A truncation criterion for compactness
in asymptotic \(L_p\) spaces
January 01, 1970
We prove a compactness criterion for asymptotic \(L_p\) spaces over arbitrary measure spaces. Total boundedness is characterized by almost equiboundedness together with total boundedness in \(L_p\) of all truncations. As a consequence, we obtain a simpler proof of the Kolmogorov–Riesz compactness theorem for asymptotic \(L_p\) spaces on \(\mathbb{R}^n\).
The asymptotic \(L_p\) space over a measure space \((X,\Sigma,\mu)\) is defined as \[\label{eq:Lambda95space} \Lambda^p(X)=\left\{ f:X\to\mathbb{R}\text{ measurable} \;\Big| \; \int_X \min(|f|,1)^p \, \mathrm{d}\mu< \infty\right\},\tag{1}\] where \(1\leq p<\infty\) and functions are identified if they agree almost everywhere. The space \(\Lambda^p(X)\) is naturally endowed with the topology generated by the functional \[\label{eq:F-norm} f\mapsto \|\!\min(|f|,1)\|_p.\tag{2}\] Here and throughout, \(\|\cdot \|_p\) denotes the norm of the standard Lebesgue space \(L^p(X)\).
The functional in 2 is an \(\mathrm F\)-norm: it satisfies the usual properties of a norm except for homogeneity, which is replaced by the following two conditions: \[\|\!\min(|\lambda f|,1) \|_p \leq \|\!\min(| f|,1) \|_p,\] for all \(|\lambda| \leq 1\) and all \(f \in \Lambda^p(X)\), and \[\lim_{\lambda \to 0} \|\!\min(|\lambda f|,1) \|_{p} = 0,\] for all \(f \in \Lambda^p(X)\); see [1]. With this topology, \(\Lambda^p(X)\) is an \(\mathrm F\)-space, that is, a complete metrizable topological vector space. We recall the proof of this fact in Section 2. For background on metric linear spaces and \(\mathrm F\)-spaces, see [2], [3].
The spaces \(\Lambda^p(X)\) were introduced in [1] as spaces of functions almost in \(L_p\), equipped with the topology of asymptotic \(L_p\)-convergence; see also [4], [5]. A real-valued measurable function \(f\) is said to be almost in \(L_p\) if, for every \(\delta>0\), there exists a measurable set \(E_\delta\subseteq X\) such that \[\mu(E_\delta)<\delta \qquad \text{and} \qquad f\chi_{E_\delta^c}\in L^p(X),\] where \(E_\delta^c = X \setminus E_\delta\). Related almost-\(L_p\) spaces had previously appeared in connection with factorization and representation questions in functional analysis; see [6], [7]. The equivalence between the original almost-\(L_p\) definition and 1 is given in Proposition 2.
Although the notation suggests a close relationship with \(L^p(X)\), the spaces \(\Lambda^p(X)\) are fundamentally different from Lebesgue spaces. For example, it was shown in [1] that, in the Euclidean setting, \(\Lambda^p(\mathbb{R}^n)\) is neither locally convex nor locally bounded, and that its continuous dual is trivial. More generally, on nonatomic measure spaces, the same type of behavior for \(\Lambda^p(X)\) can be deduced from the general theory of metric linear spaces; see [3].
The main result of this note is an abstract compactness criterion for \(\Lambda^p(X)\) over arbitrary measure spaces. It characterizes total boundedness in terms of almost equiboundedness and total boundedness in \(L^p(X)\) of all truncations; see Theorem 5. In the Euclidean case, this criterion gives, by applying the classical Kolmogorov–Riesz theorem in \(L^p(\mathbb{R}^n)\) to the truncated families, a short proof of the Kolmogorov–Riesz compactness theorem in \(\Lambda^p(\mathbb{R}^n)\) obtained in [8]. This Euclidean compactness theorem has also been used to deduce a compactness result of Rellich–Kondrachov type in \(\Lambda^p(\mathbb{R}^n)\), which in turn led to a well-posedness theory for \(p\)-Schrödinger equations with integrable data and confinement in measure; see [9].
Compactness criteria in function spaces have been extensively studied in many different settings. For the classical Kolmogorov–Riesz theorem in \(L^p(\mathbb{R}^n)\), see [10], [11]. Related criteria have been obtained for spaces of measurable functions [12], for variable exponent and variable summability spaces [13]–[15], for Banach and quasi-Banach function spaces [16]–[19], and in other settings related to Riesz–Kolmogorov compactness criteria [20]. The present paper thus fits into this line of results by providing an abstract compactness criterion for asymptotic \(L_p\) spaces over arbitrary measure spaces, reducing total boundedness in \(\Lambda^p(X)\) to total boundedness of the truncated families in \(L^p(X)\) together with a uniform control of large values.
The paper is organized as follows. In Section 2, we recall the basic structure of \(\Lambda^p(X)\), prove that it is an \(\mathrm F\)-space under the truncation definition, and establish the equivalence between this definition and the original almost-\(L_p\) formulation from [1]. Section 3 contains the main compactness criterion. In Section 4, we recover the Kolmogorov–Riesz compactness theorem in \(\Lambda^p(\mathbb{R}^n)\). Finally, Section 5 gives a short proof, based on the truncation definition, of the Vitali convergence theorem in \(\Lambda^p(X)\) obtained in [1].
In this section we collect the basic structural properties of \(\Lambda^p(X)\). Although these facts are elementary, we include the proofs in order to make clear that the truncation definition 1 gives a complete metrizable topological vector space. We also recall the relation between this definition and the original formulation from [1].
We first show that \(\Lambda^p(X)\) is an \(\mathrm F\)-space. In the terminology of Kalton, Peck, and Roberts [2], this means that it is a metrizable topological vector space which is complete with respect to a translation-invariant metric. In the present case, the metric is given by \[(f,g)\mapsto \|\!\min(|f-g|,1)\|_p.\]
Theorem 1. Let \((X,\Sigma,\mu)\) be a measure space and let \(1\le p<\infty\). Then \(\Lambda^p(X)\) is an \(\mathrm{F}\)-space. Moreover, when \(\mu(X) < \infty\), the space \(\Lambda^p(X)\) consists of all real-valued measurable functions on \(X\) with the topology of convergence in measure.
The final assertion in Theorem 1 should be understood in two parts. If \(\mu(X)<\infty\), then every real-valued measurable function belongs to \(\Lambda^p(X)\), since \[\min(|f|,1)^p\leq 1.\] The statement about the topology uses the fact that, on finite measure spaces, convergence with respect to the \(\mathrm{F}\)-norm 2 is equivalent to convergence in measure; see [1], [4], [5].
That \(\Lambda^p(X)\) is a vector space follows immediately from the following two inequalities \[\label{eq:ineq95scalar95prod} \min(|\lambda a|,1)^p \leq \max\{|\lambda|^p,1\} \min(|a|,1)^p,\tag{3}\] and \[\label{eq:ineq95triang} \min(|a+b|,1)^p \leq 2^{p-1}\big( \min(|a|,1)^p + \min(|b|,1)^p \big),\tag{4}\] valid for all \(\lambda,a,b \in \mathbb{R}\) and \(1 \leq p < \infty\).
From the triangle inequality it also follows that the operation of addition is continuous with respect to the metric of \(\Lambda^p(X)\). In the next lemma we show that the operation of scalar multiplication is also continuous in this topology, thus establishing that \(\Lambda^p(X)\) is a metrizable topological vector space.
Lemma 1. Assume that \(\{\lambda_k\}_{k \in \mathbb{N}}\) is a sequence of real numbers converging to a real number \(\lambda\), and assume that \(\{f_k \}_{k \in \mathbb{N}}\) is a sequence of functions in \(\Lambda^p(X)\) converging to a function \(f \in \Lambda^p(X)\). Then \[\lambda_k \, f_k \to \lambda \, f \qquad \text{in } \Lambda^p(X) \qquad \text{as } k \to \infty.\]
Proof. Since \(\lambda_k\to\lambda\), there exists \(C>0\) such that \(|\lambda_k|\le C\) for every \(k\in\mathbb{N}\). Using the triangle inequality and 3 we deduce \[\begin{align} \|\min(|\lambda_k f_k-\lambda f|,1)\|_p &\le \|\min(|\lambda_k(f_k-f)|,1)\|_p + \|\min(|(\lambda_k-\lambda)f|,1)\|_p \\ &\le \max\{C,1\}\|\min(|f_k-f|,1)\|_p + \|\min(|(\lambda_k-\lambda)f|,1)\|_p . \end{align}\] The first term tends to \(0\), since \(f_k\to f\) in \(\Lambda^p(X)\).
For the second term, we have \[\min(|(\lambda_k-\lambda)f(x)|,1)^p\to0\] for almost every \(x\in X\). Moreover, for all sufficiently large \(k\), \(|\lambda_k-\lambda|\le1\), and hence \[\min(|(\lambda_k-\lambda)f|,1)^p \le \min(|f|,1)^p.\] Since \(f\in\Lambda^p(X)\), the right-hand side is integrable. Therefore, by the dominated convergence theorem, \[\|\!\min(|(\lambda_k-\lambda)f|,1)\|_p\to0.\]
Consequently, \[\|\!\min(|\lambda_k f_k-\lambda f|,1)\|_p\to0,\] which means that \(\lambda_k f_k\to\lambda f\) in \(\Lambda^p(X)\). ◻
To conclude the proof of Theorem 1 it remains to prove that the metric of \(\Lambda^p(X)\) is complete. This is the content of the next lemma; the standard proof is included for convenience.
Lemma 2. Let \(\{f_k \}_{k \in \mathbb{N}}\) be a Cauchy sequence in \(\Lambda^p(X)\). Then, there exists \(f \in \Lambda^p(X)\) such that \[f_k \to f \qquad \text{in } \Lambda^p(X) \qquad \text{as } k \to \infty.\]
Proof. Since \(\{f_k\}_{k\in\mathbb{N}}\) is Cauchy in \(\Lambda^p(X)\), it is Cauchy in measure. Indeed, for \(0<\alpha<1\), \[\alpha^p\mu(\{|f_k-f_\ell|>\alpha\}) \le \int_X\min(|f_k-f_\ell|,1)^p\,\mathrm{d}\mu,\] and the right-hand side tends to \(0\) as \(k,\ell\to\infty\).
Hence, there exist a subsequence \(\{f_{k_j}\}_{j\in\mathbb{N}}\) and a real-valued measurable function \(f\) such that \[f_{k_j}\to f \qquad\text{a.e.\;on }X.\]
We show that this subsequence converges to \(f\) in \(\Lambda^p(X)\). Let \(\varepsilon>0\). Since \(\{f_k\}_{k\in\mathbb{N}}\) is Cauchy in \(\Lambda^p(X)\), there exists \(N\in\mathbb{N}\) such that \[\|\!\min(|f_k-f_\ell|,1)\|_p<\varepsilon\] whenever \(k,\ell\ge N\). Fix \(j\) such that \(k_j\ge N\). Since \(f_{k_\ell}\to f\) a.e., Fatou’s lemma gives \[\begin{align} \int_X\min(|f_{k_j}-f|,1)^p\,\mathrm{d}\mu &\le \liminf_{\ell\to\infty} \int_X\min(|f_{k_j}-f_{k_\ell}|,1)^p\,\mathrm{d}\mu\\ &\le \varepsilon^p. \end{align}\] Thus \[\|\!\min(|f_{k_j}-f|,1)\|_p\le \varepsilon\] for all sufficiently large \(j\). Thus \[\|\!\min(|f_{k_j}-f|,1)\|_p\le \varepsilon\] for all sufficiently large \(j\). Choosing one such \(j\), the triangle inequality gives \(f\in\Lambda^p(X)\). Hence \[f_{k_j}\to f \qquad\text{in }\Lambda^p(X).\] Since the original sequence is Cauchy in \(\Lambda^p(X)\) and has a subsequence converging to \(f\), the whole sequence converges to \(f\) in \(\Lambda^p(X)\), again by the triangle inequality. ◻
We next give the characterization that explains the equivalence between the original definition of \(\Lambda^p(X)\) in [1] and the one adopted in the present paper.
Proposition 2. Fix \(1\le p<\infty\), and let \(f\) be a real-valued measurable function on \(X\). The following statements are equivalent:
(i) \(f\) is almost in \(L_p\).
(ii) There exists a sequence \(\{f_k\}_{k\in\mathbb{N}}\) in \(L^p(X)\) such that \[\|\!\min(|f_k-f|,1)\|_p\to0 \qquad\text{as } k\to\infty.\]
(iii) \(\min(|f|,1)\in L^p(X)\).
Proof. We first show that (i) implies (ii). For each \(k\in\mathbb{N}\), choose \(E_k\in\Sigma\) such that \(\mu(E_k)<1/k\) and \[f_k:=f\chi_{E_k^c}\in L^p(X).\] Then \[\begin{align} \|\!\min(|f_k-f|,1)\|_p^p &= \int_{E_k}\min(|f\chi_{E_k^c}-f|,1)^p\,\mathrm{d}\mu\le \mu(E_k) < \frac{1}{k} \to 0, \end{align}\] as \(k \to \infty\).
Next, we show that (ii) implies (iii). Choose \(N\in\mathbb{N}\) such that \[\|\!\min(|f_N-f|,1)\|_p<1.\] Since \[\min(|f|,1) \le \min(|f-f_N|,1)+\min(|f_N|,1),\] the triangle inequality gives \[\|\!\min(|f|,1)\|_p \le \|\!\min(|f-f_N|,1)\|_p+\|\!\min(|f_N|,1)\|_p < 1+\|f_N\|_p < \infty.\] Hence \(\min(|f|,1)\in L^p(X)\).
Finally, we prove that (iii) implies (i). For each \(k\in\mathbb{N}\), set \[E_k:=\{|f|>k\}.\] Then \(E_{k+1}\subseteq E_k\) for every \(k\in\mathbb{N}\), and \(\mu(E_1)<\infty\), because \[\chi_{E_1}\le \min(|f|,1)^p\in L^1(X).\] Moreover, the countable intersection \[\bigcap_{k\in\mathbb{N}}E_k\] has measure zero, since \(f\) is real-valued almost everywhere. Hence \[\mu(E_k)\to0 \qquad\text{as } k\to\infty.\] Given \(\delta>0\), choose \(K_\delta\in\mathbb{N}\) such that \[\mu(E_{K_\delta})<\delta.\] On \(E_{K_\delta}^c\) we have \(|f|\le K_\delta\), and therefore \[|f|^p \le K_\delta^p\min(|f|,1)^p.\] It follows that \[f\chi_{E_{K_\delta}^c}\in L^p(X).\] Thus \(f\) is almost in \(L_p\). ◻
We finish this section with two elementary observations. First, the spaces \(\Lambda^p(X)\) are nested as \(p\) increases. Second, for bounded functions, belonging to \(\Lambda^p(X)\) is the same as belonging to \(L^p(X)\).
Proposition 3. Let \(1\le p\le q<\infty\). Then \[\Lambda^p(X)\subseteq \Lambda^q(X).\]
Proof. The result follows from the fact that, for \(1 \leq p \leq q < \infty\), one has \[\min(|a|,1)^q\le \min(|a|,1)^p,\] for every \(a \in \mathbb{R}\). ◻
Proposition 4. Let \(M>0\), and let \(f\) be a measurable function on \(X\) such that \(|f|\le M\) almost everywhere. Then \[f\in \Lambda^p(X) \qquad\text{if and only if}\qquad f\in L^p(X).\] Consequently, \[L^\infty(X)\cap\Lambda^p(X)=L^\infty(X)\cap L^p(X).\]
Proof. From the hypothesis \(|f| \leq M\) a.e., we deduce \[\|\!\min(|f|,1)\|_p \le \|f\|_p \le \max\{1,M\}\|\!\min(|f|,1)\|_p,\] from which the result clearly follows. ◻
We now state and prove the main compactness criterion of the paper. It says that total boundedness of a family in \(\Lambda^p(X)\) can be characterized by two conditions: the total boundedness in \(L_p\) of each truncated family, and the uniform control of large values, expressed through approximation by truncations.
For \(M>0\), we define \[\label{eq:truncations} T_M(a):=\max\big\{\!-M,\min\{a,M\}\big\}, \qquad a\in\mathbb{R},\tag{5}\] and write \(T_Mf:=T_M\circ f\).
Theorem 5. Let \((X,\Sigma,\mu)\) be a measure space, let \(1\leq p<\infty\), and let \(\mathcal{F}\subseteq\Lambda^p(X)\). Then \(\mathcal{F}\) is totally bounded in \(\Lambda^p(X)\) if and only if the following two conditions hold:
(i) For every fixed \(M>0\), the truncated family \[T_M(\mathcal{F}):=\big\{T_Mf:\,f\in\mathcal{F}\big\}\] is totally bounded in \(L^p(X)\).
(ii) The family \(\mathcal{F}\) is uniformly approximable by truncations, that is, \[\label{eq:approx95trunc} \lim_{M\to\infty}\sup_{f\in\mathcal{F}} \|\!\min(|f-T_M f|,1)\|_p=0.\tag{6}\]
The first lemma contains the basic properties of \(T_M\) that will be used below.
Lemma 3. Let \(M>0\). The following statements hold.
(i) If \(f\in\Lambda^p(X)\), then \(T_Mf\in L^p(X)\).
(ii) For every \(f,g\in\Lambda^p(X)\), \[\|\!\min(|T_M f -T_M g|,1) \|_p \leq \|\!\min(|f -g|,1) \|_p.\]
(iii) If \(u,v\in L^p(X)\) and \(|u|,|v|\leq M\) a.e., then \[\|\!\min(|u -v|,1) \|_p\leq \|u-v\|_p\leq C_M \|\!\min(|u -v|,1) \|_p,\] where \(C_M=\max\{1,2M\}\).
Proof. For (i), observe that \[|T_M(a)|\leq \max\{M,1\}\min(|a|,1), \qquad a\in\mathbb{R}.\] Thus \(|T_Mf|^p\leq \max\{1,M\}^p\min(|f|,1)^p\in L^1(X)\).
For (ii), the scalar truncation map \(T_M:\mathbb{R}\to\mathbb{R}\) is \(1\)-Lipschitz. Hence \[\min(|T_M(f)-T_M(g)|,1)\leq \min(|f-g|,1) \qquad \text{a.e.},\] and the claim follows.
Finally, if \(|u|,|v|\leq M\), then \(|u-v|\leq 2M\). Therefore \[\min(|u-v|,1)\leq |u-v|\leq C_M\min(|u-v|,1) \qquad \text{a.e.},\] and taking \(L^p\)-norms gives (iii). ◻
The next estimate relates truncation errors to the measure of level sets. In particular, truncations approximate each fixed element of \(\Lambda^p(X)\).
Lemma 4. Let \(f\in\Lambda^p(X)\). Then, for every \(M>0\), \[\label{eq95truncation95measure} \mu(|f|>M+1) \leq \|\!\min(| f -T_M f|,1) \|_p^p \leq \mu(|f|>M).\tag{7}\] Consequently, \[\|\!\min(| f -T_M f|,1) \|_p\to0 \qquad \text{as } M\to\infty .\]
Proof. Since \[|f-T_Mf|=(|f|-M)_+,\] we have \[\|\!\min(| f -T_M f|,1) \|_p^p = \int_X \min\big((|f|-M)_+,1\big)^p\,\mathrm{d}\mu.\] The integrand is supported in \(\{|f|>M\}\) and is bounded by \(1\). Hence \[\|\!\min(| f -T_M f|,1) \|_p^p\leq \mu(|f|>M).\] On the other hand, if \(|f|>M+1\), then \((|f|-M)_+>1\), and therefore \[\min\big((|f|-M)_+,1\big)^p=1.\] Thus \[\mu(|f|>M+1)\leq \|\!\min(| f -T_M f|,1) \|_p^p.\]
It remains only to observe that \(\mu(|f|>M)\to0\) as \(M \to \infty\). For \(M\ge1\), \[\chi_{\{|f|>M\}}\leq \min(|f|,1)^p\in L^1(X),\] and \(\chi_{\{|f|>M\}}\to0\) pointwise a.e.as \(M\to\infty\). By the dominated convergence theorem, \[\mu(|f|>M)\to0 \qquad \text{as } M \to \infty.\] The upper bound in 7 therefore gives \[\|\!\min(| f -T_M f|,1) \|_p\to0 \qquad \text{as } M \to \infty,\] as claimed. ◻
The preceding lemma gives the following useful reformulation of the second condition in Theorem 5.
Lemma 5. Let \(\mathcal{F}\subseteq\Lambda^p(X)\). Then \(\mathcal{F}\) satisfies 6 if and only if it is almost equibounded, that is, \[\label{eq:almost95equiboundedness} \lim_{M\to\infty}\sup_{f\in\mathcal{F}}\mu(|f|>M)=0.\tag{8}\]
Proof. By Lemma 4, for every \(f\in\mathcal{F}\) and every \(M>0\), \[\mu(|f|>M+1) \le \|\!\min(|f-T_Mf|,1)\|_p^p \le \mu(|f|>M).\] Taking suprema over \(f\in\mathcal{F}\) and letting \(M\to\infty\) gives the equivalence. ◻
We now proceed with the proof of the compactness criterion.
Assume first that \(\mathcal{F}\) is totally bounded in \(\Lambda^p(X)\). Fix \(M>0\), let \(\varepsilon>0\), and let \(C_M =\max\{1,2M\}\). Choose \(f_1,\ldots,f_N\in\Lambda^p(X)\) such that, for every \(f\in\mathcal{F}\), there is \(i\in\{1,\ldots,N\}\) with \[\|\!\min(|f-f_i|,1)\|_p<\frac{\varepsilon}{C_M}.\] By Lemma 3(i), \(T_Mf_i\in L^p(X)\) for each \(i=1,\ldots, N\). Moreover, using Lemma 3(iii) and then Lemma 3(ii), we obtain \[\begin{align} \|T_Mf-T_Mf_i\|_p & \leq C_M\|\!\min(|T_Mf-T_Mf_i|,1)\|_p \\ & \leq C_M\|\!\min(|f-f_i|,1)\|_p \\ & <\varepsilon, \end{align}\] which proves that \(T_M(\mathcal{F})\) is totally bounded in \(L^p(X)\).
Next, we show that \[\lim_{M\to\infty}\sup_{f\in\mathcal{F}} \|\!\min(|f-T_Mf|,1)\|_p=0.\] Let \(\varepsilon>0\). Choose \(f_1,\ldots,f_N\in\Lambda^p(X)\) such that, for every \(f\in\mathcal{F}\), there is \(i\in\{1,\ldots,N\}\) with \[\|\!\min(|f-f_i|,1)\|_p<\frac{\varepsilon}{3}.\] By Lemma 4, for each \(i=1,\ldots,N\) there exists \(M_i>0\) such that \[\|\!\min(|f_i-T_Mf_i|,1)\|_p<\frac{\varepsilon}{3}\] for every \(M\geq M_i\). Setting \[M_0:=\max\{M_1,\ldots,M_N\},\] we obtain \[\|\!\min(|f_i-T_Mf_i|,1)\|_p<\frac{\varepsilon}{3}\] for every \(M\geq M_0\) and every \(i=1,\ldots,N\). Hence, for \(M\geq M_0\) and \(f\in\mathcal{F}\), choosing \(i\) as above gives \[\begin{align} \|\!\min(|f-T_Mf|,1)\|_p &\leq \|\!\min(|f-f_i|,1)\|_p +\|\!\min(|f_i-T_Mf_i|,1)\|_p \\ &\qquad +\|\!\min(|T_Mf_i-T_Mf|,1)\|_p \\ &\leq 2\|\!\min(|f-f_i|,1)\|_p +\|\!\min(|f_i-T_Mf_i|,1)\|_p \\ &<\varepsilon, \end{align}\] where Lemma 3(ii) was used in the second inequality. Therefore \[\sup_{f\in\mathcal{F}}\|\!\min(|f-T_Mf|,1)\|_p<\varepsilon\] for every \(M\geq M_0\), as required.
Conversely, assume that \[\lim_{M\to\infty}\sup_{f\in\mathcal{F}} \|\!\min(|f-T_Mf|,1)\|_p=0\] and that \(T_M(\mathcal{F})\) is totally bounded in \(L^p(X)\) for every \(M>0\). Let \(\varepsilon>0\) be given. Choose \(M>0\) such that \[\sup_{f\in\mathcal{F}}\|\!\min(|f-T_Mf|,1)\|_p<\frac{\varepsilon}{2}.\] Since \(T_M(\mathcal{F})\) is totally bounded in \(L^p(X)\), there exist \(h_1,\ldots,h_N\in L^p(X)\) such that for every \(f\in\mathcal{F}\) there is \(i\in\{1,\ldots,N\}\) with \[\|T_Mf-h_i\|_p<\frac{\varepsilon}{2}.\] Then \[\|\!\min(|T_Mf-h_i|,1)\|_p\leq \|T_Mf-h_i\|_p<\frac{\varepsilon}{2}.\] Thus \[\begin{align} \|\!\min(|f-h_i|,1)\|_p &\leq \|\!\min(|f-T_Mf|,1)\|_p + \|\!\min(|T_Mf-h_i|,1)\|_p <\varepsilon, \end{align}\] so \(\mathcal{F}\) is totally bounded in \(\Lambda^p(X)\).
0◻
We now show that Theorem 5, combined with the classical Kolmogorov–Riesz theorem in \(L^p(\mathbb{R}^n)\), gives a short proof of the Euclidean compactness criterion for \(\Lambda^p(\mathbb{R}^n)\) obtained in [8] under the original definition of these spaces from [1].
We start by recalling the classical Kolmogorov–Riesz theorem in \(L^p(\mathbb{R}^n)\) in the form stated in [11]. There, the usual boundedness assumption on the family has been shown to be a consequence of the tail and translation conditions; see also [10].
Theorem 6 (Kolmogorov–Riesz compactness in \(L^p(\mathbb{R}^n)\)). Let \(1\le p<\infty\), and let \(\mathcal{F}\subseteq L^p(\mathbb{R}^n)\). Then \(\mathcal{F}\) is totally bounded in \(L^p(\mathbb{R}^n)\) if and only if the following two conditions hold:
(i) \(\mathcal{F}\) satisfies the tail condition \[\lim_{R\to\infty} \sup_{f\in\mathcal{F}} \int_{\{|x|>R\}}|f(x)|^p\,\mathrm{d} x=0.\]
(ii) \(\mathcal{F}\) satisfies the translation condition \[\lim_{|y|\to0} \sup_{f\in\mathcal{F}} \int_{\mathbb{R}^n}|f(x+y)-f(x)|^p\,\mathrm{d} x=0.\]
Combining Theorem 5 and Theorem 6, we recover the Kolmogorov–Riesz compactness theorem in \(\Lambda^p(\mathbb{R}^n)\).
Theorem 7 (Kolmogorov–Riesz compactness in \(\Lambda^p(\mathbb{R}^n)\)). Let \(1\le p<\infty\), and let \(\mathcal{F}\subseteq\Lambda^p(\mathbb{R}^n)\). Then \(\mathcal{F}\) is totally bounded in \(\Lambda^p(\mathbb{R}^n)\) if and only if the following three conditions hold:
(i) \(\mathcal{F}\) satisfies the truncated tail condition \[\lim_{R\to\infty} \sup_{f\in\mathcal{F}} \int_{\{|x|>R\}}\min(|f(x)|,1)^p\,\mathrm{d} x=0.\]
(ii) \(\mathcal{F}\) satisfies the truncated translation condition \[\lim_{|y|\to0} \sup_{f\in\mathcal{F}} \int_{\mathbb{R}^n}\min(|f(x+y)-f(x)|,1)^p\,\mathrm{d} x=0.\]
(iii) \(\mathcal{F}\) is almost equibounded, that is, \[\lim_{M\to\infty} \sup_{f\in\mathcal{F}} \big|\{|f|>M\}\big|=0.\]
By Theorem 5 and Lemma 5, total boundedness in \(\Lambda^p(\mathbb{R}^n)\) is equivalent to almost equiboundedness together with total boundedness in \(L^p(\mathbb{R}^n)\) of every truncated family \(T_M(\mathcal{F})\). Thus, to prove Theorem 7, it remains to relate the truncated tail and translation conditions to the total boundedness in \(L_p\) of these truncated families. This is done in the next two lemmas.
Lemma 6. Let \(\mathcal{F}\subseteq\Lambda^p(\mathbb{R}^n)\) satisfy the truncated tail condition and the truncated translation condition in Theorem 7. Then, for every \(M>0\), the truncated family \(T_M(\mathcal{F})\) is totally bounded in \(L^p(\mathbb{R}^n)\).
Proof. Fix \(M>0\). By Lemma 3(i), we have \(T_Mf\in L^p(\mathbb{R}^n)\) for every \(f\in\mathcal{F}\). We shall verify the tail and translation hypotheses of Theorem 6.
We start with the tail condition. Using the inequality \[|T_M(a)|^p\le \max\{M^p,1\}\min(|a|,1)^p, \qquad a\in\mathbb{R},\] we deduce \[\int_{\{|x|>R\}} |T_Mf(x)|^p\,\mathrm{d} x \le \max\{M^p,1\} \int_{\{|x|>R\}}\min(|f(x)|,1)^p\,\mathrm{d} x.\] Taking the supremum over \(f\in\mathcal{F}\) and then letting \(R\to\infty\), the right-hand side tends to zero by the truncated tail condition.
Next, since \(T_M\) is \(1\)-Lipschitz and \(|T_M(a)-T_M(b)|\le 2M\), we have \[|T_M(a)-T_M(b)|^p \le \max\{(2M)^p,1\}\min(|a-b|,1)^p, \qquad a,b\in\mathbb{R}.\] Thus, for every \(y\in\mathbb{R}^n\) and every \(f\in\mathcal{F}\), \[\begin{align} \int_{\mathbb{R}^n} |& T_Mf(x+y)-T_Mf(x)|^p\,\mathrm{d} x\\ & \leq \max\{(2M)^p,1\} \int_{\mathbb{R}^n}\min(|f(x+y)-f(x)|,1)^p\,\mathrm{d} x. \end{align}\] Taking the supremum over \(f\in\mathcal{F}\) and then letting \(|y|\to0\), the right-hand side tends to zero by the truncated translation condition.
Therefore, by Theorem 6, \(T_M(\mathcal{F})\) is totally bounded in \(L^p(\mathbb{R}^n)\). ◻
The converse implication uses almost equiboundedness to pass from total boundedness of the truncated families back to the truncated tail and translation conditions.
Lemma 7. Let \(\mathcal{F}\subseteq\Lambda^p(\mathbb{R}^n)\) be almost equibounded, and assume that \(T_M(\mathcal{F})\) is totally bounded in \(L^p(\mathbb{R}^n)\) for every \(M>0\). Then \(\mathcal{F}\) satisfies the truncated tail condition and the truncated translation condition in Theorem 7.
Proof. Let \(\varepsilon>0\). By almost equiboundedness, we may choose \(M>0\) such that \[\sup_{f\in\mathcal{F}}\big|\{|f|>M\}\big|<\frac{\varepsilon}{4}.\] Moreover, since \(T_M(\mathcal{F})\) is totally bounded in \(L^p(\mathbb{R}^n)\), by Theorem 6 there exist \(R > 0\) and \(r > 0\) such that \[\sup_{f\in\mathcal{F}} \int_{|x|>R} |T_Mf(x)|^p\,\mathrm{d} x< \frac{\varepsilon}{2},\] and \[\sup_{f\in\mathcal{F}} \int_{\mathbb{R}^n}|T_Mf(x+y)-T_Mf(x)|^p\,\mathrm{d} x< \frac{\varepsilon}{2},\] whenever \(|y| < r\).
We first prove the truncated tail condition. For every \(f\in\mathcal{F}\) we have: \[\begin{align} \int_{|x|>R} \min(|f(x)|,1)^p \, \mathrm{d} x& = \int_{\{|x| > R \} \cap \{|f| \leq M \}} \min(|f(x)|,1)^p \, \mathrm{d} x\\ & \quad + \int_{\{|x| > R \} \cap \{|f| > M \}} \min(|f(x)|,1)^p \, \mathrm{d} x\\ & \leq \int_{|x|>R} |T_M(f)|^p \, \mathrm{d} x+ \big| \{|f| > M \} \big| \\ & < \varepsilon. \end{align}\]
We now prove the truncated translation condition. Fix \(f\in\mathcal{F}\) and \(y\in\mathbb{R}^n\) with \(|y| < r\). Set \[A_{f,y}:=\big\{ |f|>M \big\} \cup \big\{|\tau_y f|>M \big\},\] where \(\tau_y f(x) = f(x+y)\), and note that \[|A_{f,y}| \le 2\big|\{|f|>M\}\big|.\] If \(x \in \mathbb{R}^n\setminus A_{f,y}\), then both \(|f(x)|\) and \(|f(x+y)|\) are at most \(M\), and therefore \[T_Mf(x)=f(x), \qquad T_Mf(x+y)=f(x+y).\] Thus \[\begin{align} \int_{\mathbb{R}^n}\min(|f(x+y)-f(x)|,1)^p\,\mathrm{d} x& = \int_{\mathbb{R}^n \setminus A_{f,y}}\min(|f(x+y)-f(x)|,1)^p\,\mathrm{d} x\\ & \quad + \int_{ A_{f,y}}\min(|f(x+y)-f(x)|,1)^p\,\mathrm{d} x\\ & \le \int_{\mathbb{R}^n}|T_Mf(x+y)-T_Mf(x)|^p\,\mathrm{d} x + |A_{f,y}| \\ & \le \int_{\mathbb{R}^n}|T_Mf(x+y)-T_Mf(x)|^p\,\mathrm{d} x + 2 \big| \{|f| > M \} \big| \\ & < \varepsilon. \end{align}\] Since \(f\) is arbitrary, the result follows. ◻
Proof of Theorem 7. Assume first that \(\mathcal{F}\) is totally bounded in \(\Lambda^p(\mathbb{R}^n)\). By Theorem 5, the truncated family \(T_M(\mathcal{F})\) is totally bounded in \(L^p(\mathbb{R}^n)\) for every \(M>0\), and \(\mathcal{F}\) satisfies 6 . Hence, by Lemma 5, \(\mathcal{F}\) is almost equibounded. Lemma 7 then gives the truncated tail and translation conditions.
Conversely, assume that \(\mathcal{F}\) satisfies the three conditions in Theorem 7. By Lemma 6, \(T_M(\mathcal{F})\) is totally bounded in \(L^p(\mathbb{R}^n)\) for every \(M>0\). Moreover, condition (iii) and Lemma 5 imply 6 . Hence the two conditions of Theorem 5 hold, and therefore \(\mathcal{F}\) is totally bounded in \(\Lambda^p(\mathbb{R}^n)\). ◻
We conclude this note with a simple proof of the Vitali convergence theorem in \(\Lambda^p(X)\), obtained in [1] using the almost-\(L_p\) definition of the spaces. The proof below uses the truncation definition of \(\Lambda^p(X)\) together with the classical Vitali convergence theorem in \(L^p(X)\), which we now recall; see [21].
Theorem 8 (Vitali convergence in \(L^p(X)\)). Let \((X,\Sigma,\mu)\) be a measure space, let \(1\le p<\infty\), let \(\{f_k\}_{k \in \mathbb{N}}\subseteq L^p(X)\), and let \(f\) be measurable. Then \(\{f_k\}_{k\in\mathbb{N}}\) converges to \(f\) in \(L^p(X)\) if and only if the following two conditions hold:
(i) \(\{f_k\}_{k\in\mathbb{N}}\) converges to \(f\) in measure,
(ii) for every \(\varepsilon > 0\) there exist \(E_\varepsilon \in \Sigma\) with \(\mu(E_\varepsilon) < \infty\) and \(\delta_\varepsilon > 0\) such that \[\sup_{k \in \mathbb{N}} \int_{E_\varepsilon^c} |f_k |^p \, \mathrm{d}\mu < \varepsilon^p\] and, if \(F \in \Sigma\) and \(\mu(F) < \delta_\varepsilon\), then \[\sup_{k \in \mathbb{N}} \int_{E_\varepsilon \cap F} |f_k |^p \, \mathrm{d}\mu < \varepsilon^p \, .\]
In \(\Lambda^p(X)\), the boundedness of the truncated integrands removes the absolute continuity condition on sets of small measure. The resulting criterion is the following.
Theorem 9 (Vitali convergence in \(\Lambda^p(X)\)). Let \((X,\Sigma,\mu)\) be a measure space, let \(1\le p<\infty\), let \(\{f_k\}_{k \in \mathbb{N}}\subseteq \Lambda^p(X)\), and let \(f\) be measurable. Then \(\{f_k\}_{k\in\mathbb{N}}\) converges to \(f\) in \(\Lambda^p(X)\) if and only if the following two conditions hold:
(i) \(\{f_k\}_{k\in\mathbb{N}}\) converges to \(f\) in measure,
(ii) for every \(\varepsilon>0\) there exists a measurable set \(E_\varepsilon\in\Sigma\) with \(\mu(E_\varepsilon)<\infty\) such that \[\sup_{k \in \mathbb{N}} \int_{E_\varepsilon^c}\min(|f_k|,1)^p\,\mathrm{d}\mu < \varepsilon^p.\]
In particular, when \(\mu(X) < \infty\), convergence in \(\Lambda^p(X)\) and convergence in measure coincide.
Proof. Assume first that \(\{f_k\}_{k \in \mathbb{N}}\) converges to \(f\) in \(\Lambda^p(X)\). Then \(\{f_k\}_{k \in \mathbb{N}}\) converges to \(f\) in measure and \(\{\min(|f_k|,1)\}_{k \in \mathbb{N}}\) converges to \(\min(|f|,1)\) in \(L^p(X)\). The latter implies condition (ii) by Theorem 8.
Conversely, assume (i) and (ii). We prove that \(\{f_k\}_{k\in\mathbb{N}}\) converges to \(f\) in \(\Lambda^p(X)\). Let \(\varepsilon>0\). By (ii), we may choose a measurable set \(E_\varepsilon\) with \(\mu(E_\varepsilon)<\infty\) such that \[\sup_{k \in \mathbb{N}} \int_{E_\varepsilon^c}\min(|f_k|,1)^p\,\mathrm{d}\mu < \frac{\varepsilon^p}{2^{p}}.\] Using a subsequence converging almost everywhere to \(f\) and Fatou’s lemma, we also have \[\int_{E_\varepsilon^c}\min(|f|,1)^p\,\mathrm{d}\mu \le \frac{\varepsilon^p}{2^{p}}.\] Therefore \[\begin{align} \int_X\min(|f_k-f|,1)^p \, \mathrm{d}\mu& = \int_{E_\varepsilon^c}\min(|f_k-f|,1)^p \, \mathrm{d}\mu+ \int_{E_\varepsilon}\min(|f_k-f|,1)^p \, \mathrm{d}\mu\\ & \leq 2^{p-1} \int_{E_\varepsilon^c}\min(|f_k|,1)^p\,\mathrm{d}\mu+2^{p-1} \int_{E_\varepsilon^c}\min(|f|,1)^p\,\mathrm{d}\mu\\ & \quad + \int_{E_\varepsilon}\min(|f_k-f|,1)^p \, \mathrm{d}\mu\\ & < \varepsilon^p + \int_{E_\varepsilon}\min(|f_k-f|,1)^p \, \mathrm{d}\mu. \end{align}\] Moreover, since \(f_k\to f\) in measure and \(\mu(E_\varepsilon)<\infty\), we have \[\int_{E_\varepsilon}\min(|f_k-f|,1)^p\,\mathrm{d}\mu\to0 \qquad \text{as } k \to \infty.\] Consequently \[\limsup_{k \to \infty} \int_X\min(|f_k-f|,1)^p \, \mathrm{d}\mu\leq \varepsilon^p.\] As \(\varepsilon>0\) is arbitrary, convergence in \(\Lambda^p(X)\) follows. The final assertion is obtained by taking \(E_\varepsilon=X\) in condition (ii) when \(\mu(X)<\infty\). ◻
This publication is based upon work supported by King Abdullah University of Science and Technology (KAUST) under Award No. ORFS-CRG12-2024-6430.