Collect. Math. DOI 10.1007/s13348-013-0099-y
On the essential bounded variation of L p (R, X)-functions Gen Nakamura · Kazuo Hashimoto
Received: 4 August 2013 / Accepted: 11 November 2013 © Universitat de Barcelona 2013
Abstract Let X be a Banach space. Nakamura and Hashimoto in (Proc Japan Acad Ser A 87:77–82, 2011), we showed that for every f ∈ L 1 (R), f (t + h) − f (t) dt = essV1 ( f ). h→0 h lim
R
In this paper, we are concerned with the limit f (t + h) − f (t) p dt lim h→0 h
(∗ )
R
for f ∈ L loc 1 (R, X ). We show that the limit (∗) coincides with the essential p-variation of f in the sense of F. Riesz, and we give characterizations of functions with bounded essential p-variation, i.e, essV p ( f, X ) < ∞. Keywords
Bochner spaces · Bounded variation · Essensial p-variation · Sobolev spaces
Mathematics Subject Classification (2000)
26A45 · 26A46 · 46G10
Research supported by Grant for Scientific Research of Hiroshima Jogakuin University. G. Nakamura Matsue College of Technology, 14-4 Nishi-ikuma, Matsue, Shimane 690-8518, Japan e-mail:
[email protected] K. Hashimoto (B) Hiroshima Jogakuin University, Faculty of Liberal Arts, 4-13-1 Ushita Higashi, Higashi-ku, Hiroshima 732-0063, Japan e-mail:
[email protected]
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1 Introduction and preliminaries Let 1 ≤ p < +∞. Let (R, M, m) be the usual Lebesuge measure space on the set R of real numbers, and X a Banach space. L p (R, X ) will denote the Bochner space L p (R, M, m, X ). For definitions and properties of these spaces we refer to [1]. In particular, if X = R, we simply denote by L p (R). We denote by L loc 1 (R, X ) the vector space of functions f Bochner integrable on every bounded measurable subset of R. Let W p1 (R) be a Sobolev space, i.e, f ∈ W p1 (R) if and only if f ∈ L p (R) and the derivative D f of f in the sense of distribution belongs to L p (R). In particular, if f ∈ L 1 (R) and D f is a Radon measure of bounded variation on R, f called a function of bounded variation. The class of all such functions will be denoted by BV (R). Thus, f ∈ BV (R) if and only if there is a Radon measure μ defined in R such that |μ|(R) < +∞ and f ϕ dt = − ϕdμ, ϕ ∈ C0∞ (R), R
R
where, |D f |(R) = |μ|(R) means the total variation of μ. It is obvious that a function f on R is absolutely continuous and the derivative f is in L 1 (R), then f is of bounded variation, i.e. W11 (R) ⊂ BV (R) (see [3]). Given f ∈ L loc 1 (R, X ), we write n f (ti ) − f (ti−1 ) p (ti − ti−1 ) V p ( f, X ) = sup ti − ti−1 π i=1
for the p-variation of the function f in the sense of F. Riesz ([5]), where the supremum is taken over all partition π : t0 < t1 < · · · < tn of R. We define a subset essD( f ) of R by ⎧ ⎫ t+h ⎨ ⎬ 1 ess D( f ) = t ∈ R : lim f (s) ds converges in X . ⎩ ⎭ h→0 h t
We shall call the set ess D( f ) the essential domain of f . We should note that m(R\essD( f )) = 0 and ess D( f ) does not depend on measurable functions equal to f a.e. on R. We define ⎧ t+h ⎪ ⎪ 1 ⎨ lim f (s) ds for t ∈ ess D( f ), f (t) = h→0 h ⎪ t ⎪ ⎩ 0 for t ∈ ess D( f ). Then we should note that f = f (a.e.) ( R , X ), define For f ∈ L loc 1 n p f (tk ) − f (tk−1 ) n ess V p ( f, X ) = sup (tk − tk−1 ) : π = (tk )k=1 , tk ∈ ess D( f ) , tk − tk−1 π k=1
where the supremum is taken over all partition π : t0 < t1 < · · · < tn of R. In what follows, we call ess V p ( f, X ) the essential p variation of f in the sense of F. Riesz. The symbol ess V p ( f ) will be written for ess V p ( f, R). In [4], we showed that the following holds:
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On the essential bounded variation
Theorem Let f ∈ L 1 (R). Then we have f (x + h) − f (x) d x = ess V1 ( f, R). lim h→0 h R
In this paper, we are concerned with the limit f (t + h) − f (t) p dt lim h→0 h
(∗ )
R
for f ∈ L loc 1 (R, X ). We show that the limit (∗) coincides with the essential p-variation of f in the sense of F. Riesz (Theorem 1), and we give characterizations of functions with bounded essential p-variation, i.e, ess V p ( f, X ) < ∞.
2 Results Theorem 1 Let 1 ≤ p < ∞ and f ∈ L loc 1 (R, X ). Then we have f (t + h) − f (t) p dt = ess V p ( f, X ). lim h→0 h R
Proof We first prove that lim inf h→0
f (t+h)− f (t) p dt ≥ ess V p ( f, X ). Let t0 < t1 < h
R
· · · < tn be an arbitrary sequence in ess D( f ). Then by Hölder’s inequality, we have tk f (t + h) − f (t) dt ≤ tk−1
⎧ t ⎨ k ⎩
f (t + h) − f (t) p dt
tk−1
⎫1/ p ⎬ ⎭
(tk − tk−1 )1−1/ p ,
and so tk f (t + h) − f (t) p dt ≥ tk−1
1
⎧ t ⎨ k
(tk − tk−1 ) p−1 ⎩
f (t + h) − f (t) dt
tk−1
⎫p ⎬ ⎭
.
Thus for all h = 0, we have ⎞p ⎛ t k n f (t + h) − f (t) p 1 1 dt ≥ ⎝ f (t + h) − f (t) dt ⎠ h |h| p (tk − tk−1 ) p−1 k=1
R
≥
=
1 |h| p
n k=1
tk−1
1 (tk − tk−1 ) p−1
n 1 1 |h| p (tk − tk−1 ) p−1 k=1
p t k ( f (t + h) − f (t)) dt tk−1 p tk +h tk f (t) dt − f (t) dt tk−1 +h tk−1
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p tk +h tk−1 +h 1 1 = f (t) dt − f (t) dt |h| p (tk − tk−1 ) p−1 k=1 tk−1 tk−1 p tk +h tk−1 +h n 1 1 1 = f (t) dt − f (t) dt . (tk − tk−1 ) p−1 h h k=1 tk−1 tk−1 n
It is obvious that the right-hand side of the above equation coverges to the following value as h → 0: n k=1
Thus we have lim inf h→0
R
1 f (tk ) − f (tk−1 ) p . (tk − tk−1 ) p−1
n f (t + h) − f (t) p 1 dt ≥ f (tk ) − f (tk−1 ) p , h (tk − tk−1 ) p−1 k=1
and so we see from the definition of ess V p ( f, X ) that f (t + h) − f (t) p dt ≥ ess V p ( f, X ). lim inf h→0 h
(1)
R
To show the converse inequality, f (t + h) − f (t) p dt ≤ ess V p ( f, X ), lim sup h h→0 R
it suffices to show that f (t + h) − f (t) p dt ≤ ess V p ( f, X ) h
for all h = 0.
R
Without loss of generality, we can assume that h > 0. f (t + h) − f (t) p p 1 dt = 1 f (t + h) − f (t) dt p−1 h h h R
R
∞ 1 = h
(k+1)h
k=−∞ kh
=
1 h p−1
p f (t + h) − f (t) dt
∞ h p 1 1 f (t + (k + 1)h) − f (t + kh) dt p−1 h h k=−∞ 0
=
123
1 h
h ∞ 0 k=−∞
1 h p−1
p f (t + (k + 1)h) − f (t + kh) dt.
On the essential bounded variation
∞ On the other hand, since R\ess ∞D( f ) is a null set, we have m( k=−∞ {(R\ess D( f )) − kh}) = 0. Moreover, since t ∈ k=−∞ {(R\ess D( f )) − kh} implies that t + kh ∈ ess D( f ) for every integer k, we see that ∞ k=−∞
1 h p−1
p f (t + (k + 1)h) − f (t + kh) ≤ ess V p ( f, X ) a.e. on [0, h],
and so h f (t + h) − f (t) p 1 dt ≤ ess V p ( f, X ) dt = ess V p ( f, X ). h h
R
(2)
0
Hence we have lim sup h→0
f (t + h) − f (t) p dt ≤ ess V p ( f, X ). h
R
Combining with (1), we have f (t + h) − f (t) p dt = ess V p ( f, X ). lim h→0 h R
Remark 1 In Theorem 1, we see easily from (2) that f (t + h) − f (t) p dt = ess V p ( f, X ). sup h h =0 R
Theorem 2 Let X be a Banach space and 1 ≤ p < ∞. Then for all f ∈ L loc 1 (R, X ), we have ess V p ( f, X ) = inf V p (g, X ) : g ∈ L loc 1 (R, X ), f = g a.e. . In particular, if ess V p ( f, X ) < ∞, then there exists a right-continuous function g ∈ L loc 1 (R, X ) such that f = g a.e., and ess V p ( f, X ) = V p (g, X ). Proof For any f ∈ L loc 1 (R, X ), we show that ess V p ( f, X ) ≤ inf V p (g, X ) : g ∈ L loc 1 (R, X ), f = g a.e. holds. Let h > 0 and g ∈ L loc 1 (R, X ) satisfying f = g a.e. Then p f (t + h) − f (t) p dt = g(t + h) − g(t) dt h h R
R
∞ 1 = h
kh g(t + h) − g(t) p h dt h
k=−∞(k−1)h
1 = h
h ∞ g(t + kh) − g(t + (k − 1)h) p h dt h 0 k=−∞
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1 ≤ h
h V p (g, X ) dt 0
= V p (g, X ). As h ↓ 0, we have from Theorem 1 that ess V p ( f, X ) ≤ V p (g, X ). Thus since g ∈ L loc 1 (R, X ) is arbitrary, we have ess V p ( f, X ) ≤ inf V p (g, X ) : g ∈ L loc 1 (R, X ), f = g a.e. . Conversely, for any f ∈ L loc 1 (R, X ), we show that ess V p ( f, X ) ≥ inf V p (g, X ) : g ∈ L loc 1 (R, X ), f = g a.e. holds. Without loss of gererality, we assume that ess V p ( f, X ) < ∞. Then we show that there exists a g ∈ L loc 1 (R, X ) with f = g a.e. such that ess V p ( f, X ) = V p (g, X ). Since ess D( f ) is dense in R, for any t ∈ R there exists a sequence with tn → t and t1 > t2 > t3 > · · · in ess D( f ), we have ∞ ∞ f (tn ) − f (tn+1 ) (tn − tn+1 ) f (tn ) − f (tn+1 ) = tn − tn+1 n=1 n=1 ∞ f (tn ) − f (tn+1 ) p + 1 (tn − tn+1 ) ≤ tn − tn+1 n=1 ∞ ∞ f (tn ) − f (tn+1 ) p (tn − tn+1 ) + (tn − tn+1 ) = tn − tn+1 n=1
n=1
≤ ess V p ( f, X ) + (t1 − t) < ∞. So limn→∞ f (tn ) converges. Thus we see that lim t ∈ R. Define a function g on R by g(t) =
lim h↓0
h↓0
t+h∈ess D( f )
f (t + h)
f (t + h) converges for every
for t ∈ R.
t+h∈ess D( f )
Then it is obvious that g is right continuous on R. Moreover, we see that 1 f (t) = lim h↓0 h
t+h f (s) ds = g(t) for t ∈ ess D( f ). t
Thus f = g a.e. Now let t1 < t2 < t3 < · · · < tn be arbitrary sequence in R. Since ess D( f ) is dense in R, there exists a sequence {tk,m }1≤k≤n,m≥1 ⊂ ess D( f ) such that tk,m ∈ (tk , tk+1 ) for 1 ≤ k ≤ n − 1, tn,m ∈ (tn , ∞) and limm→∞ tk,m = tk for 1 ≤ k ≤ n. For m ≥ 1 we have n−1 g(tk+1,m ) − g(tk,m ) p (tk+1,m −tk,m ) tk+1,m −tk,m k=1 p n−1 f (tk+1,m )− f (tk,m ) (tk+1,m −tk,m ) = tk+1,m −tk,m k=1
≤ ess V p ( f, X ).
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On the essential bounded variation
Since g is right continuous, as m → ∞, we have n−1 g(tk+1 ) − g(tk ) p (tk+1 − tk ) ≤ ess V p ( f, X ), t −t k=1
k+1
k
and so V p (g, X ) ≤ ess V p ( f, X ). Consequently, we conclude that V p (g, X ) = ess V p ( f, X ). Theorem 3 Let X be a Banach space. Let 1 ≤ p < ∞ and f ∈ L loc 1 (R, X ). Then the following are equivalent: (i) ess V p ( f, X ) < ∞ . f (·) (ii) f (·+h)− : h = 0, h ∈ R is bounded in L p (R, X ). h f (·) (iii) lim h→0 f (·+h)− < +∞. h p
Proof These equivalences follow immediately from Theorem 1 and Remark 1. Theorem 4 Let X be a reflexive Banach space. Let 1 < p < ∞ and f ∈ L loc 1 (R, X ). Then the following are equivalent: (i) ess V p ( f, X ) < +∞ . f (·) (ii) f (·+h)− : h = 0, h ∈ R is bounded in L p (R, X ). h f (·) (iii) lim h→0 f (·+h)− < +∞. h p
(iv)
f (·) lim h→0 f (·+h)− h
converges in L p (R, X ).
Then there exists an absolutely continuous function g ∈ L loc 1 (R, X ) such that f = g a.e., p g ∈ L p (R, X ) and ess V p ( f, X ) = V p (g, X ) = g p . Proof The equivalences from (i) to (iii) have already been shown by the previous theorem. Also, it is obvious that (iv) implies (iii), and so it suffices to show that (ii) implies (iv). We suppose that (ii) holds. We define two operators τ (s) and (s) on L loc 1 (R, X ) for s ∈ R by τ (s)g(·) = g(· + s), (s)g = τ (s)g − g for g ∈ L loc 1 (R, X ) We see obviously that τ (s) and (s) are bounded linear operators on L p (R, X ). From the condition (ii), we see that (s) f ∈ L p (R, X ) for every s ∈ R. For every s1 , s2 ∈ R, we have from Remark 1 that
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(s2 ) f − (s1 ) f p = τ (s2 ) f − τ (s1 ) f ) p = τ (s2 − s1 ) f − f p ≤ |s2 − s1 | · (ess V p ( f, X ))1/ p . Thus we see that (·) f is Lipschitz continuous on R in L p (R, X ). On the other hand, since d L p (R, X ) has the Radon–Nikodým property (see [1, p.140]), ds (s) f exist a.e. s ∈ R in (s0 +h) f −(s0 ) f converges in L p (R, X ). L p (R, X ). Choose a point s0 ∈ R such that lim h→0 h Then we have lim
h→0
f (· + h) − f (·) f (· + h) − f (·) = lim τ (−s0 )τ (s0 ) h→0 h h f (· + s0 + h) − f (· + s0 ) = lim τ (−s0 ) h→0 h (s0 + h) f − (s0 ) f = lim τ (−s0 ) h→0 h (s0 + h) f − (s0 ) f , = τ (−s0 ) lim h→0 h
in L p (R, X ). Thus we have (iv). To show the last part, we should note from Theorem 1.3 that there exists a right-continuous function g ∈ L loc 1 (R, X ) such that f = g a.e., and ess V p ( f, X ) = V p (g, X ). Consequently, it suffices to show that g is absolutely continuous and g ∈ L p (R, X ) and ess V p ( f, X ) = p g p . In fact, for every finite number of nonoverlapping intervals (sk , tk ) (k = 1, · · · , n), n g(tk ) − g(sk ) (sk , tk ) and u(t) = let I = for sk < t < tk (k = 1, · · · , n). k=1 t −s k
k
Then we have n
n g(tk ) − g(sk ) (tk − sk ) t −s k k k=1 = u(t) dt
g(tk ) − g(sk ) =
k=1
I
⎛ =⎝ =
⎞1/ p ⎛ ⎞1/q u p (t) dt ⎠ ⎝ dt ⎠
I
(1/ p + 1/q = 1)
I
1/ p n 1/q n g(tk ) − g(sk ) p (tk − sk ) (tk − sk ) t −s k=1
k
≤ V p (g, X )1/ p
k
n
1/q (tk − sk )
k=1
Hence, for every ε > 0, there exists a δ > 0 such that n k=1
123
g(tk ) − g(sk ) < ε
for
n (tk − sk ) < δ. k=1
k=1
On the essential bounded variation
Thus we see that g is absolute continuous on R. Since X has the Radon Nikodým Property, we have lim
h→0
g(t + h) − g(t) = g (t) a.e. t ∈ R. h
On the other hand, we have from (iv) lim
h→0
f (· + h) − f (·) g(· + h) − g(·) = lim h→0 h h
coverges in L p (R, X ), and hence lim
h→0
Thus we have
f (· + h) − f (·) = g (·) in L p (R, X ). h
f (· + h) − f (·) p p = ess V p ( f, X ). g p = lim h→0 h p
Theorem 5 Let 1 ≤ p < ∞ and f ∈ L p (R). Then f ∈ W p1 (R) if and only if lim
h→0
f (· + h) − f (·) h
converges in L p (R). Then the limit is a derivative D f ∈ L p (R) of f in the sense of distribution. Proof For f ∈ L p (R) we define (T (t) f )(s) = f (t + s) s, t ∈ R. Then it is easy to check that {T (t)} is a C0 semigroup of bounded linear operator on L p (R) satisying T (t) ≤ 1 for t ≥ 0. Moreover, it is well known (see [2, p.66, Proposition 1]) that the generator (A, D(A)) is given by differentiation and D(A) = W p1 (R). Remark 2 From Theorem 4 and Theorem 5, for f ∈ L p (R) ( p > 1) we see that f ∈ W p1 (R) if and only if ess V p ( f ) < ∞. But if p = 1, ess V1 ( f ) < ∞ does not imply f ∈ W11 (R) as following theorem shows: Theorem 6 Let f ∈ L 1 (R). Then the following are equivalent: (i) ess V1 ( f ) < ∞ . f (·) (ii) lim h→0 f (·+h)− < +∞. h 1 f (·+h)− f (·) (iii) is bounded in L 1 (R). : h = 0, h ∈ R h (iv) f ∈ BV (R). Then there exists a right continuous function g such that f = g a.e. and |D f |(R) = V1 (g) = ess V1 ( f ). Proof These results are immediate from [4, Theorem 3.4] , Theorem 2 and Theorem 3.
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