Section 8.1

Lp\boldsymbol{L}^{\boldsymbol{p}} Space: The Case 1⩽p<+∞\mathbf{1} \leqslant \boldsymbol{p}<+\infty

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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}.