Fréchet-Kolmogorov-Riesz-Weil s theorem on locally compact groups via Arzelà-Ascoli s theorem

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1 arxiv: v2 [math.fa] 17 Jun 2018 Fréchet-Kolmogorov-Riesz-Weil s theorem on locally comact grous via Arzelà-Ascoli s theorem Mateusz Krukowski Łódź University of Technology, Institute of Mathematics, Wólczańska 215, Łódź, Poland June 19, 2018 Abstract In the aer, we resent a functional analysis view on the Arzelà- Ascoli theorem for the Banach sace C 0 X), where X is a locally comact Hausdorff sace. The roof hinges uon the Banach-Alaoglu s theorem. This aroach is motivated by the work of abriel Nagy. In the second art of the aer, we ut forward the most natural roof in author s oinion) of Fréchet-Kolmogorov-Riesz-Weil theorem for locally comact Hausdorff grous. The method basically amounts to the fact that boundedness, equicontinuity and equivanishing are reserved by convolution with continuous and comactly suorted functions. Keywords : Arzelà-Ascoli theorem, Fréchet-Kolmogorov-Riesz-Weil theorem, convolution, locally comact grous AMS Mathematics Subject Classification : 46E15, 46E30 1 Introduction The main objective of the aer is to resent, in author s oinion, the most natural roof of the Fréchet-Kolmogorov-Riesz-Weil theorem. In its basic form, 1

2 it characterizes the relatively comact families of the Banach sace L R N ), where 1 <. The roof is attributed to Maurice Fréchet ), Andrey Kolmogorov ) and Frigyes Riesz ). It can be found in [2],. 111 Theorem 4.26): Theorem 1. Fréchet-Kolmogorov-Riesz theorem) Let F be a bounded set in L R N ) with 1 <. Assume that lim R xf f = 0 x 0 uniformly in f F where R x fy) = fy + x). Then the closure of F Ω in L Ω) is comact for every measurable set Ω R N with finite measure. Scrutinizing this theorem, two questions immediately sring to mind: 1. What is so secial about R N, could the theorem be generalized to more abstract saces? 2. Could we get rid of the finite measure assumtion on the subset Ω? In 1940 André Weil ) wrote a book L intégration dans les groues toologique com. [15]), in which he answered both questions: Theorem 2. Let be a locally comact Hausdorff grou. A family F L ) is relatively comact if and only if 1. F is bounded in L norm, 2. for every ε > 0, there exists K comact subset of ) such that f F f f1 K < ε, 3. for every ε > 0, there exists an oen identity neighbourhood V such that f F x U L x f f < ε, where L x fy) = fxy). 2

3 The roof of this theorem is rather difficult to follow. The exosition is very terse and atly avoids technical details. The fact that the book is written in French does not make matters easier. As we have stated earlier, the main goal of this aer is to rovide a simle and elegant roof of the Fréchet-Kolmogorov-Riesz-Weil theorem. In order to do that, we build uon the visionary work of abriel Nagy com. [12]). His beautiful aer characterized relatively comact families in CK), the sace of continuous functions on comact sace K. A crucial art of his aroach was the use of Banach-Alaoglu s theorem. We emloy Nagy s techniques to characterize relatively comact families in C 0 X), the sace of continuous functions, vanishing at infinity on locally comact Hausdorff sace X. This is the content of Section 2. A brief remark is in order. As far as Fréchet-Kolmogorov-Riesz-Weil theorem is concerned, roving the Arzelà-Ascoli theorem in an extravagant way is not necessary. In fact, Theorem 5 could be concluded from the revious aers that the author has written on the subject com. [8, 9, 10]). Yet another way to derive Theorem 5 is to view C 0 X), with X locally comact Hausdorff sace, as the ideal { } f CαX)) : f ) = 0 where αx) is the Alexandroff one-oint comactification of X [11],.185). From this ersective, it is enough to remark that the equivanishing of the family F C 0 X) is equivalent to equicontinuity at the oint. This aroach is well-known in the folklore. However, having said all the above, it is not our aim to resent the shortest ossible roof of the Arzelà-Ascoli theorem. Instead, it is the author s strong conviction that a new ersective on an old result desite its length) is enlightening. The crux of Section 3 lies in Lemmas 7, 8 and 9, which assert that for a suitable continuous function with comact suort φ C c ), being a locally comact Hausdorff sace, and F L ) we have L boundedness of F imlies ointwise boundedness of F φ, L equicontinuity of F imlies equicontinuity of F φ, L equivanishing of F imlies equivanishing of F φ, where F φ = { f φ : f F }. Observe that the above imlications shift the roblem of characterizing the relatively comact families of L ) to an analogous roblem in C 0 ). However, we have already tackled this issue in, 3

4 Section 2. Finally, the Fréchet-Kolmogorov-Riesz-Weil theorem Theorem 14) for locally comact Hausdorff grous is the climax of the aer. 2 Comact families in C 0 X) Let us begin with a lemma, which is well-known in the mathematical folklore. However, due to the lack of reference, we rovide a short roof. Lemma 3. com. Proosition 1 in [12]) Let X be a comlete metric sace. Set A X is relatively comact if and only if there does not exist an infinite set B A such that inf x,y B x =y dx, y) > 0. 1) Proof. For the first art, suose that A is comact and that there exists a set B A such that 1) is satisfied. Consequently, any sequence in B cannot contain a convergent subsequence. This contradicts the comactness of A com. Theorem 3.28 in [1],. 86). For the second art, we investigate the situation when A is not relatively comact. This means that there exists an ε > 0, for which there is no finite ε net. In other words, for every finite family x n ) N n=1 A there exists an element x A such that dx, x n ) ε for every n = 1,..., N. We may adjoin x as the new element x N+1, thus extending the finite list by one element. Continuing this rocedure results in roducing an infinite sequence which satisfies 1). This ends the roof. Our second result should be contrasted with Proosition 2 in [12]. Our exosition contains slight modifications of the notation but more imortantly, the roof does not exloit the notion of Moore-Smith sequences com. [4],. 49). It resorts solely to the familiar definition of continuity. Lemma 4. Let X be a normed vector sace, K be a comact subset of X and let B X 0, 1) X be a closed unit ball. Then the restriction ma Φ : B X 0, 1), τ ) χ χ K CK) is continuous. In articular, ΦB X 0, 1)) is comact in CK). 4

5 Proof. Let B CK) ξ, ε) be an oen ball in CK) of radius ε, centered at ξ CK). Choose η Φ 1 B CK) ξ, ε) ) = { χ B X 0, 1) : χ K ξ < ε }. Since K is comact, we find x n ) N n=1 such that x K n=1,...,n x x n < δ, 2) where δ > 0 is such that We consider the weak oen set 3δ + η K ξ < ε. 3) Uη, x 1,..., x N, δ) = { χ B X 0, 1) : n=1,...,n χx n ) ηx n ) < ε } and rove that Uη, x 1,..., x N, δ) Φ 1 B CK) ξ, ε) ). For x K, we choose n = 1,..., N as in 2). If χ Uη, x 1,..., x N, δ), then χx) ξx) χx) χx n ) + χx n ) ηx n ) + ηx n ) ηx) + ηx) ξx) χ X x x n + δ + η X x x n + ηx) ξx) 3δ + η K ξ 3) < ε. Consequently, we have χ K ξ < ε, i.e. χ Φ 1 B CK) ξ, ε) ). The second art of the theorem follows immediately from Banach-Alaoglu s theorem com. [1],. 235). At this oint, we resent the first main result of the aer. Theorem 5. Arzelà-Ascoli theorem for C 0 X)) Let X be a locally comact Hausdorff sace. The family F C 0 X) is relatively comact if and only if AA1) F is ointwise bounded, i.e. su f F fx) < for every x X, AA2) F is equicontinuous in the sense ε>0 x X Ux τ X f F y U x fy) fx) < ε, 5

6 AA3) F is equivanishing in the sense ε>0 K X f F x K fx) < ε. Proof. At first, suose that F C 0 X) is relatively comact. It is obviously bounded and equicontinuity follows from a classical 3 ε argument com. Theorem 2 in [7]). As for AA3), let f n ) N n=1 be an ε 2 net for F. For each f n, let K n be a comact set such that x X\Kn fx) < ε 2. Put K := N n=1 K n. Hence, for every f F there exists n = 1,..., N such that x X\K fx) fx) f n x) + f n x) < ε 2 + ε 2 = ε. We roceed with roving the converse, i.e. we rove that AA1), AA2) and AA3) together imly relative comactness of F. Any set A F induces a function Ψ A : X x fx)) f A l A). Such a function is well-defined due to AA1). Moreover, it is continuous due to AA2). Below, we rove that Ψ A X) is comact. Fix ε > 0. By AA3), we choose K X such that f A x K fx) < ε 2. 4) Since Ψ A K) is comact, then it has an ε net, Ψ 2 A x n ) ) N. We adjoin to this n=1 ε net 0 2 l A) and rove that it is an ε net. For every element φ Ψ A X), there exists x X such that If x K, then there exists n = 1,..., N such that By the triangle inequality we have φ Ψ A x) < ε 2. 5) Ψ A x) Ψ A x n ) < ε 2. 6) φ Ψ A x n ) φ Ψ A x) + Ψ A x) Ψ A x n ) 5), 6) < ε. 6

7 In the event that x K, we have Ψ A x) 0 = su f A fx) 4) < ε 2. Again by the triangle inequality, we have φ 0 < ε. In conclusion, Ψ A x n ) ) N {0} is an ε net for Ψ A X). Hence Ψ A X) is comact. Suose, for the sake of contradiction, that F is not relatively comact. By Lemma 3, there exists an infinite set A F and δ > 0 such that inf f,g A f g f g > δ. Let us consider a closed, unit ball B in l A). By Lemma 4, the set { } ΦB) = χ ΨA X) : χ B C Ψ A X) ) is comact in the norm toology). Furthermore, let us consider the rojections π f : l A) C given by π f φ) = φf). It is trivial to check that π f B for every f A. We have ) gx) ) π f Ψ A x) = π f = fx), g A which we deict in the commutative diagram below: n=1 X Ψ A l A) f C 0 X) π f B C By the choice of A, for every f, g A we can find x X such that fx) gx) > δ, or equivalently π f Ψ A x) π g Ψ A x) > δ. 7) Since ) π f ΨA X) ΨB), then 7) contradicts the comactness of ΦB). f A We conclude that F is relatively comact, which ends the roof. 7

8 3 Comact families in L saces Throughout this section, we assume that is a locally comact Hausdorff grou with left-invariant Haar measure µ. Note that the inverse ma ι : is a homeomorhism. At first, let us rove a technical lemma. Lemma 6. Let K be a comact subset of and let U be oen and relatively comact identity neighbourhood. Then, there exists a comact set D such that x D xu K =. Proof. We ut A = U ιu), which is comact due to the continuity of multilication. Observe that xu) x K is an oen cover of K, so we may choose a finite subcover x n U) N n=1. We ut D = N n=1 x n A, which is again a comact set. Suose that x D and, for the sake of contradiction, assume that there exists y xu K. Hence, there exists n = 1,..., N such that x yιu) x n U ιu) x n A D. We reached a contradiction x D, which ends the roof. In the sequel, we will need the concets of L equicontinuity and L equivanishing. A family F L ) is said to be L equicontinuous if ε>0 Ue τ f F su L x f f < ε and su R x f f < ε, x U e x U e where L x fy) = fxy) and R x fy) = fyx). A family F L ) is said to be L equivanishing if ε>0 K f F \K fy) dy < ε. 8

9 3.1 Inheritance of boundedness, equicontinuity and equivanishing In the three lemmas below, we rove that boundedness, equicontinuity and equivanishing of F are in a sense inherited when convoluted with a continuous function with comact suort φ C c ). Lemma 7. Let F L ) be bounded in L norm. F φ C b ) is bounded. If φ C c ) then Proof. Let M > 0 be a L bound on F. At first, we rove that f φ is continuous for every f F. Fix ε > 0 and x. By Proosition 2.41 in [5],. 53 there exists a symmetric U e τ such that x Ue For x x U e, we have f φx) f φx ) = Hölder f φ ιxy) φ ιy) dy < ε M. 8) φ y 1 x ) φ y 1 x ) dy = M fy)φ y 1 x ) fy)φ y 1 x ) dy y x y M φ ι x 1 x y ) φ ι y) dy 8) < ε. φ y 1 x 1 x) φ y 1) dy This roves that F φ is a family of continuous functions. In order to rove that F φ is a family of bounded functions and moreover, that it is bounded in the suremum norm, we have f F fy)φ y 1 x ) dy x M φ µ x ιsuφ)) )) 1 Hölder f = M φ µ ιsuφ)) φ y 1 x ) dy 9) )) 1, 10) where the second inequality stems from the fact that if y 1 x suφ) y x ι suφ) ), then φ y 1 x) = 0. We conclude that F φ C b ) is bounded. 9

10 Lemma 8. Let F L ) be L equicontinuous. If φ C c ), then F φ is equicontinuous. Proof. Fix ε > 0 and x. Let V = x U e, where U e is symmetric and such that f F su x U e L x f f < ε φ µ 1. 11) )) ιsuφ)) For convenience, ut K := ιsuφ)). Observe that for every f F, we have f φx) = fy)φ y 1 x ) dy y xy = fxy)φ y 1) dy. 12) Finally, for every f F and x V, we obtain f φx) f φx ) 12) Hölder y x 1 y = which ends the roof. fxy) fx y) dy fxy) fx y) φ y 1) dy fxy) fx y) dy fxx 1 y) fy) dy φ y 1) dy φ µk) φ µk) 11) < ε, Lemma 9. Let F L ) be L equivanishing. If φ C c ) is such that φe) 0, then F φ is equivanishing. Proof. Fix ε > 0 and choose K such that f F \K f dµ < ε φ µ ιsuφ)) )). 13) Put U := ι{φ 0}), which is oen and relatively comact identity neighbourhood with U = ιsuφ)). By Lemma 6, there exists D such that x D xu K =. 14) 10

11 Finally, for f F and x D we have f φx) 14) which ends the roof. Hölder φ \K fy)φ y 1 x ) dy φ f dµ xu xu f dµ µxu) φ µxu) 13) < ε, f dµ We need to show that F φ is in a sense close to family F. In order to achieve this goal, we make use of the following result: Theorem 10. Minkowski s integral inequality, com. [6],. 194 or [14],. 271) Let X, Y be σ finite measure saces, 1 < and let F : X Y C be measurable. Then X Y ) Fx, y) dy dx Y X Fx, y) dx dy. 15) Theorem 11. com. Proosition 2.42 in [5],. 53) If F L ) is L equicontinuous, then for every ε > 0 there exists φ C c ) such that f F f φ f < ε. Proof. Fix ε > 0. Pick φ C c ) such that φe) 0, suφ ι) U, where U is the oen identity neighbourhood such that φ ι dµ = 1 and f F su y U R y f f < ε. 16) For every x, we have f φx) fx) = fy)φ y 1 x ) dy fx) ) = ) fxy) fx) φ y 1 dy. φ y 1) dy 11

12 We ut Fx, y) := fxy) fx) ) φ y 1). This function is measurable as a comosition of the following measurable functions: F 1 : x, y) x, y, y), F 2 : x, y, z) x, y, z 1 ), F 3 : x, y, z) x, xy, φz)), F 4 : x, y, z) fx), fy), z), F 5 : x, y, z) y x)z. Since f, φ are integrable, then suf) and suφ) are σ comact. Hence, also the sets suf) suφ) ιsuφ)) and suf) ιsuφ)) are σ comact. Next, we follow a series of logical imlications: = x, y) {F 0} = fxy) fx) 0 AND φ y 1) 0 ) = xy suf) OR x suf) AND y 1 suφ) ) ) xy suf) AND y 1 suφ) OR x suf) AND y 1 suφ) = x, y) suf) suφ) ιsuφ)) OR x, y) suf) ιsuφ)). We conclude that {F 0} is σ comact. Finally, we are in osition to aly Minkowski s integral inequality: f F f φ f = = 15) which ends the roof. ) ) fxy) fx) φ y 1 dy fxy) fx) φ y 1) dx R y f f φ y 1) dy su y U dy dx R y f f 16) < ε, 12

13 3.2 Young s inequality and Fréchet-Kolmogorov-Riesz- Weil s theorem We begin with a version of Young s inequality for locally comact grous. In [13], one can find a version for unimodular grous, but we do not imose such restriction. As far as the notation is concerned, for [1, ) we understand to be the number satisfying = 1. Theorem 12. Young s inequality) Let, q, r [1, ) be such that 1 r = ) q For f L ) and g L q ), the convolution f 1 g exists almost everywhere. Moreover, f 1 g L r ) and we have f 1 g r f g q. 18) Proof. Observe that it suffices to rove 18). This will immediately mean that f 1 g L r ) and consequently that the convolution exists almost everywhere. At first, we note a coule of equalities: 1 r + 1 q + 1 = 1 1 r + 1q ) ) 1 ) q 17) = 1 1 ) q =, r q 1 q ) 17) = q 1 1 ) = q. r 17) = 1, 19) 13

14 With the use of Hölder s inequality, we may erform the main calculation: 19) = f 1 gx) f y) r f y) g y 1 x) q dy f y) g y 1 x) q dy y xy = = = r r g y 1 x) q r f y) g y 1 x) q dy f y) g y 1 x) q dy f y) g y 1 x) q dy ) f y) 1 r ) f y) 1 r )q dy f y) dy r q f q r f q r f q The above estimates lead to r f 1 gx) dx f g q r q x) dx f f 1 g q r q 1 f qr g q = f q g qr g q r + q g y 1 x) ) 1 q r ) 1 y 1 x) dy g y 1 x) 1 q r ) y 1 x) dy g y 1 x) q y 1 x) dy g y 1 ) q y 1 ) dy g y) q dy q q = f g q x) r f q 20) = f g q r q 1 f g qr q+ q, where the second inequality follows from Theorem in [3],. 26. Taking the r-th root, we conclude that qr g q g q q. r f g r f + q q+ qr g q = f g q, which ends the roof. We need one final theorem before the Fréchet-Kolmogorov-Riesz-Weil theorem. Theorem 13. Let F be L bounded and let φ C c ). If K is a comact set then F K φ is relatively comact in C 0 ). 14

15 Proof. Suose that f M for every f F. At first, we show that the functions in F K φ have common suort, which is comact. For every f F observe that f K x φx) = fy)φy 1 x) dy, and the integral on the right-hand side is 0 for x K suφ). Hence suf K φ) K suφ) for every f F, and we conclude that the whole family F K φ has a common comact suort. Next, we rove that F K φ is equicontinuous. Pick ε > 0 and fix x. Since φ is uniformly continuous com. [3], Lemma 1.3.6,. 11), there exists a symmetric, oen neighbourhood of the identity U e such that Since K ε u 1 v U e φu) φv) <. 21) MµK z 1 y) 1 z 1 x) U e y 1 x U e 22) then f F f K φy) f K φx) = y 1 x U e K Hölder ineq. f K φz 1 y) φz 1 x) dz ) fz) φz 1 y) φz 1 x) dz 21), 22) < ε. The above estimate roves that F K φ is equicontinuous. Since we already established that the whole family has a common comact suort, F K φ is obviously equivanishing. Finally, we rove that F K φ is bounded in C 0 ). We have f F f K φ = su x f K y)φy 1 x) dy Hölder ineq. M φ ι, which roves that F K φ is bounded in C 0 ). We conclude the roof by alying Theorem 5. 15

16 Below, we resent the crowning result of the aer. Theorem 14. Fréchet-Kolmogorov-Riesz-Weil s theorem on a locally comact grou) A family F L ) is relatively comact if and only if FKRW1) F is bounded in L ) norm, FKRW2) F is L equicontinuous, FKRW3) F is L equivanishing. Proof. At first, suose that F is relatively comact. Then FKRW1) follows immediately. For the rest of the roof, fix ε > 0. As far as FKRW2) is concerned, let f n ) N n=1 F be an ε net. By Proosition 2.41 in [5],. 53 for every n = 1,..., N there exists an oen identity 3 neighbourhood U n such that su x U n L x f n f n < ε 3 and su R x f n f n < ε x U n 3. 23) Put U = N n=1 U n, which is obviously oen. Then for every f F, there exists n = 1,..., N such that su x U L x f f su x U L x f L x f n + su x U = 2 f n f + su x U L x f n f n + f n f L x f n f n 23) < ε. An analogous reasoning works for R x f f, which roves that F is L equicontinuous. As far as FKRW3) is concerned, let g n ) N n=1 be an ε net for F. For every 2 n = 1,..., N there exists K n such that \K n g n dµ < ε 2. 24) Put K = N n=1 K n, which is also comact. Then, for every f F there exists n = 1,..., N such that \K f dµ \K f g n dµ + \K g n dµ 24) < ε 2 + ε 2 = ε. 16

17 This roves that F is L equivanishing. At this oint we rove the converse, namely that FKRW1), FKRW2) and FKRW3) imly relative comactness of F in L ). By Theorem 11, ick φ C c ) such that Now let K be a comact set such that f F f F f φ f < ε 3. 25) \K f dµ < ε 3 1 φ 1 ). 26) The family F φ is relatively comact in C 0 ) by Theorem 5 and Lemmas 7, 8 and 9. By Theorem 13, we obtain the relative comactness of F K φ in C 0 ). Hence, there exists a finite sequence h n ) N n=1 C c) with suh n ) K suφ) such that f F n=1,...,n f K φ h n < ε 3µK suφ)). 27) Finally, for every f F there exists n = 1,..., N such that f h n f f φ + f φ f K φ + f K φ h n 25), Young s ineq. < which ends the roof. References ε 3 + f f K 1 φ 1 + K suφ) ) f K φ h n x) 26), 27) < ε, [1] Alirantis C. D., Border K. C. : Infinite Dimensional Analysis. A Hitchhiker s uide, Sringer-Verlag, Berlin, 2006 [2] Brezis H. : Functional Analysis, Sobolev Saces and Partial Differential Equations, Sringer, New York, 2011 [3] Deitmar A., Echterhoff S.: Princiles of Harmonic Analysis, Sringer, New York,

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