa\phi_{a} for each 0<a<+∞0<a<+\infty.

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(a)

ϕa∈Cc∞\phi_{a} \in \mathfrak {C}_{c}^{\infty }.

(b)

ϕa\phi_{a} is nonnegative and bounded.

(c)

ϕa(x)>0\phi_{a}(\boldsymbol {x})>0 iff 0⩽∥x∥<a0 \leqslant \left\| \boldsymbol {x}\right\| <a. [These first three parts are easy.]

(d)

∫ϕa=1\int \phi_{a}=1. [Let T\boldsymbol {T} denote the k×kk \times k matrix 1aI\frac{1}{a} \boldsymbol {I}, where I\boldsymbol {I} is the identity matrix of order kk. We have by Claim 1 of Section 7.5 that

∫ϕa(x) dλk(x)=a−n∫ϕ(Tx) dλk(x)=a−ndet⁡(T−1)∫ϕ(x) dλk(x)=1. \int \phi _{a}(\boldsymbol {x}) \, d\lambda _{k}(\boldsymbol {x})=a^{-n} \int \phi (\boldsymbol {T} \boldsymbol {x}) \, d\lambda _{k}(\boldsymbol {x})=a^{-n} \operatorname {det}\left(\boldsymbol {T}^{-1}\right) \int \phi (\boldsymbol {x}) \, d\lambda _{k}(\boldsymbol {x})=1 .

Claim 1 of Section 7.5: Suppose that T\boldsymbol {T} denotes a nonsingular transformation from Rk\mathbb {R}^{k} to Rk\mathbb {R}^{k}. Let f:Rk→R‾f: \mathbb {R}^{k} \rightarrow \overline{\mathbb {R}} denote a Lebesgue measurable mapping. Then f∘Tf \circ \boldsymbol {T} is Lebesgue measurable, and if ff is nonnegative, then ∫f dλˉk=∣det⁡(T)∣∫f∘T dλˉk\int f \, d\bar{\lambda }_{k}=\left|\operatorname {det}\left(\boldsymbol {T}\right)\right| \int f \circ \boldsymbol {T} \, d\bar{\lambda }_{k}.]

Problem 8.6.9

Let f∈L1(Rk,Bk,λk)f \in L^{1}\left(\mathbb {R}^{k}, \mathscr {B}^{k}, \lambda_{k}\right). For each 0<a<+∞0<a<+\infty and x∈Rk\boldsymbol {x} \in \mathbb {R}^{k}, define

ψa(x)=∫f(y)ϕa(x−y) dλk(y), \psi _{a}(\boldsymbol {x})=\int f(\boldsymbol {y}) \phi _{a}(\boldsymbol {x}-\boldsymbol {y}) \, d\lambda _{k}(\boldsymbol {y}),

where ϕa\phi_{a} is the function of Exercise 8.6.8.

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(a)

ψa\psi_{a} is sensibly defined since the integral exists, and ψa\psi_{a} is bounded. [We know that ϕa\phi_{a} is bounded, say ϕa⩽Ma\phi_{a} \leqslant M_{a}, so ∣f(y)ϕa(x−y)∣⩽∣f(y)∣Ma\left|f(\boldsymbol {y}) \phi_{a}(\boldsymbol {x}-\boldsymbol {y})\right| \leqslant \left|f(\boldsymbol {y})\right| M_{a} for every x,y∈Rk\boldsymbol {x}, \boldsymbol {y} \in \mathbb {R}^{k}; note that fMa∈L1f M_{a} \in L^{1}.]

(b)

For each i=1,…,ki=1, \ldots , k and x∈Rk\boldsymbol {x} \in \mathbb {R}^{k}, we obtain

∂∂xiψa(x)=∫f(y)∂∂xiϕa(x−y) dλk(y), \frac{\partial }{\partial x_{i}} \psi _{a}(\boldsymbol {x})=\int f(\boldsymbol {y}) \frac{\partial }{\partial x_{i}} \phi _{a}(\boldsymbol {x}-\boldsymbol {y}) \, d\lambda _{k}(\boldsymbol {y}),

hence ψa\psi_{a} is continuous. [Use reasoning as given in the hint to (a).]

(c)

For every i,j∈{1,…,k}i, j \in \left\{ 1, \ldots , k\right\} and x∈Rk\boldsymbol {x} \in \mathbb {R}^{k} we have

∂2∂xj∂xiψa(x)=∫f(y)∂2∂xj∂xiϕa(x−y) dλk(y). \frac{\partial ^{2}}{\partial x_{j} \partial x_{i}} \psi _{a}(\boldsymbol {x})=\int f(\boldsymbol {y}) \frac{\partial ^{2}}{\partial x_{j} \partial x_{i}} \phi _{a}(\boldsymbol {x}-\boldsymbol {y}) \, d\lambda _{k}(\boldsymbol {y}).

Conclude that ∂∂xiψa(x)\frac{\partial }{\partial x_{i}} \psi_{a}(\boldsymbol {x}) is a continuous function of x\boldsymbol {x}. [The same type of argument as used in (b) may be employed here.]

(d)

Use (b), (c), and induction to obtain ψa∈C∞\psi_{a} \in \mathfrak {C}^{\infty }.

(e)

If ff has compact support, so does ψa\psi_{a}. [Say f=0f=0 on B(0,r)cB(\mathbf{0}, r)^{c} for some r>0r>0. If ψa(x)≠0\psi_{a}(\boldsymbol {x}) \neq 0, then by definition there is y∈Rk\boldsymbol {y} \in \mathbb {R}^{k} such that f(y)ϕa(x−y)≠0f(\boldsymbol {y}) \phi_{a}(\boldsymbol {x}-\boldsymbol {y}) \neq 0. If f(y)≠0f(\boldsymbol {y}) \neq 0, then y∈B(0,r)\boldsymbol {y} \in B(\mathbf{0}, r). If ϕa(x−y)≠0\phi_{a}(\boldsymbol {x}-\boldsymbol {y}) \neq 0, then x−y∈B(0,a)\boldsymbol {x}-\boldsymbol {y} \in B(\mathbf{0}, a). Observe that ∥x∥⩽∥y∥+∥x−y∥<r+a\left\| \boldsymbol {x}\right\| \leqslant \left\| \boldsymbol {y}\right\| +\left\| \boldsymbol {x}-\boldsymbol {y}\right\| <r+a follows in any case. Thus ψa(x)≠0\psi_{a}(\boldsymbol {x}) \neq 0 forces x∈B(0,r+a)\boldsymbol {x} \in B(\mathbf{0}, r+a), so ψa\psi_{a} has compact support.]

(f)

If ff has compact support, then ψa∈Cc∞\psi_{a} \in \mathfrak {C}_{c}^{\infty }. [Automatic from the previous parts.]

Problem 8.6.10

Exercises 8.6.5-8.6.9 lead to the following main result: Cc∞\mathfrak {C}_{c}^{\infty } is dense in L1(Rk,Bk,λk)L^{1}\left(\mathbb {R}^{k}, \mathscr {B}^{k}, \lambda_{k}\right). We shall achieve this by showing that Cc∞\mathfrak {C}_{c}^{\infty } is dense in Cc\mathfrak {C}_{c}. Since Cc\mathfrak {C}_{c} is dense in L1L^{1}, the claim will follow. We shall do this with the given outline.

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(a)

Let f∈Ccf \in \mathfrak {C}_{c} and ε>0\varepsilon >0; find r>0r>0 so that f=0f=0 on B(0,r)cB(\mathbf{0}, r)^{c}. Show that there is 0<a⩽10<a \leqslant 1 such that

∥y−x∥⩽a implies ∣f(y)−f(x)∣⩽ελk(B(0,r+1)). \left\| \boldsymbol {y}-\boldsymbol {x}\right\| \leqslant a \text{ implies } \left|f(\boldsymbol {y})-f(\boldsymbol {x})\right| \leqslant \frac{\varepsilon }{\lambda _{k}\left(B(\mathbf{0}, r+1)\right)} .

[Note that ff is uniformly continuous since ff is continuous with compact support.]

(b)

We have f(x)=∫f(x)ϕa(x−y) dλk(y)f(\boldsymbol {x})=\int f(\boldsymbol {x}) \phi_{a}(\boldsymbol {x}-\boldsymbol {y}) \, d\lambda_{k}(\boldsymbol {y}) for every x∈Rk\boldsymbol {x} \in \mathbb {R}^{k}. [Trivial by part (d) of Exercise 8.6.8, which states ∫ϕa=1\int \phi_{a}=1.]

(c)

Both ff and ψa\psi_{a} are zero on B(0,r+1)cB(\mathbf{0}, r+1)^{c}. [Recall that 0<a⩽10<a \leqslant 1. The rest is straightforward.]

(d)

For every x∈Rk\boldsymbol {x} \in \mathbb {R}^{k} we have

∣ψa(x)−f(x)∣⩽ελk(B(0,r+1)). \left|\psi _{a}(\boldsymbol {x})-f(\boldsymbol {x})\right| \leqslant \frac{\varepsilon }{\lambda _{k}\left(B(\mathbf{0}, r+1)\right)} .

[We have

∣ψa(x)−f(x)∣=∣∫[f(y)−f(x)]ϕa(x−y) dλk(y)∣⩽∫∣f(y)−f(x)∣ϕa(x−y) dλk(y)=∫y∈Rk:∥x−y∥⩽a∣f(y)−f(x)∣ϕa(x−y) dλk(y)⩽ελk(B(0,r+1))∫y∈Rk:∥x−y∥⩽aϕa(x−y) dλk(y)=ελk(B(0,r+1)). \begin{aligned} \left|\psi _{a}(\boldsymbol {x})-f(\boldsymbol {x})\right| & =\left|\int \left[f(\boldsymbol {y})-f(\boldsymbol {x})\right] \phi _{a}(\boldsymbol {x}-\boldsymbol {y}) \, d\lambda _{k}(\boldsymbol {y})\right| \\ & \leqslant \int \left|f(\boldsymbol {y})-f(\boldsymbol {x})\right| \phi _{a}(\boldsymbol {x}-\boldsymbol {y}) \, d\lambda _{k}(\boldsymbol {y}) \\ & =\int _{\boldsymbol {y} \in \mathbb {R}^{k}: \left\| \boldsymbol {x}-\boldsymbol {y}\right\| \leqslant a}\left|f(\boldsymbol {y})-f(\boldsymbol {x})\right| \phi _{a}(\boldsymbol {x}-\boldsymbol {y}) \, d\lambda _{k}(\boldsymbol {y}) \\ & \leqslant \frac{\varepsilon }{\lambda _{k}\left(B(\mathbf{0}, r+1)\right)} \int _{\boldsymbol {y} \in \mathbb {R}^{k}: \left\| \boldsymbol {x}-\boldsymbol {y}\right\| \leqslant a} \phi _{a}(\boldsymbol {x}-\boldsymbol {y}) \, d\lambda _{k}(\boldsymbol {y}) \\ & =\frac{\varepsilon }{\lambda _{k}\left(B(\mathbf{0}, r+1)\right)} . \end{aligned}

]

(e)

We have ∥ψa−f∥1⩽ε\left\| \psi_{a}-f\right\|_{1} \leqslant \varepsilon, and hence Cc∞\mathfrak {C}_{c}^{\infty } is dense in Cc\mathfrak {C}_{c}. [We have

∥ψa−f∥1=∫∣ψa(x)−f(x)∣ dλk(x)=∫B(0,r+1)∣ψa(x)−f(x)∣ dλk(x)⩽[ελk(B(0,r+1))]∫B(0,r+1)dλk(x)=ελk(B(0,r+1))λk(B(0,r+1))=ε. \begin{aligned} \left\| \psi _{a}-f\right\| _{1} & =\int \left|\psi _{a}(\boldsymbol {x})-f(\boldsymbol {x})\right| \, d\lambda _{k}(\boldsymbol {x}) \\ & =\int _{B(\mathbf{0}, r+1)}\left|\psi _{a}(\boldsymbol {x})-f(\boldsymbol {x})\right| \, d\lambda _{k}(\boldsymbol {x}) \\ & \leqslant \left[\frac{\varepsilon }{\lambda _{k}\left(B(\mathbf{0}, r+1)\right)}\right] \int _{B(\mathbf{0}, r+1)} d\lambda _{k}(\boldsymbol {x}) \\ & =\frac{\varepsilon }{\lambda _{k}\left(B(\mathbf{0}, r+1)\right)} \lambda _{k}\left(B(\mathbf{0}, r+1)\right) \\ & =\varepsilon . \end{aligned}

]

Section 8.7More Convergence Concepts[16/31%]
Problem 8.7.1

Prove the following simple claims.

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(a)

Let {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } denote a Cauchy sequence in LpL^{p}, where 0<p<∞0<p<\infty. Show that {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } is a Cauchy sequence in measure: for every ε>0\varepsilon >0 and δ>0\delta >0 there is N∈NN \in \mathbb {N} such that for every n,m⩾Nn, m \geqslant N we have μ({ω∈Ω:∣fn(ω)−fm(ω)∣>δ})<ε\mu \left(\left\{ \omega \in \Omega :\left|f_{n}(\omega )-f_{m}(\omega )\right|>\delta \right\} \right)<\varepsilon.

(b)

Let f,f1,f2,…∈Lpf, f_{1}, f_{2}, \ldots \in L^{p} and suppose that fn→Lpff_{n} \xrightarrow {L^{p}} f, where 0<p<∞0<p<\infty. If g∈L∞g \in L^{\infty }, then fg,f1g,f2g,…∈Lpf g, f_{1} g, f_{2} g, \ldots \in L^{p} and fng→Lpfgf_{n} g \xrightarrow {L^{p}} f g.

Problem 8.7.2

While convergence in pth mean implies convergence in measure, it is not the case that convergence in measure implies convergence in pthe mean.

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Problem 8.7.3

It is possible for a sequence {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } in LpL^{p} to converge μ\mu-a. e. to some f∈Lpf \in L^{p} but not in pth mean. That is, convergence μ\mu-a. e. does not force convergence in pp th mean.

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Problem 8.7.4

It is possible for a sequence {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } in LpL^{p} to converge in pth mean to zero, but {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } converges at no point of Ω\Omega.

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Problem 8.7.5

It is possible to have functions f,f1,f2,…∈Lp1∩Lp2f, f_{1}, f_{2}, \ldots \in L^{p_{1}} \cap L^{p_{2}} such that fn→Lp1ff_{n} \xrightarrow {L^{p_{1}}} f but fn̸→Lp2ff_{n} \not\xrightarrow {L^{p_{2}}} f.

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Problem 8.7.6

Let 1<p0<+∞1<p_{0}<+\infty. There is a sequence {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } in LpL^{p} for all 1⩽p<p01 \leqslant p<p_{0} with fn→Lp0f_{n} \xrightarrow {L^{p}} 0 for all 1⩽p<p01 \leqslant p<p_{0}, but fnf_{n} fails to converge in p0p_{0}th mean.

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Problem 8.7.7

Let 1⩽p,q<+∞1 \leqslant p, q<+\infty, and let {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } denote a sequence in Lp∩LqL^{p} \cap L^{q}. Suppose that there is g∈Lpg \in L^{p} and h∈Lqh \in L^{q} with fn→Lpgf_{n} \xrightarrow {L^{p}} g and fn→Lqhf_{n} \xrightarrow {L^{q}} h. Then g=hg=h μ\mu-a.e. on Ω\Omega.

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Problem 8.7.8

Suppose that {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } is a sequence in LpL^{p}, where 1<p<+∞1<p<+\infty. Suppose that h∈L1h \in L^{1} with fn⩽hf_{n} \leqslant h μ\mu-a.e. on Ω\Omega. If fnf_{n} converges in measure to some ff, then f∈Lpf \in L^{p} and fn→Lpff_{n} \xrightarrow {L^{p}} f.

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Problem 8.7.9

If f,f1,f2,…∈Lpf, f_{1}, f_{2}, \ldots \in L^{p} are such that fn→ff_{n} \rightarrow f uniformly, then fn→Lpff_{n} \xrightarrow {L^{p}} f.

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Remark.
?

Here 0<p<+∞0<p<+\infty.

Problem 8.7.10

Let f,f1,f2,…∈Lpf, f_{1}, f_{2}, \ldots \in L^{p}, where 1⩽p⩽+∞1 \leqslant p \leqslant +\infty, and suppose that ∥fn∥p→∥f∥p\left\| f_{n}\right\|_{p} \rightarrow \left\| f\right\|_{p}. Then either of the conditions fn→ff_{n} \rightarrow f μ\mu-a.e. on Ω\Omega and fn→ff_{n} \rightarrow f in measure is sufficient for the conclusion fn→Lpff_{n} \xrightarrow {L^{p}} f.

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Problem 8.7.11

Consider the measure space ([0,1],B,λ)\left(\left[0,1\right], \mathscr {B}, \lambda \right), where here B\mathscr {B} denotes the Borel subsets of [0,1]\left[0,1\right] and λ\lambda denotes Lebesgue measure restricted to B\mathscr {B}. Let fn=n  1[0,n−1]f_{n}=n \; \mathbb {1}_{\left[0, n^{-1}\right]} for each n∈Nn \in \mathbb {N}. Then each fnf_{n} is in LpL^{p} for 1⩽p<+∞1 \leqslant p<+\infty, fn→0f_{n} \rightarrow 0 μ\mu-a.e. on [0,1]\left[0,1\right] and in measure, but ∫fng↛∫0g\int f_{n} g \not\rightarrow \int 0 g when g=1g=1, and hence fn↛0f_{n} \not\rightarrow 0 weakly.

Definition. Consider a generic measure space (Ω,F,μ)\left(\Omega , \mathcal{F}, \mu \right), and let 1⩽p<+∞1 \leqslant p<+\infty. Let f,f1,f2,…∈Lpf, f_{1}, f_{2}, \ldots \in L^{p}. We will say that fn→ff_{n} \rightarrow f weakly [in LpL^{p}] iff

lim⁡n∫fng=∫fgfor every g∈Lp′. \lim _{n} \int f_{n} g = \int f g \quad \text{for every } g \in L^{p^{\prime }}.

Here p′p^{\prime } denotes the conjugate exponent of pp: p′=pp−1p^{\prime }=\frac{p}{p-1} for 1<p<+∞1<p<+\infty, and 1′=+∞1^{\prime }=+\infty.

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Problem 8.7.12

Consider the measure space (R,B,λ)\left(\mathbb {R}, \mathscr {B}, \lambda \right). Let fn=n−1  1[1,en]f_{n}=n^{-1} \; \mathbb {1}_{\left[1, e^{n}\right]} for every n∈Nn \in \mathbb {N}. Then each fnf_{n} is in LpL^{p} for 1⩽p<+∞1 \leqslant p<+\infty, fn→0f_{n} \rightarrow 0 uniformly, but fn↛0f_{n} \not\rightarrow 0 weakly in LpL^{p}.

?
Remark.
?

Consider the function g(ω)=ω−1  1[1,+∞)(ω)g(\omega )=\omega^{-1} \; \mathbb {1}_{[1,+\infty )}(\omega ). We have that g∈Lp′g \in L^{p^{\prime }} for each 1<p<+∞1<p<+\infty, gg is bounded, and ∫fng=1\int f_{n} g=1 for every n∈Nn \in \mathbb {N}, while ∫0g=0\int 0 g=0.

Problem 8.7.13

Consider the measure space ([0,2π],B,λ)\left(\left[0,2 \pi \right], \mathscr {B}, \lambda \right), where here B\mathscr {B} denotes the Borel subsets of [0,2π]\left[0,2 \pi \right] and λ\lambda denotes Lebesgue measure restricted to B\mathscr {B}. For each n∈Nn \in \mathbb {N} and 0⩽ω⩽2π0 \leqslant \omega \leqslant 2 \pi, define fn(ω)=cos⁡nωf_{n}(\omega )=\cos n \omega. Then each fnf_{n} is in L1L^{1}, and fn→0f_{n} \rightarrow 0 weakly in L1L^{1}. However, fn↛0f_{n} \not\rightarrow 0. In fact, we have fn̸→μ0f_{n} \not\xrightarrow {\mu } 0 and fn↛0f_{n} \not\rightarrow 0 μ\mu-a.e. on Ω\Omega.

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Remark.
?

This exercise might be difficult.

Problem 8.7.14

Refer to the Vitali Convergence Theorem. Prove that conditions (i) and (ii) imply that f∈Lpf \in L^{p} and fn→ff_{n} \rightarrow f in pth mean when μ(Ω)=+∞\mu (\Omega )=+\infty.

Vitali Convergence Theorem (finite-measure case). Let (Ω,F,μ)\left(\Omega , \mathcal{F}, \mu \right) denote a measure space with μ(Ω)<+∞\mu (\Omega )<+\infty, and let 1⩽p<+∞1 \leqslant p<+\infty. Let {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } denote a sequence of functions in LpL^{p}. Let ff denote an F/B∗\mathcal{F} / \mathscr {B}^{*}-measurable function on Ω\Omega such that fn→ff_{n} \rightarrow f μ\mu-a.e. on Ω\Omega and ff is finite μ\mu-a.e. on Ω\Omega. Then the condition f∈Lpf \in L^{p} and fn→Lpff_{n} \xrightarrow {L^{p}} f is equivalent to the dual conditions (i) for every ε>0\varepsilon >0 there is A∈FA \in \mathcal{F} [depending on ε\varepsilon] with μ(A)<+∞\mu (A)<+\infty and ∫Ac∣fn∣p<ε\int_{A^{c}} \left|f_{n}\right|^{p}<\varepsilon for every n∈Nn \in \mathbb {N}, and (ii) for every ε>0\varepsilon >0 there is δ>0\delta >0 [depending only on ε\varepsilon, not nn] such that whenever μ(E)<δ\mu (E)<\delta we have ∫E∣fn∣p<ε\int_{E} \left|f_{n}\right|^{p}<\varepsilon for every n∈Nn \in \mathbb {N}.

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Problem 8.7.15

Let (Ω,F,μ)\left(\Omega , \mathcal{F}, \mu \right) denote a measure space with 1⩽p<+∞1 \leqslant p<+\infty. Let {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } denote a sequence in LpL^{p}, and let ff denote an F/B∗\mathcal{F} / \mathscr {B}^{*}-measurable function that is finite μ\mu-a.e. on Ω\Omega with fn→ff_{n} \rightarrow f μ\mu-a.e. on Ω\Omega. For every n,k∈Nn, k \in \mathbb {N}, let Bn,k={ω∈Ω:∣fn(ω)∣p⩾k}B_{n, k}=\left\{ \omega \in \Omega :\left|f_{n}(\omega )\right|^{p} \geqslant k\right\}. Make the following additional assumption (a): for every ε>0\varepsilon >0 there is a set A∈FA \in \mathcal{F} [depending on ε\varepsilon] such that μ(A)<+∞\mu (A)<+\infty and ∫Ac∣fn∣p<ε\int_{A^{c}} \left|f_{n}\right|^{p}<\varepsilon for every n∈Nn \in \mathbb {N}.

Then the following four statements are logically equivalent:

  1. f∈Lpf \in L^{p} and fn→Lpff_{n} \xrightarrow {L^{p}} f

  2. If {Ek}k=1∞\left\{ E_{k}\right\}_{k=1}^{\infty } is a nonincreasing sequence of F\mathcal{F}-sets with limit ∅\varnothing, then for every ε>0\varepsilon >0 there is K∈NK \in \mathbb {N} such that k⩾Kk \geqslant K forces ∫Ek∣fn∣p<ε\int_{E_{k}} \left|f_{n}\right|^{p}<\varepsilon for every n∈Nn \in \mathbb {N}.

  3. For every ε>0\varepsilon >0 there is K∈NK \in \mathbb {N} such that k⩾Kk \geqslant K forces ∫Bn,k∣fn∣p<ε\int_{B_{n, k}} \left|f_{n}\right|^{p}<\varepsilon for every n∈Nn \in \mathbb {N}.

  4. For every ε>0\varepsilon >0 there is δ>0\delta >0 such that μ(E)<δ\mu (E)<\delta forces ∫E∣fn∣p<ε\int_{E} \left|f_{n}\right|^{p}<\varepsilon for every n∈Nn \in \mathbb {N}.

Vitali Convergence Theorem (finite-measure case). Let (Ω,F,μ)\left(\Omega , \mathcal{F}, \mu \right) denote a measure space with μ(Ω)<+∞\mu (\Omega )<+\infty, and let 1⩽p<+∞1 \leqslant p<+\infty. Let {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } denote a sequence of functions in LpL^{p}. Let ff denote an F/B∗\mathcal{F} / \mathscr {B}^{*}-measurable function on Ω\Omega such that fn→ff_{n} \rightarrow f μ\mu-a.e. on Ω\Omega and ff is finite μ\mu-a.e. on Ω\Omega. Then the condition f∈Lpf \in L^{p} and fn→Lpff_{n} \xrightarrow {L^{p}} f is equivalent to the dual conditions (i) for every ε>0\varepsilon >0 there is A∈FA \in \mathcal{F} with μ(A)<+∞\mu (A)<+\infty and ∫Ac∣fn∣p<ε\int_{A^{c}} \left|f_{n}\right|^{p}<\varepsilon for every n∈Nn \in \mathbb {N}, and (ii) for every ε>0\varepsilon >0 there is δ>0\delta >0 such that whenever μ(E)<δ\mu (E)<\delta we have ∫E∣fn∣p<ε\int_{E} \left|f_{n}\right|^{p}<\varepsilon for every n∈Nn \in \mathbb {N}.

?
Remark.
?

The following outline is useful. By the Vitali Convergence Theorem and Exercise 8.7.14, it is immediate that (i) and (iv) are logically equivalent.

We show that (i) implies (ii). Let ε>0\varepsilon >0. Find N∈NN \in \mathbb {N} such that n⩾Nn \geqslant N implies ∥fn−f∥p<ε\left\| f_{n}-f\right\|_{p}<\varepsilon. For n⩾Nn \geqslant N and k∈Nk \in \mathbb {N}, Minkowski's Inequality gives

(∫Ek∣fn∣p)1/p⩽(∫Ek∣f∣p)1/p+(∫Ek∣fn−f∣p)1/p<(∫Ek∣f∣p)1/p+ε. \left(\int _{E_{k}} \left|f_{n}\right|^{p}\right)^{1 / p} \leqslant \left(\int _{E_{k}} \left|f\right|^{p}\right)^{1 / p}+\left(\int _{E_{k}} \left|f_{n}-f\right|^{p}\right)^{1 / p} <\left(\int _{E_{k}} \left|f\right|^{p}\right)^{1 / p}+\varepsilon .

Apply DCT to {∣f∣p  1Ek}k=1∞\left\{ \left|f\right|^{p} \; \mathbb {1}_{E_{k}}\right\}_{k=1}^{\infty } to show that (∫Ek∣fn∣p)1/p<ε\left(\int_{E_{k}} \left|f_{n}\right|^{p}\right)^{1 / p}<\varepsilon for large enough kk, say k⩾Kk \geqslant K. Then for every n∈Nn \in \mathbb {N} and k⩾Kk \geqslant K, we have (∫Ek∣fn∣p)1/p<2ε\left(\int_{E_{k}} \left|f_{n}\right|^{p}\right)^{1 / p}<2 \varepsilon. For n∈{1,…,N−1}n \in \left\{ 1, \ldots , N-1\right\}, observe for every k∈Nk \in \mathbb {N} that we have

∫Ek∣fn∣p⩽∫Ekmax⁡{∣f1∣p,…,∣fN−1∣p}<+∞, \int _{E_{k}} \left|f_{n}\right|^{p} \leqslant \int _{E_{k}} \max \left\{ \left|f_{1}\right|^{p}, \ldots , \left|f_{N-1}\right|^{p}\right\} <+\infty ,

and therefore DCT can be used to obtain (ii).

Now assume (ii), picking ε>0\varepsilon >0. For each k∈Nk \in \mathbb {N}, define Ek=⋃n=k∞Bn,kE_{k}=\bigcup_{n=k}^{\infty } B_{n, k}. Then E1⊇E2⊇⋯E_{1} \supseteq E_{2} \supseteq \cdots and lim⁡‾n∣fn(ω)∣=+∞\overline{\lim }_{n} \left|f_{n}(\omega )\right|=+\infty on ⋂k=1∞Ek\bigcap_{k=1}^{\infty } E_{k}; from this we have μ(⋂k=1∞Ek)=0\mu \left(\bigcap_{k=1}^{\infty } E_{k}\right)=0. Next, write Fk=Ek−⋂j=1∞EjF_{k}=E_{k}-\bigcap_{j=1}^{\infty } E_{j} for every k∈Nk \in \mathbb {N}. We have that F1⊇F2⊇⋯F_{1} \supseteq F_{2} \supseteq \cdots and lim⁡kFk=∅\lim_{k} F_{k}=\varnothing. Choose KK such that k⩾Kk \geqslant K implies ∫Fk∣fn∣p<ε\int_{F_{k}} \left|f_{n}\right|^{p}<\varepsilon for every n∈Nn \in \mathbb {N}. For every n⩾Kn \geqslant K we have Bn,K⊆EKB_{n, K} \subseteq E_{K}. Therefore, if n,k⩾Kn, k \geqslant K, then

∫Bn,k∣fn∣p⩽∫Bn,K∣fn∣p⩽∫EK∣fn∣p=∫FK∣fn∣p<ε. \int _{B_{n, k}} \left|f_{n}\right|^{p} \leqslant \int _{B_{n, K}} \left|f_{n}\right|^{p} \leqslant \int _{E_{K}} \left|f_{n}\right|^{p}=\int _{F_{K}} \left|f_{n}\right|^{p}<\varepsilon .

For n∈{1,…,K−1}n \in \left\{ 1, \ldots , K-1\right\}, we have ∣fn∣p⩽max⁡{∣f1∣p,…,∣fK−1∣p}=g\left|f_{n}\right|^{p} \leqslant \max \left\{ \left|f_{1}\right|^{p}, \ldots , \left|f_{K-1}\right|^{p}\right\} =g, say, and hence ∫Bn,k∣fn∣p⩽∫Bkg<+∞\int_{B_{n, k}} \left|f_{n}\right|^{p} \leqslant \int_{B_{k}} g<+\infty, where Bk=⋃n=1K−1Bn,k={ω∈Ω:g(ω)⩾k}B_{k}=\bigcup_{n=1}^{K-1} B_{n, k}=\left\{ \omega \in \Omega : g(\omega ) \geqslant k\right\}. Use DCT to derive (iii) in direct fashion.

Now assume (iii). Pick K∈NK \in \mathbb {N} such that k⩾Kk \geqslant K forces ∫Bn,k∣fn∣p<εp\int_{B_{n, k}} \left|f_{n}\right|^{p}<\varepsilon^{p} for every n∈Nn \in \mathbb {N}. Argue that μ(E)<K−1εp\mu (E)<K^{-1} \varepsilon^{p} implies

(∫E∣fn∣p)1/p⩽(∫E∩Bn,K∣fn∣p)1/p+(∫E∩Bn,Kc∣fn∣p)1/p<2ε. \left(\int _{E} \left|f_{n}\right|^{p}\right)^{1 / p} \leqslant \left(\int _{E \cap B_{n, K}} \left|f_{n}\right|^{p}\right)^{1 / p}+\left(\int _{E \cap B_{n, K}^{c}} \left|f_{n}\right|^{p}\right)^{1 / p}<2 \varepsilon .

Therefore we may deduce (iv) from (iii).

Problem 8.7.16

Assume everything in Exercise 8.7.15, including (a). Assume further that (ii) holds. Argue that (i) and (ii) of the Vitali Convergence Theorem follow as consequences.

Vitali Convergence Theorem (finite-measure case). Let (Ω,F,μ)\left(\Omega , \mathcal{F}, \mu \right) denote a measure space with μ(Ω)<+∞\mu (\Omega )<+\infty, and let 1⩽p<+∞1 \leqslant p<+\infty. Let {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } denote a sequence of functions in LpL^{p}. Let ff denote an F/B∗\mathcal{F} / \mathscr {B}^{*}-measurable function on Ω\Omega such that fn→ff_{n} \rightarrow f μ\mu-a.e. on Ω\Omega and ff is finite μ\mu-a.e. on Ω\Omega. Then the condition f∈Lpf \in L^{p} and fn→Lpff_{n} \xrightarrow {L^{p}} f is equivalent to the dual conditions (i) for every ε>0\varepsilon >0 there is A∈FA \in \mathcal{F} with μ(A)<+∞\mu (A)<+\infty and ∫Ac∣fn∣p<ε\int_{A^{c}} \left|f_{n}\right|^{p}<\varepsilon for every n∈Nn \in \mathbb {N}, and (ii) for every ε>0\varepsilon >0 there is δ>0\delta >0 such that whenever μ(E)<δ\mu (E)<\delta we have ∫E∣fn∣p<ε\int_{E} \left|f_{n}\right|^{p}<\varepsilon for every n∈Nn \in \mathbb {N}.

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Remark.
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The particular form of the Vitali Convergence Theorem as presented in this text and the previous two exercises [and the outlines of the solution] owe their existence and form to the most excellent presentation of Hewitt and Stromberg, pp. 203-205.

Section 8.8Prelude to the Riesz Representation Theorem[4/0%]
Problem 8.8.1

Let TT be defined on LpL^{p} [where 1<p<+∞1<p<+\infty] by T(f)=∫fhT(f)=\int f h, where h∈Lp′h \in L^{p^{\prime }} is held fixed. Then ∥T∥=∥h∥p′\left\| T\right\| =\left\| h\right\|_{p^{\prime }}.

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Remark.
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It should be easy to obtain ∥T∥⩽∥h∥p′\left\| T\right\| \leqslant \left\| h\right\|_{p^{\prime }}. To obtain the reverse inequality, let f=∣h∣p′−1sign⁡(h)f=\left|h\right|^{p^{\prime }-1} \operatorname {sign}\left(h\right). Then ∣f∣p=∣h∣p′\left|f\right|^{p}=\left|h\right|^{p^{\prime }}, hence f∈Lpf \in L^{p}. Show next that ∥f∥p=∥g∥p′p′−1\left\| f\right\|_{p}=\left\| g\right\|_{p^{\prime }}^{p^{\prime }-1} and L(f)=∥f∥p∥h∥p′L(f)=\left\| f\right\|_{p}\left\| h\right\|_{p^{\prime }}, so that ∥L∥⩾∥h∥p′\left\| L\right\| \geqslant \left\| h\right\|_{p^{\prime }}.

Problem 8.8.2

In the specific example mentioned in this section's discussion on linear transformations, restate Claims 3 and 4 in terms of the specific setup.

Claim 3: Let AA and BB denote normed linear spaces and let T:A→BT: A \rightarrow B denote a linear transformation. The following are equivalent: (a) TT is bounded. (b) TT is uniformly continuous on AA. (c) TT is continuous at some x∈Ax \in A.

Claim 4: B(A,R)\mathcal{B}(A, \mathbb {R}) is a complete normed linear space over R\mathbb {R} relative to the operator norm. That is, B(A,R)\mathcal{B}(A, \mathbb {R}) is a Banach space. [Here B(A,R)={T:A→R:T is a bounded linear transformation}\mathcal{B}(A, \mathbb {R}) = \left\{ T: A \rightarrow \mathbb {R} : T \text{ is a bounded linear transformation}\right\}, equipped with the pointwise-defined vector-space operations and the operator norm ∥T∥=sup⁡{∣T(x)∣:x∈A,∥x∥=1}\left\| T\right\| = \sup \left\{ \left|T(x)\right| : x \in A, \left\| x\right\| =1\right\}.]

The specific setup referenced is: A=LpA=L^{p} [where 1<p<+∞1<p<+\infty], B=RB=\mathbb {R}, and T:Lp→RT: L^{p} \rightarrow \mathbb {R} is given by T(f)=∫fhT(f)=\int f h for every f∈Lpf \in L^{p}, where h∈Lp′h \in L^{p^{\prime }} is fixed.

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Problem 8.8.3

Use the Clarkson Inequalities to show that LpL^{p} [where 1<p<+∞1<p<+\infty] is locally uniformly convex: if {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty } is a sequence in LpL^{p}, f∈Lpf \in L^{p}, ∥f∥p=∥f1∥p=∥f2∥p=⋯=1\left\| f\right\|_{p}=\left\| f_{1}\right\|_{p}=\left\| f_{2}\right\|_{p}=\cdots =1, then lim⁡n∥12(fn+f)∥p=1\lim_{n}\left\| \frac{1}{2}\left(f_{n}+f\right)\right\|_{p}=1 implies lim⁡n∥fn−f∥p=0\lim_{n}\left\| f_{n}-f\right\|_{p}=0.

Clarkson's Inequality (case 2⩽p<+∞2 \leqslant p<+\infty). Let 2⩽p<+∞2 \leqslant p<+\infty and f,g∈Lpf, g \in L^{p}. Then

∥f+g2∥pp+∥f−g2∥pp⩽∥f∥pp2+∥g∥pp2. \left\| \frac{f+g}{2}\right\| _{p}^{p}+\left\| \frac{f-g}{2}\right\| _{p}^{p} \leqslant \frac{\left\| f\right\| _{p}^{p}}{2}+\frac{\left\| g\right\| _{p}^{p}}{2} .

Clarkson's Inequality (case 1<p<21<p<2). Let 1<p<21<p<2 and f,g∈Lpf, g \in L^{p}. Then

∥f+g2∥pp′+∥f−g2∥pp′⩽[∥f∥pp2+∥g∥pp2]1p−1. \left\| \frac{f+g}{2}\right\| _{p}^{p^{\prime }}+\left\| \frac{f-g}{2}\right\| _{p}^{p^{\prime }} \leqslant \left[\frac{\left\| f\right\| _{p}^{p}}{2}+\frac{\left\| g\right\| _{p}^{p}}{2}\right]^{\frac{1}{p-1}} .
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Problem 8.8.4

Suppose that (Ω,F,μ)(\Omega , \mathcal{F}, \mu ) denotes a measure space, where F\mathcal{F} contains at least two disjoint sets of finite positive measure.

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(a)

There is a positive real number cc such that

1c⩽∥f+g∥p2+∥f−g∥p22(∥f∥p2+∥g∥p2)⩽c \frac{1}{c} \leqslant \frac{\left\| f+g\right\| _{p}^{2}+\left\| f-g\right\| _{p}^{2}}{2\left(\left\| f\right\| _{p}^{2}+\left\| g\right\| _{p}^{2}\right)} \leqslant c

for every f,g∈Lpf, g \in L^{p} such that at least one of ∥f∥p\left\| f\right\|_{p} and ∥g∥p\left\| g\right\|_{p} is positive.

(b)

The unique smallest possible cc in the above is 2∣2−p∣p2^{\frac{\left|2-p\right|}{p}}, and the lower and upper bounds in (a) are realized.

Section 8.9The Riesz Representation Theorem[2/0%]
Problem 8.9.1

Let the sets A(t)A(t) and CC be as in the proof of the Riesz Representation Theorem, and let tnt_{n} denote a sequence with tn→0t_{n} \rightarrow 0. We will show that lim⁡n  1A(tn)∩C=  1A(0)∩C\lim_{n} \; \mathbb {1}_{A(t_{n}) \cap C}=\; \mathbb {1}_{A(0) \cap C} on Ω\Omega.

Setup (from the proof of the Riesz Representation Theorem). Let 1<p<21<p<2, let g∈Lpg \in L^{p} be such that ∥g∥p=1\left\| g\right\|_{p}=1, and fix f∈Lpf \in L^{p}. For each real t0≠0t_{0} \neq 0, the Mean Value Theorem provides, for each ω∈Ω\omega \in \Omega, a number t0∗(ω)t_{0}^{*}(\omega ) between 00 and t0t_{0} [depending on ω\omega] such that

∣t0f(ω)+g(ω)∣p−∣g(ω)∣pt0=∂∂t∣tf(ω)+g(ω)∣p∣t=t0∗(ω). \frac{\left|t_{0} f(\omega )+g(\omega )\right|^{p}-\left|g(\omega )\right|^{p}}{t_{0}}=\left.\frac{\partial }{\partial t}\left|t f(\omega )+g(\omega )\right|^{p}\right|_{t=t_{0}^{*}(\omega )} .

For every real t0≠0t_{0} \neq 0, define

A(t0)={ω∈Ω:t0∗(ω)f(ω)+g(ω)≠0}; A(t_{0})=\left\{ \omega \in \Omega : t_{0}^{*}(\omega ) f(\omega )+g(\omega ) \neq 0\right\} ;

also set A(0):=lim⁡nA(tn)A(0):=\lim_{n} A(t_{n}) [i.e., A(0)A(0) is defined as this limit set, whose existence and value the exercise's parts (a)-(e) establish]. Finally, let

C={ω∈Ω:f(ω)≠0}. C=\left\{ \omega \in \Omega : f(\omega ) \neq 0\right\} .
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(a)

It is sufficient to show that lim⁡n[A(tn)∩C]=A(0)∩C\lim_{n}\left[A(t_{n}) \cap C\right]=A(0) \cap C.

(b)

lim⁡‾nA(tn)c⊆A(0)c\overline{\lim }_{n} A(t_{n})^{c} \subseteq A(0)^{c}, that is, A(0)⊆lim⁡‾nA(tn)A(0) \subseteq \underline{\lim }_{n} A(t_{n}). [Let ω∈lim⁡nA(tn)c\omega \in \lim_{n} A(t_{n})^{c}. Then there is an increasing sequence of indices n1<n2<⋯n_{1}<n_{2}<\cdots with the property that g(ω)=−tnk∗(ω)f(ω)g(\omega )=-t_{n_{k}}^{*}(\omega ) f(\omega ) for each k∈Nk \in \mathbb {N}. If f(ω)≠0f(\omega ) \neq 0, then g(ω)f(ω)=−tnk∗(ω)\frac{g(\omega )}{f(\omega )}=-t_{n_{k}}^{*}(\omega ) for each k∈Nk \in \mathbb {N}. That is, tn1∗(ω)=tn2∗(ω)=⋯t_{n_{1}}^{*}(\omega )=t_{n_{2}}^{*}(\omega )=\cdots. Since tn→0t_{n} \rightarrow 0, we have that tnk∗(ω)→0t_{n_{k}}^{*}(\omega ) \rightarrow 0 as k→∞k \rightarrow \infty. It follows that tnk∗(ω)=0t_{n_{k}}^{*}(\omega )=0 for each k∈Nk \in \mathbb {N}. But then g(ω)=0g(\omega )=0, so that ω∈A(0)c\omega \in A(0)^{c}. If we instead have f(ω)=0f(\omega )=0, then clearly g(ω)=0g(\omega )=0, so again we have ω∈A(0)c\omega \in A(0)^{c}.]

(c)

A(0)∩C⊆lim⁡‾n[A(tn)∩C]A(0) \cap C \subseteq \underline{\lim }_{n}\left[A(t_{n}) \cap C\right]. [Nearly trivial.]

(d)

lim⁡‾n[A(tn)∩C]⊆A(0)∩C\overline{\lim }_{n}\left[A(t_{n}) \cap C\right] \subseteq A(0) \cap C. [Suppose that there is an increasing sequence of indices n1<n2<⋯n_{1}<n_{2}<\cdots such that ω∈A(tnk)∩C\omega \in A(t_{n_{k}}) \cap C for each k∈Nk \in \mathbb {N}. Then we have g(ω)≠−tnk∗(ω)f(ω)g(\omega ) \neq -t_{n_{k}}^{*}(\omega ) f(\omega ) for each k∈Nk \in \mathbb {N} and f(ω)≠0f(\omega ) \neq 0. Therefore, g(ω)f(ω)=−tnk∗(ω)\frac{g(\omega )}{f(\omega )}=-t_{n_{k}}^{*}(\omega ) for each k∈Nk \in \mathbb {N}, and the same reasoning as in the outline for (b) gives tnk∗(ω)=0t_{n_{k}}^{*}(\omega )=0 for each k∈Nk \in \mathbb {N}. But then g(ω)≠0g(\omega ) \neq 0, and since f(ω)≠0f(\omega ) \neq 0, we have the desired result.]

(e)

By putting (c) and (d) together, show that lim⁡n[A(tn)∩C]=A(0)∩C\lim_{n}\left[A(t_{n}) \cap C\right]=A(0) \cap C.

Problem 8.9.2

Let 1<p<+∞1<p<+\infty. Define T:Lp′→B(Lp,R)T: L^{p^{\prime }} \rightarrow \mathcal{B}\left(L^{p}, \mathbb {R}\right) [where B(Lp,R)\mathcal{B}\left(L^{p}, \mathbb {R}\right) denotes the collection of bounded linear transformations from LpL^{p} to R\mathbb {R}] by writing T(g)=LgT(g)=L_{g} for every g∈Lp′g \in L^{p^{\prime }}. [The notation LgL_{g} is as in Claim 1 of this section.]

Claim 1 (of this section). Let g∈Lp′g \in L^{p^{\prime }}, and define Lg:Lp→RL_{g}: L^{p} \rightarrow \mathbb {R} by writing Lg(f)=∫fgL_{g}(f)=\int f g for every f∈Lpf \in L^{p}. Then LgL_{g} is a bounded linear functional on LpL^{p} and ∥Lg∥=∥g∥p′\left\| L_{g}\right\| =\left\| g\right\|_{p^{\prime }}.

Riesz Representation Theorem. Let LL denote a bounded linear functional on LpL^{p}, where 1<p<+∞1<p<+\infty. Then there is h∈Lp′h \in L^{p^{\prime }} [depending on LL] such that L(f)=∫fhL(f)=\int f h for every f∈Lpf \in L^{p}.

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(a)

TT is a linear transformation from Lp′L^{p^{\prime }} to B(Lp,R)\mathcal{B}\left(L^{p}, \mathbb {R}\right). [Direct.]

(b)

TT is norm-preserving: ∥T(g)∥=∥g∥p′\left\| T(g)\right\| =\left\| g\right\|_{p^{\prime }} for every g∈Lp′g \in L^{p^{\prime }}. [Use Claim 1.]

(c)

TT is onto B(Lp,R)\mathcal{B}\left(L^{p}, \mathbb {R}\right). [Use the Riesz theorem.]

(d)

TT is one-to-one. [This follows from some basic properties of the integral.]

(e)

Lp′L^{p^{\prime }} and B(Lp,R)\mathcal{B}\left(L^{p}, \mathbb {R}\right) are isomorphic Banach spaces. [Put the previous parts together.]

Remark.
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Comment: By part (e), the spaces Lp′L^{p^{\prime }} and B(Lp,R)\mathcal{B}\left(L^{p}, \mathbb {R}\right), while consisting of formally different entities -- the former space consists of functions (or more properly equivalence classes of functions) while the latter space consists of bounded linear functionals on LpL^{p} -- are identical when viewed as Banach spaces. This interesting fact is not used anywhere in this text.