Northeast. Math. J., 19:2(2003), 133-138
47
F -Sobolev Inequality for General Symmetric Forms
∗
Wang Feng and Zhang Yuhui
(Department of Mathematics, Beijing Normal University, Beijing, 100875)
Abstract. Some sufficient conditions for the F -Sobolev inequality for symmetric forms are presented in terms of new Cheeger’s constants. Meanwhile, an estimate of the F -Sobolev constants is
obtained.
1
Introduction
The F -Sobolev inequality is a generalization of the logarithmic Sobolev inequality. The latter is initiated
by [1] in 1976 and it has attracted a great deal of research in the past two decades. Refer to the survey
articles [2] and [3]. Very recently, the inequality has been studied in [4] and [5] for general symmetric
forms. The method used in the two papers is the Cheeger’s technique for unbounded operators developed
in [6]. But the proofs are different and a detailed comparison about them is presented in [4]. The purpose
of this paper is to study the F -Sobolev inequality for general symmetric forms by combining the proofs
of [4] with some techniques in [5].
Let (E, E , π) be a measurable probability space satisfying {(x, x) : x ∈ E} ∈ E × E and denote by
L2 (π) the usual real L2 -space with norm k · k. The symmetric form (D, D(D)) considered here on L2 (π)
is as follows
Z
1
D(f, g) =
J(dx, dy)[f (x) − f (y)][g(x) − g(y)],
(1.1)
2
f, g ∈ D(D) := {f ∈ L2 (π) : D(f, f ) < ∞},
where J is a symmetric measure on E × E: J(dx, dy) = J(dy, dx). Without loss of generality, assume
that J({(x, x) : x ∈ E}) = 0.
Let F ∈ C(0, ∞) be such that inf t≥0 tF (t) > −∞ and F (t) > 0 for large t. We say that the F -Sobolev
inequality with dimension p ∈ (2, ∞] (denoted by F S(p)) holds for the symmetric form (D, D(D)) if there
exist two constants C1 > 0, C2 ≥ 0 such that
¡
¢(p−2)/p
π |f |2p/(p−2) F (f 2 )
≤ C1 D(f, f ) + C2 ,
f ∈ D(D), kf k = 1,
(1.2)
Z
holds, where π(f ) = f dπ. In particular, we call the inequality tight (denoted by T F S(p)) if (1.2) holds
for C2 = 0 and the reciprocal of the smallest possible C1 is denoted by σ which is called the F -Sobolev
constant.
The inequality is a general version of Sobolev inequality and logarithmic Sobolev inequality. When
F = 1, (1.2) is just the classical Sobolev inequality with dimension p
¡
¢(p−2)/p
π |f |2p/(p−2)
≤ C1 D(f, f ) + C2 ,
f ∈ D(D), kf k = 1.
(1.3)
When F = log and p = ∞, (1.2) becomes the defective logarithmic Sobolev inequality
π(f 2 log f 2 ) ≤ C1 D(f, f ) + C2 ,
f ∈ D(D), kf k = 1.
which is equivalent to the usual logarithmic Sobolev inequality
π(f 2 log f 2 ) ≤ CD(f, f ),
∗ Foundation
f ∈ D(D), kf k = 1,
item: 973 project, NNSFC (10121101, 10025105, 10101003) and RFDP (20010027007).
(1.4)
48
whenever Poincar´e inequality
π(f 2 ) − π(f )2 ≤ λ−1
1 D(f, f ),
f ∈ D(D),
holds where λ1 is the spectral gap.
We now look for a new Cheeger’s constant which is hoped to best describe F S(p). Following [4] and
[5], take and fix a non-negative and symmetric function r ∈ E × E such that
J (1) (dx, E)/π(dx) ≤ 1,
π-a.s.,
(1.5)
(α)
where J (dx, dy) = I{r(x,y)α >0} J(dx, dy)/r(x, y)α , α ≥ 0. We adopt the convention that r0 = 1 for all
r ≥ 0. Replacing J with J (α) in (1.1), we have the symmetric forms (D(α) , D(D(α) )) generated by J (α) .
Suppose that F satisfies the following conditions
F ∈ C[0, ∞), increasing, F (0) = 1, F 0 is piecewise continuous
tF 0 (t)
(1.6)
< ∞,
and c1 := sup ±
t≥0 F (t)
where F±0 denote the right- and left-derivatives of F . Define
J (1/2) (A × Ac ) + δπ(A)
ξ δ = inf
,
δ > 0,
π(A)>0 π(A)(p−1)/p F (π(A)−1 )(p−2)/(2p)
J(A × Ac )
κ = inf
,
(p−2)/p
π(A)>0 π(A)
F (π(A)−1 )(p−2)/p
³ p − 1 ´2(p−1)/p ³ p − 1 p − 1 ´2/p ³ p
´2(p−1)/p
C(p, c1 ) =
+
c1
+ c1
.
p
p−2
p
p−2
Put ξ 0 := limδ→0 ξ δ = inf δ>0 ξ δ . Some sufficient conditions are presented as follows.
Theorem 1.1. Under (1.6). Then the following conclusions hold.
(1) If there exists some δ > 0 such that ξ δ > 0, then F S(p) holds for C1 = 4C(p, c1 )(2 + δ)(ξ δ )−2
and C2 = C1 δ;
(2) If ξ 0 > 0, then T F S(p) holds and κ ≥ σ ≥ (ξ 0 )2 /(8C(p, c1 )).
Remark 1.1. In particular, if F (r) = 1 + log+ r = 1 + max{0, log r}, p = ∞ and ξ δ > 0 for some δ > 0
±£
in Theorem 1.1, then the logarithmic Sobolev inequality (1.4) holds and κ ≥ σ ≥ λ1 1 + 16 inf δ>0 (2 +
¤
δ)(λ1 + δ)(ξ δ )−2 . See [4, Theorem 1.1].
To prove Theorem 1.1, consider
the following general symmetric form
as in [4]
Z
Z
1
J(dx, dy)[f (x) − f (y)][g(x) − g(y)] + K(dx)f (x)g(x),
D(f, g) =
(1.7)
2
f, g ∈ D(D) := {f ∈ L2 (π) : D(f, f ) < ∞},
where J is the same as before and K is a measure. Choose a non-negative, symmetric function r¯ ∈ E × E
and a non-negative function s ∈ E such that
[J¯(1) (dx, E) + K (1) (dx)]/π(dx) ≤ 1,
π-a.s,
(1.8)
where J¯(α) (dx, dy) is defined as J (α) (dx, dy) but replacing r with r¯, and K (α) (dx) = I{s(x)α >0} K(dx)/s(x)α,
(α)
(α)
α ≥ 0. As in (1.7), we have the symmetric forms (D , D(D )) generated by (J¯(α) , K (α) ). Define
∗(α)
λ0
= inf{D
(α)
(f, f ) : kf k = 1}. Next, set
J¯(1/2) (A × Ac ) + K (1/2) (A)
.
π(A)>0 π(A)(p−1)/p F (π(A)−1 )(p−2)/(2p)
Now we can state another main result of the paper.
¡
¢(p−2)/p
Theorem 1.2. Set σ(F ) = inf{D(f, f )/π |f |2p/(p−2) F (f 2 )
: kf k = 1}. Then
J(A × Ac ) + K(A)
ξ ∗2
inf
≥
σ(F
)
≥
.
∗(1)
π(A)>0 π(A)(p−2)/p F (π(A)−1 )(p−2)/p
4C(p, c1 )(2 − λ0 )
ξ∗ =
inf
(1.9)
Remark 1.2. Theorem 1.2 is an extension of [4, Theorem 2.1] which is in the case of p = ∞. As pointed
out in [4], Theorem 1.2 is true in the general case, i.e. π is a reference measure (See [5]).
49
2
Proofs
In this section, the proofs of the two theorems are presented in details.
¡
¢
Proof of Theorem 1.2. Put EF (f ) = π |f |2p/(p−2) F (f 2 ) . The first inequality is obtained directly
p
by computing D(f, f )/EF (f )(p−2)/p for f = IA / π(A) with π(A) > 0.
a) To prove the second inequality, we need more notations (this is the part a) of the proof of [4,
Theorem 2.1]).
When K(dx) 6= 0, it is convenient to enlarge
define f ∗ = f IE ∈ E ∗ . Next, define a symmetric

(α)

J (C),
∗(α)
J
(C) = K (α) (A),


0,
the space E by letting E ∗ = E ∪ {∞}. For any f ∈ E ,
measure J ∗(α) on E ∗ × E ∗ by
C ∈ E ×E,
C = A × {∞} or {∞} × A, A ∈ E ,
C = {∞} × {∞}.
Then, we have J ∗(α) (dx, dy) = J ∗(α) (dy, dx) and
Z
Z
(α)
2
(α) 2
¯
J ∗(α) (dx, E ∗ )f ∗ (x)2 ,
J (dx, E)f (x) + K (f ) =
E
E∗
Z
1
(α)
D (f, f ) =
J ∗(α) (dx, dy)[f ∗ (y) − f ∗ (x)]2 ,
2 E ∗ ×E ∗
Z
Z
1
J¯(α) (dx, dy)|f (y) − f (x)| +
K (α) (dx)|f (x)|
2 E×E
E
Z
1
∗(α)
=
J
(dx, dy)|f ∗ (y) − f ∗ (x)|.
2 E ∗ ×E ∗
Note that if we set r∗ (x, y) = r¯(x, y), r∗ (x, ∞) = r∗ (∞, x) = s(x) for all x, y ∈ E and r∗ (∞, ∞) = 0,
then J ∗(α) (dx, dy) can be also expressed by I{r∗ (x,y)α >0} J ∗ (dx, dy)/r∗ (x, y)α .
p
b) Following [4] and [5], take ϕ(t) = t(p−1)/(p−2) F (t) and η(t) = ϕ(t2 ). Note that ϕ is a strictly
increasing function and so is η. Moreover,
·
¸
·
¸
p
p − 1 tF 0 (t)
2η(t) p − 1 t2 F 0 (t2 )
0
0
1/(p−2)
F (t)
+
, η (t) =
+
.
ϕ (t) = t
p−2
2F (t)
t
p−2
2F (t2 )
except a finite number of points on each finite interval. By (1.6), we have η 0 (t) ≤ c2 η(t)/t = c2 tp/(p−2)
p
· F (t2 ) where c2 := 2(p − 1)/(p − 2) + c1 . Given s < t, label the discontinuous points in [s, t] by
s = t1 < · · · < tm = t. Then, by the Mean Value Theorem and the monotonicity of η(t)/t, there exist
θi ∈ (ti , ti+1 ) such that
0 ≤ η(t) − η(s) =
m−1
X
[η(ti+1 ) − η(ti )] =
i=1
≤ c2
m−1
X
i=1
η(θi )
η(t)
(ti+1 − ti ) ≤ c2
θi
t
m−1
X
η 0 (θi )(ti+1 − ti )
i=1
m−1
X
i=1
(ti+1 − ti ) = c2
η(t)
(t − s).
t
c) Let f ≥ 0, kf k = 1 and set g ∗ = ϕ(f ∗2 ). Then by b), we have |g ∗ (y) − g ∗ (x)| ≤ c2 |f ∗ (y) −
η(f ∗ (x) ∨ f ∗ (y))
f ∗ (x)| ∗
. Thus, by Cauchy-Schwarz inequality and (1.8), we have
f (x) ∨ f ∗ (y)
Z
1
I ∗ :=
J ∗(1/2) (dx, dy)|g ∗ (y) − g ∗ (x)|
2
Z
i1/2 · Z
h
p
c2 h
∗
∗
∗
2
J ∗(1) (dx, dy) f ∗ (y)p/(p−2) F (f ∗ (y)2 )
≤
J (dx, dy)|f (y) − f (x)|
2
i2 ¸1/2
p
∗
p/(p−2)
∗
2
F (f (x) )
+f (x)
50
c2
=√
2
q
·Z
h
D(f, f )
J ∗(1) (dx, dy) 2f ∗ (y)2p/(p−2) F (f ∗ (y)2 ) + 2f ∗ (x)2p/(p−2) F (f ∗ (x)2 )
³
´2 i¸1/2
p
p
∗
p/(p−2)
∗
p/(p−2)
∗
2
∗
2
− f (y)
F (f (y) ) − f (x)
F (f (x) )
q
h
i1/2
p
p
c2
(1)
=√
D(f, f ) 4EF (f ) − 2D (f p/(p−2) F (f 2 ), f p/(p−2) F (f 2 ))
2
q
∗(1)
≤ c2 (2 − λ0 )D(f, f )EF (f ).
(2.1)
d) Define h(t) = π(f ∗2 > t), At = {g ∗ > t} = {ϕ(f ∗2 ) > t}. Then h(t) ≤ 1 ∧ t−1 , π(At ) = π(f ∗2 >
ϕ−1 (t)) = h ◦ ϕ−1 (t), where ϕ−1 denotes the inverse function of ϕ. By definition of ξ ∗ , we have
·Z ∞
¸p/(p−1)
I ∗p/(p−1) =
J ∗(1/2) (At × Act )dt
0
≥ξ
·Z
∗p/(p−1)
¸p/(p−1)
∞
(p−1)/p
π(At )
−1 (p−2)/(2p)
F (π(At )
)
0
·Z
=ξ
∗p/(p−1)
∞
(p−1)/p
h(s)
−1 (p−2)/(2p)
F (h(s)
)
dt
¸p/(p−1)
ϕ (s)ds
0
0
¸p/(p−1)
h(t)(p−1)/p F (t)(p−2)/(2p) ϕ0 (t)dt
0
Z
p ∗p/(p−1) ∞
=
ξ
h(t)(p−1)/p F (t)(p−2)/(2p) ϕ0 (t)
p−1
0
·Z t
¸1/(p−1)
(p−1)/p
(p−2)/(2p) 0
·
h(s)
F (s)
ϕ (s)ds
dt
0
Z ∞
p ∗p/(p−1)
ξ
h(t)(p−1)/p F (t)(p−2)/(2p) ϕ0 (t)
≥
p−1
0
·
¸1/(p−1)
Z t
· h(t)(p−1)/p
F (s)(p−2)/(2p) ϕ0 (s)ds
dt
·Z
≥ ξ ∗p/(p−1)
∞
0
·Z t
¸1/(p−1)
p ∗p/(p−1)
(p−2)/(2p) 0
(p−2)/(2p) 0
ξ
=
h(t)F (t)
ϕ (t)
F (s)
ϕ (s)ds
dt.
p−1
0
0
Next, by (1.6) and the absolute continuity of F , we can obtain
Z t
F (s)(p−2)/(2p) ϕ0 (s)ds ≥ c3 t(p−1)(p−2) F (t)(p−1)/p ,
t ≥ 0,
³ p − 1 10 ´.³ p − 1 p − 1 ´
p
+ c1
+
c1 >
. Hence,
where c3 =
p−2 2
p−2
p
2(pZ− 1)
p
p 1/(p−1) ∗p/(p−1) ∞
I ∗p/(p−1) ≥
c3
ξ
h(t)t1/(p−2) F (t)ϕ0 (t)dt
p−1
0
Z
Z f ∗2
p
p 1/(p−1) ∗p/(p−1)
dπ
=
c3
ξ
t1/(p−2) F (t)ϕ0 (t)dt.
p−1
0
By (1.6) and the absoluteZcontinuity of F once again, we can derive
r
p
t1/(p−2) F (t)ϕ0 (t)dt ≥ c4 rp/(p−2) F (r),
r ≥ 0,
0
³ p − 1 1 ´.³
´
p
1
where c4 =
+ c1
+ c1 > . Thus,
p−2 2
p−2
2
Z
p 1/(p−1) ∗p/(p−1)
∗p/(p−1)
I
dπf ∗2p/(p−2) F (f ∗2 )
≥
c
c4 ξ
p−1 3
p 1/(p−1) ∗p/(p−1)
=
c
c4 ξ
EF (f ).
(2.2)
p−1 3
e) Combining (2.1) with (2.2), we get
q
h p
i(p−1)/p
1/(p−1)
∗(1)
∗p/(p−1)
c
c4 ξ
EF (f )
≤ c2 (2 − λ0 )D(f, f )EF (f ).
p−1 3
Z
∞
51
That is,
³ p ´2(p−1)/p c2/p c2(p−1)/p ξ ∗2
D(f, f )
ξ ∗2
≥
· 3 4
=
.
(p−2)/p
∗(1)
∗(1)
p−1
EF (f )
c22 (2 − λ0 )
4C(p, c1 )(2 − λ0 )
So the second inequality of Theorem 1.2 holds. The proof is comleted.
Proof of Theorem 1.1. Let (1.5) holds for some symmetric function r. Fix δ > 0 and take K(dx) =
δπ(dx). Next, take r¯ = (1 + δ)r and s = 1 + δ so that (1.8) holds. Then
δ
ξδ
∗(α)
,
λ0 =
ξ∗ = √
.
(1 + δ)α
1+δ
Therefore, for the symmetric form (D, D(D)) given in (1.1), by Theorem 1.2, we obtain
D(f, f ) + δ
(ξ δ )2
(ξ δ )2 /(1 + δ)
σ(F ) = inf
=
,
≥
4C(p, c1 )(2 − δ/(1 + δ))
cδ
kf k=1 EF (f )(p−2)/p
where cδ = 4C(p, c1 )(2 + δ). Then we have
π(|f |2p/(p−2) F (f 2 ))(p−2)/p ≤ cδ (ξ δ )−2 [D(f, f ) + δ],
f ∈ D(D), kf k = 1.
(2.3)
δ −2
By (2.3), it is easily seen that Theorem 1.1(i) holds and C1 = cδ (ξ ) , C2 = C1 δ. Let δ → 0 in (2.3).
p
Then it follows that T F S(p) holds and σ ≥ (ξ 0 )2 /(8C(p, c1 )). Take f = IA / π(A) (π(A) > 0) and
compute D(f, f )/EF (f )(p−2)/p . We have σ ≤ κ. The proof is completed. By the way, the estimate for
the F -Sobolev constant can be obtained by taking K be zero-measure in (1.9) directly.
Remark 2.1. Note that it is easily proven that
ξ0 =
inf
J (1/2) (A × Ac )
π(A)>0 π(A)(p−1)/p F (π(A)−1 )(p−2)/(2p)
.
So the qualitative conclusion of Theorem 1.1(ii) is obtained by [5, Theorem 1.1].
Acknowledgement The authors would like to acknowledge Prof. Chen Mufa for his valuable advices
during the writing of the paper.
References
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