Math Problems Vault

Problem 10.4.1

Generalized Minkowski Inequality: Let $\left(\Omega_{1}, \mathcal{F}_{1}, \mu_{1}\right)$ and $\left(\Omega_{2}, \mathcal{F}_{2}, \mu_{2}\right)$ denote $\sigma$ -finite measure spaces, let $1 \leq p < +\infty$ , and let $f: \Omega_{1} \times \Omega_{2} \rightarrow \overline{\mathbb {R}}$ be $\mathcal{F}_{1} \otimes \mathcal{F}_{2} / \mathcal{B}^{*}$ -measurable and such that

\int _{\Omega _{2}} \left[\int _{\Omega _{1}} \left|f\left(\omega _{1}, \omega _{2}\right)\right|^{p} \, d\mu _{1}\left(\omega _{1}\right)\right]^{1 / p} d\mu _{2}\left(\omega _{2}\right) < +\infty .

Then

\left[\int _{\Omega _{1}} \left|\int _{\Omega _{2}} f\left(\omega _{1}, \omega _{2}\right) \, d\mu _{2}\left(\omega _{2}\right)\right|^{p} d\mu _{1}\left(\omega _{1}\right)\right]^{1 / p} \leq \int _{\Omega _{2}} \left[\int _{\Omega _{1}} \left|f\left(\omega _{1}, \omega _{2}\right)\right|^{p} \, d\mu _{1}\left(\omega _{1}\right)\right]^{1 / p} d\mu _{2}\left(\omega _{2}\right) .

Consider the $\sigma$ -finite measure space $\left(\Omega , \mathcal{F}, \mu \right)$ and consider the finite measure space $\left(\Omega_{2}, \mathcal{F}_{2}, \mu_{2}\right)$ , where $\Omega_{2} = \left\{ a, b\right\}$ , $\mathcal{F}_{2} = 2^{\Omega_{2}}$ , and $\mu_{2}$ denotes counting measure. Let $f, g \in L^{p}\left(\Omega , \mathcal{F}, \mu \right)$ , where $1 \leq p < +\infty$ . Define $h: \Omega \times \Omega_{2} \rightarrow \overline{\mathbb {R}}$ by writing $h\left(\omega , a\right) = f(\omega )$ and $h\left(\omega , b\right) = g(\omega )$ for each $\omega \in \Omega$ . Here, the Generalized Minkowski Inequality becomes as follows: Let $f, g \in L^{p}\left(\Omega , \mathcal{F}, \mu \right)$ . Then $f + g \in L^{p}\left(\Omega , \mathcal{F}, \mu \right)$ and $\left\| f + g\right\|_{p} \leq \left\| f\right\|_{p} + \left\| g\right\|_{p}$ .

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Problem 10.4.2

Let $0 < q \leq p < +\infty$ . Without worrying about measurability conditions, obtain the inequality

\left[\int _{\Omega _{1}} \left(\int _{\Omega _{2}} \left|f\left(\omega _{1}, \omega _{2}\right)\right|^{q} \, d\mu _{2}\left(\omega _{2}\right)\right)^{p / q} d\mu _{1}\left(\omega _{1}\right)\right]^{1 / p} \leq \left[\int _{\Omega _{2}} \left(\int _{\Omega _{1}} \left|f\left(\omega _{1}, \omega _{2}\right)\right|^{p} \, d\mu _{1}\left(\omega _{1}\right)\right)^{q / p} d\mu _{2}\left(\omega _{2}\right)\right]^{1 / q} .

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Remark.

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We have $1 \leq p / q < +\infty$ . Play with exponents and use the Generalized Minkowski Inequality.

Problem 10.4.3

(a)(b)(c)

Let $\mathcal{L}$ denote the Lebesgue subsets of $(0, +\infty )$ , and let $\bar{\lambda }$ denote Lebesgue measure on the Lebesgue subsets of $(0, +\infty )$ . Consider the $\sigma$ -finite measure space $\left((0, +\infty ), \mathcal{L}, \bar{\lambda }\right)$ . Given a function $f \in L^{1}\left((0, +\infty ), \mathcal{L}, \bar{\lambda }\right)$ , associate the function $F: (0, +\infty ) \rightarrow [0, +\infty )$ given by $F(x) = \frac{1}{x} \int_{0}^{x} f(y) \, d\bar{\lambda }(y)$ .

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(a)

For $1 < p < +\infty$ we have $F \in L^{p}\left((0, +\infty ), \mathcal{L}, \bar{\lambda }\right)$ and $\left\| F\right\|_{p} \leq p^{\prime } \left\| f\right\|_{p}$ .

(b)

The constant $p^{\prime } = \frac{p}{p - 1}$ is the best [smallest] possible constant for the inequality in (a).

(c)

We have excluded the $p = 1$ case so far. Show that if $f \in L^{1}\left((0, +\infty ), \mathcal{L}, \bar{\lambda }\right)$ is positive, then $F \notin L^{1}\left((0, +\infty ), \mathcal{L}, \bar{\lambda }\right)$ .

Remark.

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(a) Replace $y$ by $xt$ , and then use the main result of this section.

(b) Consider $f(x) = x^{\epsilon - \frac{1}{p}} \; \mathbb {1}_{(0, 1)}(x)$ for each $x \in (0, +\infty )$ .

Problem 10.4.4

This exercise is a lemma for the next exercise. Let $\left(\Omega , \mathcal{F}, \mu \right)$ denote a measure space with $\mu (\Omega ) < +\infty$ . Every $\mathcal{F} / \mathcal{B}^{*}$ -measurable function $f: \Omega \rightarrow [0, +\infty ]$ satisfies

\lim _{p \rightarrow \infty } \left[\int _{\Omega } f(\omega )^{p} \, d\mu (\omega )\right]^{1 / p} = \operatorname {ess} \sup _{\omega \in \Omega } f(\omega ) .

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Remark.

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Suppose $f$ is not essentially bounded. Then $\mu \left(\left\{ \omega \in \Omega : f(\omega ) \geq M\right\} \right) > 0$ has positive $\mu$ -measure for each $M > 0$ . Show for each $M > 0$ and $p > 1$ that

0 < M \mu \left(\left\{ \omega \in \Omega : f(\omega ) \geq M\right\} \right)^{1 / p} \leq \left[\int _{\Omega } \left|f(\omega )\right|^{p} \, d\mu (\omega )\right]^{1 / p}

by splitting the integral into the integral over the set of $\omega$ where $f(\omega ) \geq M$ and the set of $\omega$ where $f(\omega ) < M$ . Next, $\lim_{p \rightarrow \infty } \mu \left(\left\{ \omega \in \Omega : f(\omega ) \geq M\right\} \right)^{1 / p} = 1$ since $\mu (\Omega ) < +\infty$ . Therefore, given any large $M$ , we may choose $p$ so large that $\left(\int_{\Omega } \left|f(\omega )\right|^{p} \, d\mu (\omega )\right)^{1 / p}$ exceeds, say, $M / 2$ . This gives the result when $f$ is not essentially bounded. If on the other hand $f$ is essentially bounded, then $f \in L^{p}\left(\Omega , \mathcal{F}, \mu \right)$ for every $p \geq 1$ by the fact that $\mu (\Omega ) < +\infty$ . We have shown earlier that $\left\| f\right\|_{p} \rightarrow \left\| f\right\|_{\infty }$ in this case; see Exercise 4 in Section 8.5.

Problem 10.4.5

(a)(b)(c)(d)

We proved the Generalized Minkowski Inequality for $1 \leq p < +\infty$ . There is a version for the case $p = +\infty$ . Let $\left(\Omega_{1}, \mathcal{F}_{1}, \mu_{1}\right)$ and $\left(\Omega_{2}, \mathcal{F}_{2}, \mu_{2}\right)$ denote $\sigma$ -finite measure spaces, and let $f: \Omega_{1} \times \Omega_{2} \rightarrow [0, +\infty ]$ be $\mathcal{F}_{1} \otimes \mathcal{F}_{2} / \mathcal{B}^{*}$ -measurable.

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(a)

The mapping $\omega_{2} \mapsto \operatorname {ess} \sup_{\omega_{1} \in \Omega_{1}} f\left(\omega_{1}, \omega_{2}\right)$ is $\mathcal{F}_{2} / \mathcal{B}^{*}$ -measurable, hence we may meaningfully discuss $\int_{\Omega_{2}} \operatorname {ess} \sup_{\omega_{1} \in \Omega_{1}} f\left(\omega_{1}, \omega_{2}\right) \, d\mu_{2}\left(\omega_{2}\right)$ .

From now on, assume that $\int_{\Omega_{2}} \operatorname {ess} \sup_{\omega_{1} \in \Omega_{1}} f\left(\omega_{1}, \omega_{2}\right) \, d\mu_{2}\left(\omega_{2}\right) < +\infty$ .

(b)

For $\mu_{1}$ -a.e. $\omega_{1} \in \Omega_{1}$ the mapping $\omega_{2} \mapsto f\left(\omega_{1}, \omega_{2}\right)$ is in $L^{1}\left(\Omega_{2}, \mathcal{F}_{2}, \mu_{2}\right)$ . That is, for $\mu_{1}$ -a.e. $\omega_{1} \in \Omega_{1}$ we have $\int_{\Omega_{2}} f\left(\omega_{1}, \omega_{2}\right) \, d\mu_{2}\left(\omega_{2}\right) < +\infty$ .

(c)

The [nonnegative] mapping $\omega_{1} \mapsto \int_{\Omega_{2}} f\left(\omega_{1}, \omega_{2}\right) \, d\mu_{2}\left(\omega_{2}\right)$ defined on $\Omega_{1}$ is $\mathcal{F}_{1} / \mathcal{B}^{*}$ -measurable, hence we may meaningfully discuss $\operatorname {ess} \sup_{\omega_{1} \in \Omega_{1}} \int_{\Omega_{2}} f\left(\omega_{1}, \omega_{2}\right) \, d\mu_{2}\left(\omega_{2}\right)$ .

(d)

We have the inequality

\operatorname {ess} \sup _{\omega _{1} \in \Omega _{1}} \int _{\Omega _{2}} f\left(\omega _{1}, \omega _{2}\right) \, d\mu _{2}\left(\omega _{2}\right) \leq \int _{\Omega _{2}} \operatorname {ess} \sup _{\omega _{1} \in \Omega _{1}} f\left(\omega _{1}, \omega _{2}\right) \, d\mu _{2}\left(\omega _{2}\right) .