Math Problems Vault

Problem 10.6.1

Framework for this section: $f: \mathbb {R} \rightarrow [0, +\infty ]$ is $\mathcal{L} / \mathcal{B}^{*}$ -measurable and such that the integral $\int_{F} f \, d\bar{\lambda }$ is finite for every compact $F \subseteq \mathbb {R}$ . For each $x \in \mathbb {R}$ , define

f^{\delta (r)}(x) = \sup _{x < u < +\infty } \frac{1}{u - x} \int _{x}^{u} f \, d\bar{\lambda }, \qquad f^{\delta (l)}(x) = \sup _{-\infty < u < x} \frac{1}{x - u} \int _{u}^{x} f \, d\bar{\lambda }, \qquad f^{\delta }(x) = \max \left\{ f^{\delta (r)}(x), f^{\delta (l)}(x)\right\} .

For each $t \in \mathbb {R}$ , define $G_{t} = \left\{ x \in \mathbb {R} : f(x) > t\right\}$ , $M_{t}^{(r)} = \left\{ x \in \mathbb {R} : f^{\delta (r)}(x) > t\right\}$ , $M_{t}^{(l)} = \left\{ x \in \mathbb {R} : f^{\delta (l)}(x) > t\right\}$ , and $M_{t} = \left\{ x \in \mathbb {R} : f^{\delta }(x) > t\right\} = M_{t}^{(r)} \cup M_{t}^{(l)}$ .

Claim 1: Let $x \in \mathbb {R}$ . Then (a) the mapping $s \mapsto \frac{1}{s - x} \int_{x}^{s} f \, d\bar{\lambda }$ on $(x, +\infty )$ is continuous, and (b) the mapping $s \mapsto \frac{1}{x - s} \int_{s}^{x} f \, d\bar{\lambda }$ on $(-\infty , x)$ is continuous.

Prove Claim 1(b) by mimicking the proof of Claim 1(a).

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Problem 10.6.2

Claim 2: We have the following: (a) for every $t \in \mathbb {R}$ , the sets $M_{t}^{(r)}$ and $M_{t}^{(l)}$ are open, hence in $\mathcal{L}$ . In particular, the proof that $M_{t}^{(r)}$ is open proceeds by picking $t \geq 0$ (the case $t < 0$ being trivial, since then $M_{t}^{(r)} = \mathbb {R}$ by the nonnegativity of $f$ ) and, in the nonempty case, taking $x_{0} \in M_{t}^{(r)}$ : since $f^{\delta (r)}\left(x_{0}\right) > t$ , there is $u_{0} \in \left(x_{0}, +\infty \right)$ with $\frac{1}{u_{0} - x_{0}} \int_{x_{0}}^{u_{0}} f \, d\bar{\lambda } > t$ , and continuity (Claim 1(b), applied to the mapping $s \mapsto \frac{1}{u_{0} - s} \int_{s}^{u_{0}} f \, d\bar{\lambda }$ on $\left(-\infty , u_{0}\right)$ ) then produces $\delta > 0$ with $\left(x_{0} - \delta , x_{0} + \delta \right) \subseteq M_{t}^{(r)}$ .

In Claim 2(a), mimic the proof of the fact that $M_{t}^{(r)}$ is open to show that $M_{t}^{(l)}$ is open.

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Problem 10.6.3

Claim 3: Let $0 < t < +\infty$ . The following equalities and inequalities obtain: (a) $\bar{\lambda }\left(M_{t}^{(r)}\right) = \frac{1}{t} \int_{M_{t}^{(r)}} f \, d\bar{\lambda }$ ; (b) $\bar{\lambda }\left(M_{t}^{(l)}\right) = \frac{1}{t} \int_{M_{t}^{(l)}} f \, d\bar{\lambda }$ ; (c) $\bar{\lambda }\left(M_{t}\right) \leq \frac{2}{t} \int_{M_{t}} f \, d\bar{\lambda }$ .

Prove Claim 3(b) by mimicking the proof of Claim 3(a).

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Problem 10.6.4

Claim 4: Let $0 < k < 1$ and $0 < t < +\infty$ . We have the following: (a) $\bar{\lambda }\left(M_{t}^{(r)}\right) \leq \frac{1}{(1 - k) t} \int_{G_{kt}} f \, d\bar{\lambda }$ ; (b) $\bar{\lambda }\left(M_{t}^{(l)}\right) \leq \frac{1}{(1 - k) t} \int_{G_{kt}} f \, d\bar{\lambda }$ ; (c) $\bar{\lambda }\left(M_{t}\right) \leq \frac{2}{(1 - k) t} \int_{G_{kt}} f \, d\bar{\lambda }$ .

Prove Claim 4(b) by mimicking the proof of Claim 4(a).

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Problem 10.6.5

Hardy-Littlewood Maximal Theorem: Let $1 < p < +\infty$ . We have the following inequalities: (a) $\left[\int_{\mathbb {R}} \left[f^{\delta (r)}\right]^{p} \, d\bar{\lambda }\right]^{1 / p} \leq \frac{p}{p - 1} \left[\int_{\mathbb {R}} f^{p} \, d\bar{\lambda }\right]^{1 / p}$ . Therefore, $f \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ forces $f^{\delta (r)} \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ and $\left\| f^{\delta (r)}\right\|_{p} \leq \frac{p}{p - 1} \left\| f\right\|_{p}$ . (b) $\left[\int_{\mathbb {R}} \left[f^{\delta (l)}\right]^{p} \, d\bar{\lambda }\right]^{1 / p} \leq \frac{p}{p - 1} \left[\int_{\mathbb {R}} f^{p} \, d\bar{\lambda }\right]^{1 / p}$ . Therefore, $f \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ forces both $f^{\delta (l)} \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ and $\left\| f^{\delta (l)}\right\|_{p} \leq \frac{p}{p - 1} \left\| f\right\|_{p}$ . (c) $\left[\int_{\mathbb {R}} \left[f^{\delta }\right]^{p} \, d\bar{\lambda }\right]^{1 / p} \leq \frac{2p}{p - 1} \left[\int_{\mathbb {R}} f^{p} \, d\bar{\lambda }\right]^{1 / p}$ . Therefore, $f \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ forces both $f^{\delta } \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ and $\left\| f^{\delta }\right\|_{p} \leq \frac{2p}{p - 1} \left\| f\right\|_{p}$ .

Prove (b) of the Hardy-Littlewood Maximal Theorem for $1 < p < +\infty$ by a fashion similar to that for the proof of part (a) there.

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Problem 10.6.6

In the Hardy-Littlewood Maximal Theorem, we had $1 < p < +\infty$ . There is a version of the theorem for the case $p = 1$ . The framework is the same as in the $1 < p < +\infty$ case: $f: \mathbb {R} \rightarrow [0, +\infty ]$ is $\mathcal{L} / \mathcal{B}^{*}$ -measurable and such that $\int_{F} f \, d\bar{\lambda } < +\infty$ for every compact $F \subseteq \mathbb {R}$ . The theorem is this: Let $E \in \mathcal{L}$ and $0 < k < 1$ . Then

\int _{E} f^{\delta (r)} \, d\bar{\lambda } \leq \frac{1}{k} \bar{\lambda }(E) + \frac{1}{1 - k} \int _{\mathbb {R}} f \log ^{+} f \, d\bar{\lambda }

and

\int _{E} f^{\delta (l)} \, d\bar{\lambda } \leq \frac{1}{k} \bar{\lambda }(E) + \frac{1}{1 - k} \int _{\mathbb {R}} f \log ^{+} f \, d\bar{\lambda } .

Also, we have

\int _{E} f^{\delta } \, d\bar{\lambda } \leq \frac{1}{k} \bar{\lambda }(E) + \frac{2}{1 - k} \int _{\mathbb {R}} f \log ^{+} f \, d\bar{\lambda } .

Here, $\log^{+} f$ denotes $\max \left\{ \log f, 0\right\}$ .

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Remark.

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Verify each of the following steps by appealing to the appropriate results of this section:

\begin{aligned} \int _{E} f^{\delta (r)} \, d\bar{\lambda } & = \int _{0}^{\infty } \bar{\lambda }\left(\left\{ y \in \mathbb {R} : \; \mathbb {1}_{E}(y) f^{\delta (r)}(y) > t\right\} \right) \, d\bar{\lambda }(t) \\ & = \int _{0}^{\infty } \bar{\lambda }\left(M_{t}^{(r)} \cap E\right) \, d\bar{\lambda }(t) \\ & = \int _{0}^{1 / k} \bar{\lambda }\left(M_{t}^{(r)} \cap E\right) \, d\bar{\lambda }(t) + \int _{1 / k}^{\infty } \bar{\lambda }\left(M_{t}^{(r)} \cap E\right) \, d\bar{\lambda }(t) \\ & \overset {(*)}{\leq } \frac{1}{k} \bar{\lambda }(E) + \int _{1 / k}^{\infty } \frac{1}{(1 - k) t} \int _{G_{kt}} f(x) \, d\bar{\lambda }(x) \, d\bar{\lambda }(t) \\ & = \frac{1}{k} \bar{\lambda }(E) + \frac{1}{1 - k} \int _{1 / k}^{\infty } \frac{1}{t} \int _{\mathbb {R}} \; \mathbb {1}_{G_{kt}}(x) f(x) \, d\bar{\lambda }(x) \, d\bar{\lambda }(t) \\ & = \frac{1}{k} \bar{\lambda }(E) + \frac{1}{1 - k} \int _{\mathbb {R}} f(x) \int _{1 / k}^{\infty } \frac{1}{t} \; \mathbb {1}_{G_{kt}}(x) \, d\bar{\lambda }(t) \, d\bar{\lambda }(x) . \end{aligned}

Next, show that $\int_{1 / k}^{\infty } \frac{1}{t} \; \mathbb {1}_{G_{kt}}(x) \, d\bar{\lambda }(t)$ is zero or $\log f(x)$ as $0 \leq f(x) \leq 1$ or $f(x) > 1$ . That is, show that $\int_{1 / k}^{\infty } \frac{1}{t} \; \mathbb {1}_{G_{kt}}(x) \, d\bar{\lambda }(t) = \log^{+} f(x)$ for each $x \in \mathbb {R}$ . This gives the $\delta (r)$ -inequality, and the $\delta (l)$ -inequality is achieved in virtually the same fashion. For the $\delta$ -inequality, the same argument as above also works upon replacing $\delta (r)$ by $\delta$ , $M_{t}^{(r)}$ by $M_{t}$ , and using the inequality $\bar{\lambda }\left(M_{t}\right) \leq \frac{2}{(1 - k) t} \int_{G_{kt}} f \, d\bar{\lambda }$ in $(*)$ .

Problem 10.6.7

Recall the Hardy-Littlewood Maximal Theorem, where $1 < p < +\infty$ : if $f \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ , then $f^{\delta (r)}, f^{\delta (l)}, f^{\delta } \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ and

\left\| f^{\delta (r)}\right\| _{p}, \; \left\| f^{\delta (l)}\right\| _{p} \leq \frac{p}{p - 1} \left\| f\right\| _{p} \qquad \text{and} \qquad \left\| f^{\delta }\right\| _{p} \leq \frac{2p}{p - 1} \left\| f\right\| _{p} .

One question that may arise is whether the constants $\frac{p}{p - 1}$ and $\frac{2p}{p - 1}$ are the sharpest possible constants. It will turn out that the answer is in the affirmative. Fix $1 < p < +\infty$ , and for each $n \in \mathbb {N}$ define $f_{n}$ on $\mathbb {R}$ by the rule $f_{n}(x) = x^{\left(\frac{1}{n} - 1\right) \frac{1}{p}} \; \mathbb {1}_{(0, 1)}(x)$ . Let $A_{n, p}$ be such that $\left\| f_{n}^{\delta (r)}\right\|_{p} = A_{n, p} \left\| f_{n}\right\|_{p}$ . Show that $\lim_{n} A_{n, p} = \frac{p}{p - 1}$ , and deduce that the constant $\frac{p}{p - 1}$ in the Hardy-Littlewood Inequality is the best possible constant.

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Remark.

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It is straightforward calculus to compute $\left\| f_{n}\right\|_{p}$ for each $n$ . To obtain the form of $f_{n}^{\delta (r)}(x)$ for each $n$ and $x$ , consider the cases $x \geq 1$ and $x < 1$ and use calculus; this won't be hard but it won't be trivial. From this $\left\| f_{n}^{\delta (r)}\right\|_{p}$ may be obtained for each $n$ , so that the ratio $\left\| f_{n}^{\delta (r)}\right\|_{p} / \left\| f_{n}\right\|_{p}$ may be computed for each $n$ . That this quotient tends to $\frac{p}{p - 1}$ as $n \rightarrow +\infty$ should be seen in straightforward fashion.

Problem 10.6.8

For a function $f$ as given in the framework for this section [ $f: \mathbb {R} \rightarrow [0, +\infty ]$ is $\mathcal{L} / \mathcal{B}^{*}$ -measurable and such that $\int_{F} f \, d\bar{\lambda } < +\infty$ for every compact $F \subseteq \mathbb {R}$ ], we defined $f^{\delta }(x)$ to be the maximum of $f^{\delta (r)}(x)$ and $f^{\delta (l)}(x)$ . We may equivalently define $f^{\delta }(x)$ for each $x \in \mathbb {R}$ by the rule

f^{\delta }(x) = \sup \left\{ \frac{1}{u - t} \int _{t}^{u} f \, d\bar{\lambda } : -\infty < t \leq x \leq u < +\infty , \, t \neq u\right\} .

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Problem 10.6.9

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

Comment: Our presentation in the text and in the exercises is based on one form of the Hardy-Littlewood framework. We now present a series of exercises dealing with a more general framework. The measure space will be $\left(\mathbb {R}^{k}, \mathcal{L}^{k}, \bar{\lambda }_{k}\right)$ , where $k \in \mathbb {N}$ is fixed. The notation $B(\boldsymbol {x}, r)$ denotes the ball of radius $r$ centered at $\boldsymbol {x} \in \mathbb {R}^{k}$ . Given $0 < a < +\infty$ , we will write $a B(\boldsymbol {x}, r)$ for $B(\boldsymbol {x}, ar)$ . The following exercises give the background leading up to Jones' presentation of an alternative version of the Hardy-Littlewood Maximal Theorem.

Prove the Vitali Covering Theorem: Let $E \subseteq \mathbb {R}^{k}$ be bounded. Let $\mathcal{F}$ denote a collection of open balls centered at points of $E$ such that every $\boldsymbol {x} \in E$ is the center of some ball in $\mathcal{F}$ . Then there exists an at most countable subcollection $B_{1}, B_{2}, \cdots$ of balls in $\mathcal{F}$ such that (i) $B_{1}, B_{2}, \cdots$ are disjoint, and (ii) $E \subseteq \bigcup_{n \geq 1} 3 B_{n}$ . Follow the given outline.

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(a)

The theorem is immediate if the radii of the balls are not bounded above.

(b)

There is $n$ such that $B \cap B_{n} \neq \varnothing$ .

(c)

We have $B \cap \bigcup_{m < n} B_{m} = \varnothing$ , hence $\operatorname {rad} B \leq d_{n} < 2 \operatorname {rad} B_{n}$ .

(d)

$\boldsymbol {x} \in 3 B_{n}$ , hence $E \subseteq \bigcup_{m \geq 1} 3 B_{m}$ , completing the proof.

Remark.

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(a) In this case, we can find a ball $B$ centered at some point of $E$ with radius so large that $E \subseteq B$ , trivially giving (i) and (ii). Therefore, we may assume that the radii of the balls is bounded above. And of course we may assume that $E$ is nonempty.

Perform the following inductive scheme: Assume that $B_{1}, \cdots , B_{n - 1}$ have been selected. Let $d_{n} = \sup \left\{ \operatorname {rad} B : B \in \mathcal{F}, \, B \cap \bigcup_{m < n} B_{m} = \varnothing \right\}$ . If there are no $B \in \mathcal{F}$ with $B \cap \bigcup_{m < n} B_{m} = \varnothing$ , the scheme terminates with the sets $B_{1}, \cdots , B_{n - 1}$ . Otherwise, choose $B_{n} \in \mathcal{F}$ such that $\operatorname {rad} B_{n} > \frac{1}{2} d_{n}$ and $B_{n} \cap \bigcup_{m < n} B_{m} = \varnothing$ .

The above scheme is problem-free, and the resulting collection $B_{1}, B_{2}, \cdots$ is clearly disjoint. We must show that $E \subseteq \bigcup_{n \geq 1} 3 B_{n}$ . Fix $\boldsymbol {x} \in E$ and find $B \in \mathcal{F}$ that is centered at $\boldsymbol {x}$ .

(b) Suppose that $B \cap B_{n} = \varnothing$ for each $n$ . Then the selection process never terminates and $\operatorname {rad} B \leq d_{n}$ for each $n$ . Therefore, we have a countably infinite collection $B_{1}, B_{2}, \cdots$ of balls centered at points of $E$ with $\operatorname {rad} B_{n} > \frac{1}{2} d_{n} \geq \frac{1}{2} \operatorname {rad} B > 0$ . Now $\bigcup_{n} B_{n}$ is bounded since $E$ is bounded and the radii of the balls in $\mathcal{F}$ are bounded above, and hence $\bar{\lambda }_{k}\left(\bigcup_{n} B_{n}\right) < +\infty$ . But at the same time $\bar{\lambda }_{k}\left(\bigcup_{n} B_{n}\right) = \sum_{n} \bar{\lambda }_{k}\left(B_{n}\right) = +\infty$ by the earlier inequality. This contradiction gives the result.

By (b) there is $n$ with $B \cap B_{n} \neq \varnothing$ . Let $n$ henceforth denote the smallest such $n$ with $B \cap B_{n} \neq \varnothing$ . Pick $\boldsymbol {y} \in B \cap B_{n}$ .

(d) Let $\boldsymbol {z}$ denote the center of $B_{n}$ . Then $\left\| \boldsymbol {x} - \boldsymbol {z}\right\| \leq \left\| \boldsymbol {x} - \boldsymbol {y}\right\| + \left\| \boldsymbol {y} - \boldsymbol {z}\right\| < \operatorname {rad} B + \operatorname {rad} B_{n} < 3 \operatorname {rad} B_{n}$ . Therefore, we have that $\boldsymbol {x} \in 3 B_{n}$ .

Problem 10.6.10

Definition: Let $f: \mathbb {R}^{k} \rightarrow \overline{\mathbb {R}}$ be such that $\int_{F} f \, d\bar{\lambda }_{k} < +\infty$ for each compact $F \subseteq \mathbb {R}^{k}$ . The Hardy-Littlewood maximal function for $f$ is the function $Mf$ on $\mathbb {R}^{k}$ given for all $\boldsymbol {x} \in \mathbb {R}^{k}$ by

Mf(\boldsymbol {x}) = \sup _{0 < r < +\infty } \frac{1}{\bar{\lambda }_{k}\left(B(\boldsymbol {x}, r)\right)} \int _{B(\boldsymbol {x}, r)} \left|f(\boldsymbol {y})\right| \, d\bar{\lambda }_{k}(\boldsymbol {y}) .

$Mf$ is lower semicontinuous, hence is $\mathcal{B}^{k} / \mathcal{B}^{*}$ -measurable. In particular, $Mf$ is $\mathcal{L}^{k} / \mathcal{B}^{*}$ -measurable.

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Remark.

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Suppose that $Mf(\boldsymbol {x}) > t$ . There is $0 < r < +\infty$ such that $t < \frac{1}{\bar{\lambda }_{k}\left(B(\boldsymbol {x}, r)\right)} \int_{B(\boldsymbol {x}, r)} \left|f(\boldsymbol {y})\right| \, d\bar{\lambda }_{k}(\boldsymbol {y})$ . We may choose $r^{\prime } > r$ such that the inequality $t < \frac{1}{\bar{\lambda }_{k}\left(B(\boldsymbol {x}, r^{\prime })\right)} \int_{B(\boldsymbol {x}, r)} \left|f(\boldsymbol {y})\right| \, d\bar{\lambda }_{k}(\boldsymbol {y})$ obtains. Now $\left\| \boldsymbol {x} - \boldsymbol {x}^{\prime }\right\| \leq r^{\prime } - r$ forces $B(\boldsymbol {x}, r) \subseteq B\left(\boldsymbol {x}^{\prime }, r^{\prime }\right)$ , and the translation invariance of $\bar{\lambda }_{k}$ gives

\begin{aligned} t & < \frac{1}{\bar{\lambda }_{k}\left(B\left(\boldsymbol {x}, r^{\prime }\right)\right)} \int _{B\left(\boldsymbol {x}^{\prime }, r^{\prime }\right)} \left|f(\boldsymbol {y})\right| \, d\bar{\lambda }_{k}(\boldsymbol {y}) \\ & = \frac{1}{\bar{\lambda }_{k}\left(B\left(\boldsymbol {x}^{\prime }, r^{\prime }\right)\right)} \int _{B\left(\boldsymbol {x}^{\prime }, r^{\prime }\right)} \left|f(\boldsymbol {y})\right| \, d\bar{\lambda }_{k}(\boldsymbol {y}) \\ & \leq M f\left(\boldsymbol {x}^{\prime }\right) . \end{aligned}

Therefore, for all $\boldsymbol {x}^{\prime }$ sufficiently close to $\boldsymbol {x}$ we have $Mf\left(\boldsymbol {x}^{\prime }\right) > t$ .

Problem 10.6.11

Let $f \in L^{1}\left(\mathbb {R}^{k}, \mathcal{L}^{k}, \bar{\lambda }_{k}\right)$ . Then $\bar{\lambda }_{k}\left(\left\{ \boldsymbol {x} \in \mathbb {R}^{k} : Mf(\boldsymbol {x}) > t\right\} \right) \leq 3^{k} t^{-1} \left\| f\right\|_{1}$ for each $0 < t < +\infty$ . This is another version of the Hardy-Littlewood Maximal Theorem.

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Remark.

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Let $E$ denote $\left\{ \boldsymbol {x} \in \mathbb {R}^{k} : Mf(\boldsymbol {x}) > t\right\}$ , where $0 < t < +\infty$ is given and fixed. If $E = \varnothing$ the result is trivial. Otherwise, pick $\boldsymbol {x} \in E$ , so that $Mf(\boldsymbol {x}) > t$ and there is $0 < r < +\infty$ with $\frac{1}{\bar{\lambda }_{k}\left(B(\boldsymbol {x}, r)\right)} \int_{B(\boldsymbol {x}, r)} \left|f(\boldsymbol {y})\right| \, d\bar{\lambda }_{k}(\boldsymbol {y}) > t$ . Therefore, for any $\boldsymbol {x} \in E$ there is a ball $B$ with center $\boldsymbol {x}$ such that $\bar{\lambda }_{k}(B) < \frac{1}{t} \int_{B} \left|f(\boldsymbol {y})\right| \, d\bar{\lambda }_{k}(\boldsymbol {y})$ . Pick positive real $q$ , and let $\mathcal{F}$ denote the collection of open balls $B$ with centers in $E \cap B(\boldsymbol {0}, q)$ and satisfying the just-mentioned inequality. If $E \cap B(\boldsymbol {0}, q) \neq \varnothing$ , there exist disjoint balls $B_{1}, B_{2}, \cdots \in \mathcal{F}$ such that $E \cap B(\boldsymbol {0}, q) \subseteq \bigcup_{n \geq 1} 3 B_{n}$ . (Vitali Covering Theorem.) Verify that

\begin{aligned} \bar{\lambda }_{k}\left(E \cap B(\boldsymbol {0}, q)\right) & \leq \sum _{n \geq 1} \bar{\lambda }_{k}\left(3 B_{n}\right) = \sum _{n = 1}^{\infty } 3^{k} \bar{\lambda }_{k}\left(B_{n}\right) < \sum _{n \geq 1} \frac{3^{k}}{t} \int _{B_{n}} \left|f(\boldsymbol {y})\right| \, d\bar{\lambda }_{k}(\boldsymbol {y}) \\ & = \frac{3^{k}}{t} \int _{\bigcup _{n \geq 1} B_{n}} \left|f(\boldsymbol {y})\right| \, d\bar{\lambda }_{k}(\boldsymbol {y}) \\ & \leq 3^{k} t^{-1} \int _{\mathbb {R}^{k}} \left|f(\boldsymbol {y})\right| \, d\bar{\lambda }_{k}(\boldsymbol {y}) . \end{aligned}

Now let $q \rightarrow +\infty$ for the result.

Problem 10.6.12

Reconsider Exercise 3 of Section 10.4 in the light of this section.

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