Section 8.3

LpL^{p} Space: The Case 0<p<10<p<1

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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.