Return to Answer

I shared your confusion until a year ago, when Markus Haase explained to me that (2) can actually be rewritten as assertion (ii) in the following theorem, which I find much clearer.

(i) The operator family $(T_i)_{i \in I}$ satisfies a maximal inequality, i.e., there is a constant $C \ge 0$ with the following property: for every $f \in E$ there exists $0 \le h \in L^p$ of norm $\|h\|_{L^p} \le C \|f\|_{E}$ such that $|T_if| \le h$ for all $i \in I$ (where the inequality is to be understood pointwise almost everywhere).

(ii) There exists a constant $C$ with the following property: for every finite measurable partition $\Omega = A_1 \, \dot \cup \, \dots \, \dot \cup \, A_n$, all indices $i_1, \dots, i_n \in I$, and every $f \in E$ one has $\Big\| \sum_{k=1}^n 1_{A_k} T_{i_k} f \Big\|_{L^p} \le C \|f\|_{E}$.

For each $F \in \mathcal{F}$ the vector $0 \le \bigvee_{i \in F} |T_if| \in L^p$ (where $|\cdot|$ denotes the pointwise almost everywhere modulus and $\bigvee$ denotes the pointwise almost everywhere supremum) is norm bounded by $C\|f\|_{E}$. To see this, enumerate the elements of $F$ as $i_1, \dots, i_n$ and choose a measurable partition of $\Omega$ into sets $A_1, \dots, A_n$ such that $$ \bigvee_{i \in F} |T_if| = \sum_{k=1}^n 1_{A_k} |T_{i_k} f| = \Big| \sum_{k=1}^n 1_{A_k} T_{i_k} f \Big| . $$ Since the net $\big(\bigvee_{i \in F} |T_if|\big)_{F \in \mathcal{F}}$ in $L^p$ is increasing and norm bounded by $C\|f\|_{E}$, it is norm convergent to a vector $0 \le h \in L^p$ of norm $\le C\|f\|_{E}$ (here was used that $p \in [1,\infty)$). Clearly, $h$ satisfies the property claimed in (i). $\quad \square$

Remarks. (a) Assertion (i) is a convenient way to write down a maximal inequality if the index set $I$ is not assumed to be countable and if one does not make any regularity assumptions regarding the dependence of $T_i$ on $i$. Writing down the maximal operator before knowing whether a maximal inequality holds leads to measurability problems in this case. (But once the maximal inequality is established, one can define the maximal operator by using the supremum within the ordered space $L^p$ rather than the almost everyhwere supremum.)

(b) In the situation of the question, assertion (ii) of the theorem is precisely the estimate (2) for simple functions $t: \mathbb{R}^n \to (0,1)$. So this shows, in particular, that it suffices to consider simple functions in (2).

(e) Fun fact: The existence of $C$ is actually redundant in (i). Indeed, if one only assumes that for each $f \in E$ there exists $0 \le h \in L^p$ such that $|T_i f| \le h$ for all $i \in I$, then there automatically exists a number $C \ge 0$ such that $h$ can always be chosen to satisfy $\|h\|_{L^p} \le C \|f\|_{E}$. This can be shown by means of the closed graph theorem, see for instance (warning: shameless self-promotion ahead) Theorem 4.1 in this preprint.

$\newcommand\dotcup{\mathbin{\dot\cup}}$I shared your confusion until a year ago, when Markus Haase explained to me that \eqref{2} can actually be rewritten as assertion (ii) in the following theorem, which I find much clearer.

(i) The operator family $(T_i)_{i \in I}$ satisfies a maximal inequality, i.e., there is a constant $C \ge 0$ with the following property: for every $f \in E$ there exists $0 \le h \in L^p$ of norm $\lVert h\rVert_{L^p} \le C \lVert f\rVert_{E}$ such that $\lvert T_if\rvert \le h$ for all $i \in I$ (where the inequality is to be understood pointwise almost everywhere).

(ii) There exists a constant $C$ with the following property: for every finite measurable partition $\Omega = A_1 \dotcup \dotsb \dotcup A_n$, all indices $i_1, \dotsc, i_n \in I$, and every $f \in E$ one has $\Bigl\lVert \sum_{k=1}^n 1_{A_k} T_{i_k} f \Bigr\rVert_{L^p} \le C \lVert f\rVert_{E}$.

For each $F \in \mathcal{F}$ the vector $0 \le \bigvee_{i \in F} \lvert T_if\rvert \in L^p$ (where $\lvert\cdot\rvert$ denotes the pointwise almost everywhere modulus and $\bigvee$ denotes the pointwise almost everywhere supremum) is norm bounded by $C\lVert f\rVert_{E}$. To see this, enumerate the elements of $F$ as $i_1, \dotsc, i_n$ and choose a measurable partition of $\Omega$ into sets $A_1, \dotsc, A_n$ such that $$ \bigvee_{i \in F} \lvert T_if\rvert = \sum_{k=1}^n 1_{A_k} \lvert T_{i_k} f\rvert = \Bigl\lVert \sum_{k=1}^n 1_{A_k} T_{i_k} f \Bigr\rVert . $$ Since the net $\bigl(\bigvee_{i \in F} |T_if|\bigr)_{F \in \mathcal{F}}$ in $L^p$ is increasing and norm bounded by $C\lVert f\rVert_{E}$, it is norm convergent to a vector $0 \le h \in L^p$ of norm $\le C\lVert f\rVert_{E}$ (here was used that $p \in [1,\infty)$). Clearly, $h$ satisfies the property claimed in (i). $\quad \square$

Remarks. (a) Assertion (i) is a convenient way to write down a maximal inequality if the index set $I$ is not assumed to be countable and if one does not make any regularity assumptions regarding the dependence of $T_i$ on $i$. Writing down the maximal operator before knowing whether a maximal inequality holds leads to measurability problems in this case. (But once the maximal inequality is established, one can define the maximal operator by using the supremum within the ordered space $L^p$ rather than the almost everywhere supremum.)

(b) In the situation of the question, assertion (ii) of the theorem is precisely the estimate \eqref{2} for simple functions $t: \mathbb{R}^n \to (0,1)$. So this shows, in particular, that it suffices to consider simple functions in \eqref{2}.

(e) Fun fact: The existence of $C$ is actually redundant in (i). Indeed, if one only assumes that for each $f \in E$ there exists $0 \le h \in L^p$ such that $\lvert T_i f\rvert \le h$ for all $i \in I$, then there automatically exists a number $C \ge 0$ such that $h$ can always be chosen to satisfy $\lVert h\rVert_{L^p} \le C \lVert f\rVert_{E}$. This can be shown by means of the closed graph theorem, see for instance (warning: shameless self-promotion ahead) Theorem 4.1 in the preprint Order boundedness and order continuity properties of positive operator semigroups with Michael Kaplin.

For each $F \in \mathcal{F}$ the vector $0 \le \bigvee_{i \in F} |T_if| \in L^p$ (where $|\cdot|$ denotes the pointwise almost everywhere modulus and $\bigvee$ denotes the pointwise almost everywhere supremum) is norm bounded by $C\|f\|_{E}$. Indeed, enumerate the elements of $F$ as $i_1, \dots, i_n$ and choose a measurable partition of $\Omega$ into sets $A_1, \dots, A_n$ such that $$ \bigvee_{i \in F} |T_if| = \sum_{k=1}^n 1_{A_k} |T_{i_k} f| = \Big| \sum_{k=1}^n 1_{A_k} T_{i_k} f \Big| . $$ Since the net $\big(\bigvee_{i \in F} |T_if|\big)_{F \in \mathcal{F}}$ in $L^p$ is increasing and norm bounded by $C\|f\|_{E}$, it is norm convergent to a vector $0 \le h \in L^p$ of norm $\le C\|f\|_{E}$ (here was used that $p \in [1,\infty)$). Clearly, $f$ satisfies the property claimed in (i). $\quad \square$

For each $F \in \mathcal{F}$ the vector $0 \le \bigvee_{i \in F} |T_if| \in L^p$ (where $|\cdot|$ denotes the pointwise almost everywhere modulus and $\bigvee$ denotes the pointwise almost everywhere supremum) is norm bounded by $C\|f\|_{E}$. To see this, enumerate the elements of $F$ as $i_1, \dots, i_n$ and choose a measurable partition of $\Omega$ into sets $A_1, \dots, A_n$ such that $$ \bigvee_{i \in F} |T_if| = \sum_{k=1}^n 1_{A_k} |T_{i_k} f| = \Big| \sum_{k=1}^n 1_{A_k} T_{i_k} f \Big| . $$ Since the net $\big(\bigvee_{i \in F} |T_if|\big)_{F \in \mathcal{F}}$ in $L^p$ is increasing and norm bounded by $C\|f\|_{E}$, it is norm convergent to a vector $0 \le h \in L^p$ of norm $\le C\|f\|_{E}$ (here was used that $p \in [1,\infty)$). Clearly, $h$ satisfies the property claimed in (i). $\quad \square$

I shared your confusion until a year ago, when Markus Haase explained to me that (2) can actually be rewritten as assertion (ii) in the following theorem, which was I find much clearer.

Theorem. Let X be a Banach space, let $(\Omega,\mu)$ be a measure space, let $p \in [1,\infty)$ and let $I$ be a non-empty index set. For each $i \in I$ let $T_i: X \to L^p := L^p(\Omega,\mu)$ be a bounded linear operator. The following are equivalent:

(i) The operator family $(T_i)_{i \in I}$ satisfies a maximal inequality, i.e., there is a constant $C \ge 0$ with the following property: for every $x \in X$ there exists $0 \le f \in L^p$ of norm $\|f\| \le C \|x\|$ such that $|T_ix| \le f$ for all $i \in I$ (where the inequality is to be understood pointwise almost everywhere).

(ii) There exists a constant $C$ with the following property: for every finite measurable partition $\Omega = A_1 \, \dot \cup \, \dots \, \dot \cup \, A_n$, all indices $i_1, \dots, i_n \in I$, and every $x \in X$ one has $\Big\| \sum_{k=1}^n 1_{A_k} T_{i_k} x \Big\| \le C \|x\|$.

"(ii) $\Rightarrow$ (i)" Let $\mathcal{F}$ denote the set of all non-empty finite subsets of $I$, which is directed with respect to set inclusion. Fix $x \in X$.

For each $F \in \mathcal{F}$ the vector $0 \le \bigvee_{i \in F} |T_ix| \in L^p$ (where $|\cdot|$ denotes the pointwise almost everywhere modulus and $\bigvee$ denotes the pointwise almost everywhere supremum) is norm bounded by $C\|x\|$. Indeed, enumerate the elements of $F$ as $i_1, \dots, i_n$ and choose a measurable partition of $\Omega$ into sets $A_1, \dots, A_n$ such that $$ \bigvee_{i \in F} |T_ix| = \sum_{k=1}^n 1_{A_k} |T_{i_k} x| = \Big| \sum_{k=1}^n 1_{A_k} T_{i_k} x \Big| . $$ Since the net $\big(\bigvee_{i \in F} |T_ix|\big)_{F \in \mathcal{F}}$ in $L^p$ is increasing and norm bounded by $C\|x\|$, it is norm convergent to a vector $0 \le f \in L^p$ of norm $\le C\|x\|$ (here was used that $p \in [1,\infty)$). Clearly, $f$ satisfies the property claimed in (i). $\quad \square$

In the same vein, note that assertion (ii) makes sense in this general setting, while for an object such as $T_{i(\omega)}f(\omega)$ for a non-simple function $i: \Omega \to I$ it is not even clear how to define it if the $T_i$ do not have any particular structure.

(e) Fun fact: The existence of $C$ is actually redundant in (i). Indeed, if one only assume that for each $x \in X$ there exists $0 \le f \in L^p$ such that $|T_i x| \le f$ for all $i \in I$, then there automatically exists a number $C \ge 0$ such that $f$ can always be chosen to satisfy $\|f\| \le C \|x\|$. This can be shown by means of the closed graph theorem, see for instance (warning: shameless self-promotion ahead) Theorem 4.1 in this preprint.

I shared your confusion until a year ago, when Markus Haase explained to me that (2) can actually be rewritten as assertion (ii) in the following theorem, which I find much clearer.

Theorem. Let E be a Banach space, let $(\Omega,\mu)$ be a measure space, let $p \in [1,\infty)$ and let $I$ be a non-empty index set. For each $i \in I$ let $T_i: E \to L^p := L^p(\Omega,\mu)$ be a bounded linear operator. The following are equivalent:

(i) The operator family $(T_i)_{i \in I}$ satisfies a maximal inequality, i.e., there is a constant $C \ge 0$ with the following property: for every $f \in E$ there exists $0 \le h \in L^p$ of norm $\|h\|_{L^p} \le C \|f\|_{E}$ such that $|T_if| \le h$ for all $i \in I$ (where the inequality is to be understood pointwise almost everywhere).

(ii) There exists a constant $C$ with the following property: for every finite measurable partition $\Omega = A_1 \, \dot \cup \, \dots \, \dot \cup \, A_n$, all indices $i_1, \dots, i_n \in I$, and every $f \in E$ one has $\Big\| \sum_{k=1}^n 1_{A_k} T_{i_k} f \Big\|_{L^p} \le C \|f\|_{E}$.

"(ii) $\Rightarrow$ (i)" Let $\mathcal{F}$ denote the set of all non-empty finite subsets of $I$, which is directed with respect to set inclusion. Fix $f \in E$.

For each $F \in \mathcal{F}$ the vector $0 \le \bigvee_{i \in F} |T_if| \in L^p$ (where $|\cdot|$ denotes the pointwise almost everywhere modulus and $\bigvee$ denotes the pointwise almost everywhere supremum) is norm bounded by $C\|f\|_{E}$. Indeed, enumerate the elements of $F$ as $i_1, \dots, i_n$ and choose a measurable partition of $\Omega$ into sets $A_1, \dots, A_n$ such that $$ \bigvee_{i \in F} |T_if| = \sum_{k=1}^n 1_{A_k} |T_{i_k} f| = \Big| \sum_{k=1}^n 1_{A_k} T_{i_k} f \Big| . $$ Since the net $\big(\bigvee_{i \in F} |T_if|\big)_{F \in \mathcal{F}}$ in $L^p$ is increasing and norm bounded by $C\|f\|_{E}$, it is norm convergent to a vector $0 \le h \in L^p$ of norm $\le C\|f\|_{E}$ (here was used that $p \in [1,\infty)$). Clearly, $f$ satisfies the property claimed in (i). $\quad \square$

In the same vein, note that assertion (ii) makes sense in this general setting, while for an object such as $T_{i(x)}f(x)$ for a non-simple function $i: \Omega \to I$ it is not even clear how to define it if the $T_i$ do not have any particular structure.

(e) Fun fact: The existence of $C$ is actually redundant in (i). Indeed, if one only assumes that for each $f \in E$ there exists $0 \le h \in L^p$ such that $|T_i f| \le h$ for all $i \in I$, then there automatically exists a number $C \ge 0$ such that $h$ can always be chosen to satisfy $\|h\|_{L^p} \le C \|f\|_{E}$. This can be shown by means of the closed graph theorem, see for instance (warning: shameless self-promotion ahead) Theorem 4.1 in this preprint.