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Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 45, 2020, 991–1002

CHARACTERIZING COMPACT FAMILIES VIA THE LAPLACE TRANSFORM

Mateusz Krukowski

Łódź University of Technology, Institute of Wólczańska 215, 90-924 Łódź, Poland; [email protected]

Abstract. In 1985, Pego characterized compact families in L2(R) in terms of the . It took nearly 30 years to realize that Pego’s result can be proved in a more general setting of locally compact abelian groups (works of Górka and Kostrzewa). In the current paper, we argue that the Fourier transform is not the only integral transform that is efficient in characterizing compact families and suggest the Laplace transform as a possible alternative.

1. Introduction The main objective of the article is to use Laplace transform to characterize 2 compact families in L (R+). This task is achieved in Theorem 9, after we introduce and study the concepts of Laplace equicontinuity and equivanishing. In the author’s opinion, the culminating result of the paper elegantly rhymes with the ideas of Pego and later Górka and Kostrzewa. However, before we dive into technical details, let us first lay the historical background. Characterizing compact families has been a vital topic in function spaces’ theory at least since the end of the 19th century. Around 1883, two Italian Cesare Arzelà (1847–1912) and Giulio Ascoli (1843–1896) provided the necessary and sufficient conditions under which every sequence of a given family of real-valued continuous functions (defined on a closed and bounded interval), has a uniformly convergent subsequence (this is called sequential compactness). A couple of decades later (in 1931), (1903–1987) succeeded in characterizing the compact families in Lp(RN ), when 1

https://doi.org/10.5186/aasfm.2020.4553 2010 Mathematics Subject Classification: Primary 44A10, 42A38. Key words: Laplace transform, Pego theorem, compactness. 992 Mateusz Krukowski

In the current paper, we argue that the Fourier transform is not the only one that can be used to characterize compact families. In Section 2 we introduce the basic definitions and discuss the necessary notation. We also prove the fundamental theorems, which are very well-known in the context of the Fourier transform, and probably less known in the context of the Laplace transform. In Section 3 we prove the main results. Theorem 9, which is a counterpart of the Pego’s result, is the climax of the paper.

2. Preliminary results

For a measurable, complex-valued function f : R+ C and a real number x > 0, we denote → −xt fx(t)= f(t)e . We say that f : R C is a Laplace–Pego function of order x > 0 if + → 1 2 fx L (R ) L (R ). ∈ + ∩ + 1 2 The norms in L (R+) and L (R+) are denoted by 1 and 2, respectively. Moreover, if is a subfamily of Laplace–Pego functionsk·k with a commonk·k order x > 0, then we denoteA x = fx : f . A { ∈ A} Let f be a Laplace–Pego function of order x > 0. The Laplace transform f of the function f is defined by L{ } ∞ f (z)= f(t)e−zt dt. L{ } ˆ0 A natural question arises: when does the above integral exist? To answer this ques- tion, observe that if Re(z) > x, then ∞ ∞ f (z) = f(t)e− Re(z)te−i Im(z)t dt 6 f(t) e− Re(z)t dt |L{ } | ˆ ˆ | | 0 0 ∞ −xt (x−Re(z))t = f(t) e e dt 6 fx 1 < . ˆ0 | | k k ∞ In other words, the Laplace transform f exists in the half-plane Re(z) > x. An important special case of theL{ Laplace} transform is the Fourier transform, which we define by f(y)= f (2πiy). L{ } Let us formulate a crucial theorem regarding the Laplace transform, which we will use multiple times throughoutb the paper: Theorem 1. (Plancherel theorem for the Laplace transform) If f is a Laplace- Pego function of order x > 0, then 1 ∞ ∞ (1) f (x + iy) 2 dy = e−2xt f(t) 2 dt. 2π ˆ−∞ |L{ } | ˆ0 | |

Proof. At first, observe that ∞ −xt −iyt y∈R f (x + iy)= f(t)e e dt ∀ L{ } ˆ0 (2) ∞ y y −xt −2πi 2 t = f(t)e e π dt = fx . ˆ0 2π   b Characterizing compact families via the Laplace transform 993

By the classical Plancherel theorem [4, Theorem 3.5.2, p. 53] or [24, Theorem 1.1, p. 208] we have ∞ 2 ∞ ∞ y y 2 −2xt 2 fx d = fx(t) dt = e f(t) dt. ˆ−∞ 2π 2π ˆ−∞ | | ˆ0 | |   Upon observing that b ∞ 2 ∞ y y (2) 1 2 fx d = f (x + iy) dy ˆ−∞ 2π 2π 2π ˆ−∞ |L{ } |   we conclude the proof.  b The theorem, which we present below, is a counterpart of a well-know result in the theory of Fourier transform: Theorem 2. (Riemann–Lebesgue lemma for the Laplace transform) If f is a Laplace–Pego function of order x > 0, then (3) lim f (x + iy)=0. y→±∞ L{ }

Proof. At first, let f = 1 a,b where (a, b) R . Then, for every y R we have ( ) ⊂ + ∈ ∞ −(x+iy)t y∈R f (x + iy)= 1(a,b)(t)e dt ∀ L{ } ˆ0 b e−(x+iy)a e−(x+iy)b = e−(x+iy)t dt = − , ˆa x + iy so (3) holds. By linearity of the Laplace transform, the result is also true for all simple functions. Finally, let f be an arbitrary Laplace–Pego function and let ε> 0. Since simple 1 functions are dense in L (R+), there exists a simple function g such that ∞ (4) f(t)e−xt g(t) dt < ε. ˆ0 | − | Hence ∞ lim f (x + iy) = lim f(t)e−xte−iyt dt y→±∞ |L{ } | y→±∞ ˆ 0 ∞ ∞ −xt −iyt 6 f (t)e g(t) dt + lim g(t)e dt ˆ − y→±∞ ˆ 0 0 (4) y < ε + lim g (iy) = ε + lim g = ε, y→±∞ |L{ } | y→±∞ 2π   where the last equality follows from the classical Riemann-Lebesgue lemma for the b Fourier transform (comp. [16, Theorem 1.7, p. 136]. Since ε> 0 was chosen arbitrar- ily, we conclude the proof.  We will now recall the prominent fact that the Laplace trasnform ’changes the convolution of two functions to multiplication’. A convolution of two Laplace–Pego functions f,g with a common order x > 0 is defined by t t>0f⋆g(t)= f(s)g(t s) ds. ∀ ˆ0 − Theorem 3. If f,g are Laplace–Pego functions with a common order x > 0, 1 then (f⋆g)x L (R ). In particular, it exists almost everywhere. ∈ + 994 Mateusz Krukowski

Proof. At first, let us observe that t −xt −xt t>0e f⋆g(t)= e f(s)g(t s) ds ∀ ˆ0 − (5) t = e−xsf(s)e−x(t−s)g(t s) ds. ˆ0 −

Furthermore, by [24, Proposition 3.9, p. 86], we note that the function F : R+ R C defined by × + −→ F (t, s)= e−xsf(s)e−x(t−s)g(t s) − is measurable, so we are in position to apply Tonelli’s theorem. Consequently, we obtain ∞ (5) ∞ t e−xtf⋆g(t) dt 6 e−xs f (s)e−x(t−s) g (t s) dsdt ˆ ˆ ˆ | | | | − 0 0 0 ∞ ∞ Tonelli’s thm −xs −x(t−s) = e f (s)e g (t s) dtds ˆ0 ˆs | | | | − = fx gx < , k k1k k1 ∞ which ends the proof.  The convolution theorem for the Laplace transform, which we present below, should be juxtaposed with [22, Theorem 2.39, p. 92]. Theorem 4. (Convolution theorem for the Laplace transform) If f and g are Laplace–Pego functions with a common order x > 0, then f⋆g (z)= f (z) g (z) L{ } L{ } · L{ } for Re(z) > x. Proof. Note that the function F : R R C defined by + × + −→ F (t, s)= f(s)g(t s)e−zt − is measurable, so by Tonelli’s theorem we have ∞ ∞ t f⋆g (z)= f⋆g(t)e−zt dt = f(s)g(t s)e−zt dsdt L{ } ˆ0 ˆ0 ˆ0 − ∞ ∞ Tonelli’s thm = f(s)g(t s)e−zt dtds = f (z) g (z), ˆ0 ˆs − L{ } · L{ } which ends the proof. 

3. Main results A subfamily of Laplace–Pego functions with a common order x > 0 is said to be exponentially AL2-equivanishing at x, if ∞ −2xt 2 (6) ε>0 T>0 f∈A e f(t) dt < ε. ∀ ∃ ∀ ˆT | | Furthermore, we say that a family is Laplace equicontinuous at x, if ∞ A 1 2 (7) ε>0 δ>0 f∈A f (x + iy + δ) f (x + iy) dy < ε. ∀ ∃ ∀ 2π ˆ−∞ |L{ } − L{ } | We will now relate the concepts of Laplace equicontinuity and exponential L2- equivanishing. Characterizing compact families via the Laplace transform 995

Theorem 5. Let be a subfamily of Laplace–Pego functions with a common order x > 0. If isA Laplace equicontinuous at x, then it is exponentially L2- A 2 equivanishing at x. Furthermore, if x is L -bounded, then the implication can be reversed. A Proof. We divide the proof into two steps: Step 1. At first, we assume that is Laplace equicontinuous at x, so for a fixed ε> 0 we may choose δ > 0 accordingA to (7). Let T > 0 be such that

2 1 (8) e−δT 1 > . − 2 Consequently, for every f we obtain ∈A 1 ∞ ε> f (x + iy + δ) f (x + iy) 2 dy 2π ˆ−∞ |L{ } − L{ } | 1 ∞ ∞ 2 = f(t) e−δt 1 e−(x+iy)t dt dy 2π ˆ ˆ − −∞ 0 T  ∞ Theorem 1 −2xt 2 −δt 2 −2xt 2 −δt 2 = e f(t) e 1 dt + e f(t) e 1 dt ˆ0 | | − ˆT | | − (8) 1 ∞ > e−2xt f(t) 2 dt, 2 ˆT | | which ends the first part of the proof. 2 Step 2. At this point, we assume that x is L -bounded, so there exists M > 0 such that A ∞ −2xt 2 f∈A e f(t) dt 6 M. ∀ ˆ0 | | We will show that if is exponentially L2-equivanishing at x, then it is Laplace equicontinuous at x. A Fix ε> 0 and choose T > 0 as in the definition of the exponential L2-equivanishing (6). Let δ > 0 be such that 2 (9) e−δT 1 M < ε. − We have 1 ∞ f (x + iy + δ) f (x + iy) 2 dy 2π ˆ−∞ |L{ } − L{ } | T ∞ Theorem 2 2 = 1 e−2xt f(t) 2 e−δt 1 dt + e−2xt f(t) 2 e−δt 1 dt ˆ0 | | − ˆT | | − ∞ (9) 2 (6) 6 e−δT 1 M + e−2xt f (t) 2 dt < 2ε, − ˆT | | which ends the proof.  A subfamily of Laplace–Pego functions with a common order x > 0 is said to be exponentially AL2-equicontinuous at x, if 1 ∞ 2 −2xt 2 (10) ε> δ> e f(t) f(t s) dt < ε. ∀ 0 ∃ 0 ∀s∈(0,δ) ˆ | − − | f∈A  0  Let us make one technical remark at this point. Although we defined the domain of the Laplace–Pego functions as R+, we may actually treat these functions as functions 996 Mateusz Krukowski on R with the property f(t)=0 if t< 0. Consequently, in (10) we have f(t s)=0 whenever s > t. − Furthermore, we say that a family is Laplace equivanishing at x, if A 2 (11) ε>0 T>0 f∈A f (x + iy) dy < ε. ∀ ∃ ∀ ˆR\[−T,T ] |L{ } | We study the relationship between the novel notion of the exponential L2-equi- continuity of and the classical equicontinuity of x in the lemma below: A A Lemma 6. Let be a subfamily of Laplace–Pego functions with a common A 2 2 order x > 0. If x is L -bounded then is exponentially L -equicontinuous at x if A 2 A and only if x is L -equicontinuous, i.e. A ∞ −x(t+s) −xt 2 (12) ε>0 δ>0 s∈(0,δ), e f(t + s) e f(t) dt < ε. ∀ ∃ ∀ f∈A ˆ0 −

2 Proof. Since x is L -bounded, there exists M > 0 such that A 1 ∞ 2 −2xt 2 f e f(t) dt 6 M. ∀ ∈A ˆ | |  0  We divide the proof of the lemma into two steps: Step 1. In the first part of the proof, we assume that the family is exponentially L2-equicontinuous at x. We fix ε> 0 and choose δ > 0 such that A (12) is satisfied, and • for every s (0, δ) we have • ∈ s (13) e−2xt f(t) 2 dt < ε, ˆ0 | | which is possible due to Theorem 8 in [18], p. 148, and for every s (0, δ) we have • ∈ (14) e−xs 1 M < ε. − Consequently, for every s (0, δ) and f we have ∈ 1 ∈A 1 ∞ 2 ∞ 2 e−2xt f(t) f(t s) 2 dt = e−2x(t+s) f(t + s) f(t) 2 dt ˆ0 | − − | ˆ−s | − |    1  1 ∞ 2 ∞ 2 6 e−2x(t+s) f(t + s) exsf(t) 2 dt + e−2x(t+s) exsf(t) f(t) 2 dt ˆ−s | − | ˆ−s | − |    1  0 ∞ 2 = e−2x(t+s) f(t + s) 2 dt + e−2x(t+s) f(t + s) exsf(t) 2 dt ˆ−s | | ˆ0 | − |  1  ∞ 2 + e−2x(t+s) f(t) 2 exs 1 2 dt ˆ0 | | | − |   1 s ∞ 2 6 e−2xt f(t) 2 dt + e−x(t+s)f(t + s) e−xtf(t) 2 dt ˆ | | ˆ | − |  0 0  (12),(13),(14) 1 + e−xs 1 M 6 (2ε) 2 + ε. | − | Since ε> 0 was chosen arbitrarily, the above estimates end the first part of the proof. Characterizing compact families via the Laplace transform 997

2 Step 2. In this part of the proof, we assume that x is L -equicontinuous. Again, we fix ε> 0 and let δ > 0 be such that A (10) is satisfied, and • for every s (0, δ) we have • ∈ (15) 1 exs M <ε and exs 6 2. | − | For every s (0, δ) and f we have ∈ ∈A 1 1 ∞ 2 ∞ 2 2 2 e−x(t+s)f(t + s) e−xtf(t) dt = e−xtf(t) e−x(t−s)f(t s) dt ˆ0 − ˆs − −  1   1 ∞ 2 ∞ 2 2 2 6 e−xtf(t) e−x(t−s)f(t) dt + e−x(t−s)f(t) e−x(t−s)f(t s) dt ˆ − ˆ − −  s   s  1 ∞ 2 (15) 6 1 exs e−2xt f(t) 2dt + exsε 6 1 exs M +2ε< 3ε, | − | ˆ | | | − |  0  which ends the proof.  We will now study the relationship between the exponential L2-equicontinuity and the Laplace equivanishing. Theorem 7. Let be a subfamily of Laplace–Pego functions with a common order x > 0. ExponentialA L2-equicontinuity at x implies Laplace equivanishing at x. 2 Furthermore, if x is L -bounded, then the implication can be reversed. A Proof. We divide the proof into two steps: Step 1. We assume that the family is exponentially L2-equicontinuous at x, so for a fixed ε > 0 we can choose δA > 0 according to the exponential L2- equicontinuity (10). Let g be a nonnegative and continuous function on R+ such ∞ that supp(g) (0, δ) and 0 g(s) ds = 1. Naturally, g is a Laplace–Pego function of order x. ⊂ ´ By Theorem 2, let T > 0 be such that 1 (16) R g (x + iy) 6 . ∀y∈ \[−T,T ]|L{ } | 2 Consequently, we have

1 2 2 f∈A f (x + iy) dy ∀ ˆR\[−T,T ] |L{ } |   1 2 2 6 f (x + iy) 1 g (x + iy) dy ˆR L{ } − L{ }  \[−T,T ]   1 2 + f (x + iy) g (x + iy) 2 dy ˆR\[−T,T ] |L{ } L{ } |   1 (16) 2 2 6 f (x + iy) 1 g (x + iy) dy ˆR L{ } − L{ }  \[−T,T ]  1  1 2 + f (x + iy) 2 dy , 2 ˆR |L{ } |  \[−T,T ]  998 Mateusz Krukowski which implies 1 2 2 f∈A f (x + iy) dy ∀ ˆR\[−T,T ] |L{ } |   1 2 2 6 2 f (x + iy) 1 g (x + iy) dy ˆR L{ } − L{ } \[−T,T ] !  1 Theorem 4 2 6 2 f (x + iy) f⋆g (x + iy) 2 dy ˆR |L{ } − L{ } |  1  ∞ 2 Theorem =1 2√2π e−2xt f(t) f⋆g(t) 2 dt ˆ0 | − |   1 ∞ ∞ 2 2 =2√2π e−2xt f(t) f(t s)g(s) ds dt ˆ − ˆ − 0 0 ! 1 2 2 ∞ ∞ =2√2π e−xt(f(t) f(t s))g(s) ds dt ˆ ˆ − − 0 0 !

Minkowski integral 1 inequality ∞ ∞ 2 6 2√2π e−2xt f(t) f(t s) 2 g(s) 2 dt ds ˆ0 ˆ0 | − − | | |  1  ∞ ∞ 2 =2√2π g(s) e−2xt f(t) f(t s) 2 dt ds ˆ0 ˆ0 | − − |  1 δ ∞ 2 =2√2π g(s) e−2xt f(t) f(t s) 2 dt ds ˆ ˆ | − − | 0  0  δ 6 2√2πε g(s) ds =2√2πε. ˆ0 Let us remark that the use of Minkowski inequality in the above estimates is justified, because the function F (t, s)= e−2xt f(t) f(t s) 2 g(s) 2 is measurable due to [24, Proposition 3.9, p. 86]. Since ε > 0|was− chosen− arbitrarily,| | | the above estimates end the first part of the proof. 2 Step 2. For this part of the proof, we assume that x is L -bounded, so there A exists M1 > 0 such that ∞ −2xt 2 (17) f∈A e f(t) dt 6 M1. ∀ ˆ0 | | We will show that if is Laplace equivanishing at x then it is exponentially L2- A equicontinuous at x. For convenience, we denote T−sf(t)= f(t s). We observe the following equalities − ∞ ∞ −2xt 2 −2xt 2 s>0 e f(t) f(t s) dt = e f(t) T−sf(t) dt ∀f∈A ˆ0 | − − | ˆ0 | − | ∞ Theorem 1 1 2 (18) = f T−sf (x + iy) dy 2π ˆ−∞ |L{ − } | 1 ∞ = f (x + iy) e−s(x+iy) f (x + iy) 2 dy 2π ˆ−∞ |L{ } − L{ } | Characterizing compact families via the Laplace transform 999

∞ 1 2 = 1 e−s(x+iy) f (x + iy) 2 dy. 2π ˆ−∞ − |L{ } |

Fix ε> 0 and choose T > 0 according to Laplace equivanishing (11). Let δ > 0 be such that −s(x+iy) 2 (19) s∈(0,δ) 1 e < ε, ∀y∈[−T,T ] − and put

−s(x+iy) 2 (20) M2 = max 1 e , s∈[0,δ], − R y∈ which is finite due to −sx 2 M2 6 max 1+ e . s∈[0,δ] Finally, for every s (0, δ) and f , we have  ∈ ∈A ∞ T (18) 1 2 e−2xt f(t) f(t s) 2 dt = 1 e−s(x+iy) f (x + iy) 2 dy ˆ0 | − − | 2π ˆ−T − |L{ } | 1 2 + 1 e−s(x+iy ) f (x + iy) 2 dy 2π ˆR\[−T,T ] − |L{ } | (19), (20) T ε M2 6 f (x + iy) 2 dy + f (x + iy) 2 dy 2π ˆ−T |L{ } | 2π ˆR\[−T,T ] |L{ } | Theorem 1, (11) ∞ M M 6 ε e−2xt f(t) 2 dt + 2 ε 6 M + 2 ε. ˆ | | 2π 1 2π 0   Since ε> 0 was chosen arbitrarily, we conclude the proof.  Before we present the final theorem of the paper, let us recall the celebrated Riesz–Kolmogorov theorem: 2 Theorem 8. (Riesz–Kolmogorov theorem, comp. [14]) A family L (R+) is relatively compact if and only if A ⊂ is L2-bounded; •A is L2-equicontinuous; •A is L2-equivanishing. •A The final theorem, which is the climax of the paper, should be juxtaposed with Pego theorem in [7], [8] and [19]. Theorem 9. Let be a subfamily of Laplace–Pego functions with a common A 2 order x and such that x is L -bounded. The family x is relatively compact in 2 A A L (R+) if and only if is Laplace equicontinuous at x; •A is Laplace equivanishing at x. •A Proof. The proof is divided into two steps: Step 1. We assume that is Laplace equicontinuous and equivanishing at x. At first, we note that Laplace equicontinuityA of at x implies that this family is expo- 2 A 2 nentially L -equivanishing at x (Theorem 5). In other words, x is L -equivanishing. A 1000 Mateusz Krukowski

Furthermore, Laplace equivanishing of at x implies that this family is expo- 2 A 2 nentially L -equicontinuous (Theorem 7). In other words, x is L -equicontinuous A 2 (Lemma 6). By Theorem 8, we conclude that x is relatively compact in L (R ). A + Step 2. For the second part of the proof, we assume that x is relatively compact 2 2 A 2 in L (R+). By Theorem 8, the family x is L -equicontinuous and L -equivanishing. 2 A L -equicontinuity of x implies that is Laplace equivanishing at x (Lemma 6 and A 2 A Theorem 7). Moreover, L -equivanishing of x implies that is Laplace equicon- tinuous at x (Theorem 5), which ends the proof.A A  Let us conclude the paper with a simple example exhibiting how Theorem 9 αt works: let β R and let be the family of functions of the form e 1[0,∞)(t), where α 6 β.∈ Every x > β Ais a common order for the family . However,· it is easy to see that the family is not of order β. A Let us fix x > β. We remark that ∞ (α−z)t 1 f∈A f (z)= e dt = − . ∀ z >xL{ } ˆ α z Re( ) 0 − Furthermore, we observe that

(α−x)t (21) x = e 1 (t): α 6 β A · [0,∞)   is L2-bounded. In order to apply Theorem 9 we first check that is Laplace equicon- tinuous at x. For a fixed ε> 0 we choose δ > 0 such that A δ2 (22) < ε. 2(β x δ)2 β x − − ·| − | Consequently, for every f (or every α 6 β equivalently) we have ∈A 1 ∞ f (x + iy + δ) f (x + iy) 2 dy 2π ˆ−∞ |L{ } − L{ } | 1 ∞ 1 1 2 = − + dy 2π ˆ α (x + iy + δ) α (x + iy) −∞ − − δ2 ∞ dy = 2π ˆ α (x + iy + δ) 2 α (x + iy) 2 −∞ | − | ·| − | δ2 −∞ dy = 2π ˆ ((α x δ)2 + y2) ((α x)2 + y2) ∞ − − · − δ2 ∞ dy 6 2π(α x δ)2 ˆ (α x)2 + y2 − − −∞ − δ2 δ2 (22) = 6 < ε, 2(α x δ)2 α x 2(β x δ)2 β x − − ·| − | − − ·| − | where we used the formula dy 1 y b>0 = arctan + const. ∀ ˆ y2 + b √b √b   Since ε> 0 was chosen arbitrarily, we conclude that is Laplace equicontinuous at x. A Characterizing compact families via the Laplace transform 1001

2 To prove that is Laplace equivanishing it suffices to pick T > ε and observe that A dy f x iy 2 dy ( + ) = 2 ˆR |L{ } | ˆR α (x + iy) \[−T,T ] \[−T,T ] | − | dy ∞ dy 2 6 < ε. 2 =2 2 = ˆR\[−T,T ] y ˆT y T To sum up, we have verified that is Laplace equicontinuous and Laplace equivan- ishing. By Theorem 9 we concludeA that (21) is relatively compact in L2. Acknowledgements. I would like to express my deep gratitude towards Prze- mysław Górka for the careful study of my paper. He provided invaluable comments and remarks, which improved the article tremendously. Furthermore, I wish I could find the right words of appreciation for the anonymous Reviewer. He or she has painstakingly criticized my paper pointing out various incongruencies in the text. Without such an assistance, the paper could not have reached its final form. I feel honoured to have had my article reviewed by such a supportive and proficient person.

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Received 9 June 2019 Accepted 13 December 2019 •