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ON THE SCALING LIMIT OF A SINGULAR INTEGRAL OPERATOR 1

I.M. Krichever and D.H. Phong

Department of Mathematics

arXiv:0708.4157v1 [math.CV] 30 Aug 2007

Columbia University, New York, NY 10027

1

Introduction

Recently an alternative approach to Birkho?’s theory of di?erence equations [1] has been proposed in [4]. This approach leads naturally to local monodromies of di?erence equations, which should converge in principle to monodromy matrices of di?erential equations, thus providing a missing link in the theory of isomonodromic transformations of systems of linear di?erence equations (see e.g. [2, 3, 7] and references therein). The key to the convergence process in [4] is the scaling limit of a certain singular integral operator I, arising from a Riemann-Hilbert problem. The operator I acts on functions φ de?ned on the vertical line with ?xed abscisse at a, and its kernel k(z, ξ) is given explicitly by eπi(z?a) + e?πi(z?a) k(z, ξ) = iπ(ξ?a) (e + e?πi(ξ?a) )(eiπ(ξ?z) + e?πi(ξ?z) ) (1.1)

If we set πz = πa + iy, πξ = πa + iη, y, η ∈ R, and view φ as a function of η, we can de?ne the following re-scaled versions Iλ of I, Iλ (φ)(y) = P.V.

∞ ?∞

1 e?λy + eλy φ(η) dη, eλ(y?η) ? e?λ(y?η) e?λη + eλη

(1.2)

where P.V. denotes principal values. As noted in [4], an essential property of the operators Iλ is their formal limit,

y

Iλ (φ)(y) →

φ(ξ)dξ,

0

λ → +∞.

(1.3)

The purpose of the present paper is to provide a detailed study of the boundedness properties of the operators I and Iλ in suitable spaces of Schauder type, and to establish a precise version of the formal limit (1.3). Near the diagonal, the singularities of the kernels of Iλ are the same as for the Hilbert transform, and the techniques for handling the local behavior of such kernels are well-known. The main novel feature in our case is rather their global behavior near ∞. This global behavior prevents their boundedness on scale-invariant spaces, and accounts for the existence of non-trivial limits such as (1.3).

1

Research supported in part by National Science Foundation grants DMS-02-45371 and DMS-04-05519

2

Schauder estimates with exponential growth

We introduce the following norms of Schauder type for functions on R. Fix κ ∈ R, m ∈ Z, 0 < α < 1, and let Λα (m,κ) be the space of functions φ on R satisfying the conditions |φ(x)| ≤ C (1 + |x|)m eκ|x| , |φ(x) ? φ(y)| ≤ C |x ? y|α {(1 + |x|)m eκ|x| + (1 + |y|)meκ|y| }, (2.1) to be the in?mum of the constants C for which these for all x, y ∈ R. We de?ne ||φ||Λα (m,κ) α inequalities hold. We also require the space Λα (log,κ) and the corresponding norm ||φ||Λ(log,κ) de?ned by the conditions |φ(x)| ≤ C log (1 + |x|) eκ|x|, |φ(x) ? φ(y)| ≤ C |x ? y|α { log (1 + |x|) eκ|x| + log (1 + |y|)eκ|y|}, The singular integral operator I can be expressed as I(φ)(y) = (e?y + ey )H( e?(·) 1 φ(·)) + e(·) (2.3) (2.2)

where H is the following exponentially decaying version of the classical Hilbert transform, (Hψ)(y) = P.V. Set

∞ ?∞ ey?η

1 ψ(η) dη ≡ lim? → 0 ? e?(y?η)

|y?η|>? ey?η

1 ψ(η) dη. ? e?(y?η) (2.4) (2.5)

1 . ? e?z Then the kernel K(z) is C ∞ (R \ 0), odd, and satis?es K(z) = ez |K(z)| ≤ C |z|?1 , if |z| ≤ 1; e?|z| , if |z| > 1. |?z K(z)| ≤ C |z|?2 , if |z| ≤ 1; e?|z| , if |z| > 1.

(2.6)

In particular, these are better estimates than for the standard Hilbert transform kernel K0 (z) = z ?1 , and it follows at once that the operator H is bounded on the standard Schauder spaces (see e.g. [6, 5]). To obtain estimates for the operator I, we need the boundedness of H on the above spaces Λα , and this is provided by the following theorem: (m,κ) Theorem 1 Fix 0 < α < 1, m ∈ Z. The operator H is bounded on the following Schauder spaces, ||Hψ||Λα ≤ Cm,α,k ||ψ||Λα , ?1 < κ < 1. (2.7) (m,κ) (m,κ) For κ = ?1, we have the following bounds, for m ∈ Z, m ≥ ?1, ≤ ||Hψ||Λα (m+1,?1) ≤ ||Hψ||Λα (log,?1) , Cm,α ||ψ||Λα (m,?1) , Cα ||ψ||Λα (m,?1) 2 m ≥ 0, m = ?1. (2.8)

Proof. The method of proof is the standard method for Schauder estimates for singular integral operators. The only new feature here is the control of Hψ(x) for x large. In view of the fact that K(z) is odd and exponentially decreasing, we can write Hψ(x) =

∞ ?∞

K(x ? y) (ψ(y) ? ψ(x))dy,

(2.9)

where the integrals on the right hand side are now convergent for ψ ∈ Λα (m,κ) with 0 < α < 1, κ < 1. In particular, |Hψ(x)| ≤

|x?y|<1

|x ? y|?1+α((1 + |x|)m eκ|x| + (1 + |y|)meκ|y| ) dy e?|x?y| ((1 + |x|)m eκ|x| + (1 + |y|)meκ|y| ) dy. (2.10)

+

|x?y|≥1

These are clearly bounded for |x| bounded, so we may assume that |x| ≥ 3. In this case, 1 ≤ |x| ? 1 ≤ |y| ≤ |x| + 1 ≤ 2|x| in the integral over the region |x ? y| < 1, and 2|x| (1 + |y|)meκ|y| ≤ Cκ (1 + |x|)m eκ|x| . Thus the ?rst integral is bounded by C (1 + |x|)m eκ|x| . The same upper bound for the second integral follows from the following lemma: Lemma 1 For any ?1 < κ < 1, and any m ∈ Z, we have for all |x| > 3

R

e?|z| (1 + |x ? z|)m eκ|x?z| dz ≤ Cm,κ (1 + |x|)m eκ|x| .

(2.11)

For κ = ?1, we have for m ∈ Z, m ≥ ?1, e?|z| (1 + |x ? z|)m e?|x?z| dz ≤ Cm e?|x| (1 + |x|)m+1 , e?|x| log (1 + |x|), if m ≥ 0 if m = ?1. (2.12)

R

Proof of Lemma 1. We consider separately the cases of 0 ≤ κ < 1, ?1 < κ < 0, and κ = ?1. When 0 ≤ κ < 1, we write eκ|x?z| ≤ eκ|x| eκ|z| , and hence the integral on the left hand side of the above inequality can be bounded by eκ|x|

1 |x?z|≥ 2 |x|

e?(1?κ)|z| (1 + |x ? z|)m dz + eκ|x|

1 |x?z|< 2 |x|

e?(1?κ)|z| (1 + |x ? z|)m dz. (2.13)

In the ?rst integral we can write (1 + |x ? z|)m ≤ Cm (1 + |x|)m (1 + |z|)m . (2.14)

This is certainly true with Cm = 1 if m ≥ 0. If m < 0, then we use the condition 1 |x ? z| ≥ 2 |x| to write (1 + |x ? z|)m ≤ 2?m (1 + |x|)m , and the inequality still holds. Since κ < 1, the desired bound follows for the ?rst integral. Next, in the second integral, we 3 have 1 |x| < |z| < 2 |x|, and we can write 2

1 |x?z|< 2 |x|

e?(1?κ)|z| (1 + |x ? z|)m dz ≤ e?

1?κ |x| 2 3 |z|≤ 2 |x|

(1 + |x|)|m| (1 + |z|)|m| dz (2.15)

≤ CN (1 + |x|)?N , 3

for arbitrary N. This proves the lemma when 0 ≤ κ < 1. When ?1 < κ < 0, we write instead e?|z| eκ|x?z| = e?(1+κ)|z| eκ(|z|+|x?z|) ≤ e?(1+κ)|z| eκ|x| (2.16) and bound the integral on the left hand side of the lemma by eκ|x|

1 |x?z|≥ 2 ||x|

e?(1+κ)|z| (1 + |x ? z|)m dz + eκ|x|

1 |x?z|< 2 ||x|

e?(1+κ)|z| (1 + |x ? z|)mdz. (2.17)

The bounds for these integrals are now the same as in the previous case. This establishes the estimate (2.11). Finally, consider the case κ = ?1. In the region of integration 1 |x?z| > 4|x|, we have the integrand can be crudely bounded by e?2|x| (1+|x?z|)m e? 2 |x?z| , and hence the contribution of this region is O(e?2|x| ), which is better than we actually need. Thus it su?ces to consider the region |x ? z| ≤ 4|x|. We write then

|x?z|<4|x|

e?|z| (1 + |x ? z|)m e?|x?z| dz ≤ e?|x|

|x?z|<4|x|

(1 + |x ? z|)m

(2.18)

from which the desired estimate follows at once. The proof of the lemma is complete. Q.E.D. We return to the proof of the theorem. Let x, x′ ∈ R and set δ = |x ? x′ |. The next step is to estimate Hψ(x) ? Hψ(x′ ), which can be expressed as

|y?x|<3δ

K(x ? y)(ψ(y) ? ψ(x))dy ?

|y?x′ |<3δ

K(x′ ? y)(ψ(y) ? ψ(x′ ))dy (2.19) K(x′ ? y)(ψ(y) ? ψ(x′ ))dy.

+

|y?x|≥3δ

K(x ? y)(ψ(y) ? ψ(x))dy ?

|y?x′ |≥3δ

The ?rst two integrals can be estimated as in the bounds for |Hψ(x)|. For example, |

|y?x|<3δ

K(x ? y)(ψ(y) ? ψ(x))dy| ≤ ||ψ||Λα (m,κ)

|x?y|<3δ

|x ? y|?1+α {(1 + |x|)m eκ|x|

+(1 + |y|)meκ|y| } ≤ C ||ψ||Λα δ α (1 + |x|)m eκ|x| (m,κ)

(2.20)

since (1 + |y|)meκ|y| ≤ C (1 + |x|)m eκ|x| for |x| ≥ 3 and δ << 1. To estimate the remaining two integrals, write

|y?x′ |≥3δ

K(x′ ? y)(ψ(y) ? ψ(x′ ))dy = =

|y?x′ |≥3δ

K(x′ ? y)(ψ(y) ? ψ(x))dy

|y?x′ |≥3δ,|y?x|<3δ

(2.21)

|y?x|≥3δ

+

?

|y?x′ |<3δ,|y?x|>3δ

The last two integrals on the right hand side satisfy the desired bounds, because in their ranges of integration, we have |y ? x| ? |y ? x′ | ? δ, and the same arguments above apply. The remaining integral can be combined with the third integral in (2.19) to give

|y?x|>3δ

(K(x ? y) ? K(x′ ? y))(ψ(y) ? ψ(x))dy. 4

(2.22)

Since we have |K(x ? y) ? K(x′ ? y)| ≤ |x ? y| · |?x K(z)| (2.23) for some z in the segment between x ? y and x′ ? y, and hence |z| ? |x ? y| when |y ? x| > 3|x ? x′ |, we can write, in view of the bounds for the |?z K(z)|, |

|y?x|>3δ

(K(x ? y) ? K(x′ ? y))(ψ(y) ? ψ(x))dy|

3δ<|x?y|<1

≤ δ ||ψ||Λα m,κ +δ ||ψ||Λα m,κ

|x ? y|?2+α{(1 + |x|)m eκ|x| + (1 + |y|)meκ|y| } dy (2.24)

|x?y|>1

e?|x?y| {(1 + |x|)m eκ|x| + (1 + |y|)meκ|y| } dy.

The ?rst integral on the right hand side is bounded by δ ||ψ||Λα m,κ

3δ<|x?y|<1

|x ? y|?2+α{(1 + |x|)m eκ|x| + (1 + |y|)meκ|y| } dy

3δ<|x?y|<1

≤ δ ||ψ||Λα (1 + |x|)m eκ|x| m,κ

|x ? y|?2+α ≤ C δ α (1 + |x|)m eκ|x| . (2.25)

Applying Lemma (1), we obtain similar bounds for the second integral. Altogether, we have shown that |Hψ(x) ? Hψ(x′ )| ≤ C ||ψ||Λα |x ? x′ |α (1 + |x|)m eκ|x| (m,κ) (2.26)

for |x ? x′ | small, and the theorem is proved when ?1 < κ < 1. The case κ = ?1 is established exactly in the same way, using the corresponding estimates in Lemma 1 for κ = ?1 and m ≥ ?1. The proof of Theorem 1 is complete. Q.E.D. Theorem 2 For 0 < κ < 2, m ∈ Z, and 0 < α < 1, the operator I is bounded on the following Schauder spaces, ||I(φ)||Λα ≤ Cm,κ,α ||φ||Λα , (m,κ) (m,κ) ≤ ||I(φ)||Λα (m+1,κ) ≤ ||I(φ)||Λα (log,κ) 0 < κ < 2. (2.27)

For κ = 0, m ∈ Z, m ≥ ?1, the operator I satis?es the following bounds, Cm,α ||φ||Λα , (m,κ) Cα ||φ||Λα , (m,κ) m≥0 m = ?1. (2.28)

Proof. This is an easy consequence of Theorem 1, the fact that the map φ → ψ(y) = 1 α φ(y) is a one-to-one and onto map M from Λα (m,κ) → Λ(m,κ?1) , with equivalent ey +e?y norms (2.29) ? ||φ||Λα . ||ψ||Λα (m,κ) (m,κ?1) and the relation I(φ) = M ?1 HMφ. Q.E.D. We observe that these bounds always require some space which is not scale-invariant. Thus bounds for Iλ cannot be obtained by scaling the bounds for I, and this explains partly the possibility of the scaling limits discussed in the next section. 5

3

The scaling limit of Iλ

We come now to the operators Iλ . The estimates for I in the previous section show that Iλ cannot be treated by simple scaling arguments from I. Instead, we shall study the bounds and limits for Iλ as λ → +∞ directly. It is simplest to carry this out for functions φ satisfying conditions of the form, |? l φ(x)| ≤ Ck (1 + |x|)m , 0 ≤ l ≤ k, (3.1)

for ?xed m ∈ N, k ∈ N, and norms ||φ||Λk de?ned to be the best constant Ck for which (m) the above condition holds. The following theorem describes the limit of Iλ in these spaces, although it should be clear from the proof and from the previous section that other more precise versions can be formulated as well: Theorem 3 Fix m ∈ N. Then we have the following bounds, uniform in λ and in φ ∈ Λ1 , (m)

y

||Iλ (φ)(y) ?

0

φ(ξ)dξ||Λ0 ≤ Cm λ? 2 ||φ||Λ1 . (m) (m+1)

1

(3.2)

Proof. Formally, if we write Iλ φ(y) = with Kλ (y, η)φ(η) dη (3.3)

e?λy + eλy 1 e?λη + eλη eλ(y?η) ? e?λ(y?η) then for, say, y > 0, we have the pointwise limit Kλ (y, η) = Kλ (y, η) → 1, 0, if 0 < η < y if η < 0 or η > y.

(3.4)

(3.5)

Thus, formally, the left hand side of the expression in the theorem tends to 0 as λ → +∞. However, none of the integrals involved is uniformly nor absolutely convergent, and we have to proceed with care. Fix y > 0 (the case of y < 0 being similar). The key to the estimates is the following break-up of the principal value integral de?ning Iλ (φ), (Iλ (φ))(y) = eλy + e?λy ( ψ(y ? t) ? ψ(y + t) )dt + 0 eλt ? e?λt ≡ (A) + (B)

y

|t|>y

eλy + e?λy ψ(y ? t) dt eλt ? e?λt (3.6)

with ψ(η) = eλη

1 φ(η). + e?λη 6

(3.7)

To estimate (A), we apply Taylor’s formula ψ(y ? t) ? ψ(y + t) = t which gives in this particular case, ψ(y ? t) ? ψ(y + t) = t

1 ?1 1 ?1

ψ ′ (y ? ρt)dρ

(3.8)

dρ (

1 φ′ (y ? ρt) ?λ(η?ρt) +e λ(y?ρt) e ? e?λ(y?ρt) ?λ λ(y?ρt) φ(y ? ρt) ) (e + e?λ(y?ρt) )2 eλ(y?ρt)

(3.9)

Thus (A) can be rewritten as (A) = 1 λ ?

y 1

dt

0 y ?1 1

dρχλ (ρ, t)φ′ (y ? ρt) dρχλ (ρ, t) eλ(y?ρt) ? e?λ(y?ρt) φ(y ? ρt) eλ(y?ρt) + e?λ(y?ρt) (3.10)

dt

0 ?1

≡ A1 + A0 where the function χλ (ρ, t) is de?ned by χλ (ρ, t) = λt eλy + e?λy . eλt ? e?λt eλ(y?ρt) + e?λ(y?ρt)

(3.11)

The following sharp estimates for χλ (ρ, t) play an essential role in the sequel: Lemma 2 For all 0 < t < y, the functions χλ (ρ, t) satisfy the following properties (a) (b) λt 1 λt e?λt(1?ρ) < χλ (ρ, t) ≤ 2 e?λt(1?ρ) , ?2λt 2 1?e 1 ? e?2λt 1 1 ≤ χλ (ρ, t)dρ ≤ 2. 2 ?1 |ρ| < 1, 0 < t < y (3.12)

Proof. In the region 0 < t < y, we have y ? ρt > 0 for all |ρ| < 1, and thus 1 λρt eλy + e?λy e ≤ λ(y?ρt) ≤ 2 eλρt . ?λ(y?ρt) 2 e +e (3.13)

The upper bound implies (a), while the lower bound implies (b), when combined with the following explicit formula 1 1 eλρt dρ = ( eλt ? e?λt ). (3.14) λt ?1 The proof of Lemma 2 is complete. We can now show that A1 → 0 with a precise rate: 7

Lemma 3 The term involving φ′ above tends to 0 at the following rate, |A1 | ≤ Cm Proof. It su?ces to write |A1 | ≤ 1 ||φ||Λ1 (m) λ

y 0

1 ||φ||Λ1 (1 + y)m+1 . (m,0) λ

1 ?1

(3.15)

dt (1 + |y|m + |t|m )

dρ χλ (ρ, t)

(3.16)

and the desired estimate follows from the statement (b) of Lemma 2. Q.E.D. The estimates in Lemma 2 show that χλ (ρ, t) provide an approximation of the Dirac measure concentrated at ρ = 1,

1 ?1

1 χλ (ρ, t) → δ(? ? 1) χλ (?, t)d?

(3.17)

A precise version of this statement with sharp estimates is given in the next lemma. Set

1 ?1

dρ χλ (ρ, t)

1 ?1

=

e?λ(y?ρt) ? eλ(y?ρt) φ(y ? ρt) ? φ(y ? t) e?λ(y?ρt) + eλ(y?ρt) e?λ(y?ρt) ? eλ(y?ρt) dρ χλ (ρ, t) ( ?λ(y?ρt) ? 1) φ(y ? ρt) e + eλ(y?ρt) dρ χλ (ρ, t) (φ(y ? ρt) ? φ(y ? t)) + (

1 ?1

(3.18)

+

1 ?1

dρ χλ (ρ, t) ? 1)φ(y ? t).

Lemma 4 For all 0 < t < y, and any δ > 0 and small, we have the following estimates, with absolute constants, (a) (b) (c)

1 ?1 1 ?1 1 ?1

dρ χλ (ρ, t) ? 1 · |φ(y ? t)| ≤ ||φ||Λ0 (1 + y)m (e?λy + e?2λ(y?t) ) (m) dρ χλ (ρ, t) (φ(y ? ρt) ? φ(y ? t)) ≤ ||φ||Λ1 (1 + y)m ( δt + e?λδt ) (m)

(3.19) (3.20)

dρ χλ (ρ, t)

eλ(y?ρt) ? e?λ(y?ρt) ? 1 |φ(y ? ρt)| ≤ C e?2λ(y?t) ||φ||Λ0 (1 + y)m . (m) eλ(y?ρt) + e?λ(y?ρt) (3.21)

Proof. To prove (a), we write e?λy + eλy ? eλρt ?λ(y?ρt) ? eλ(y?ρt) e In particular,

1 ?1

e?2λy ? e?2λ(y?ρt) (3.22) 1 + e?2λ(y?ρt) ≤ eλρt (e?2λy + e?2λ(y?ρt) ) ≤ eλρt (e?2λy + e?2λ(y?t) ). = eλρt

dρ χλ (ρ, t) ?

λt λt ? e?λt e

1 ?1

dρ eλρt ≤

λt λt ? e?λt e 8

1 ?1

dρ eλρt (e?2λy + e?2λ(y?t) ). (3.23)

Since

|ρ|<1

dρ eλρt = (λt)?1 (eλt ? e?λt ), the statement (a) follows.

To establish the statement (c), we begin by noting that e?2λ(y?ρt) ≤ 1 ? e?2λ(y?ρt) eλ(y?ρt) ? e?λ(y?ρt) =2 ≤ 2 e?2λ(y?ρt) . eλ(y?ρt) + e?λ(y?ρt) 1 + e?2λ(y?ρt) (3.24)

Using the estimate for χλ in Lemma 2 and carrying out explicitly the integral in ρ gives

1 ?1

dρ χλ (ρ, t)

eλ(y?ρt) ? e?λ(y?ρt) ?1 eλ(y?ρt) ? e?λ(y?ρt)

≤

1 λt e?λ(t+2y) dρe3λρt ?2λt 1?e ?1 1 ?2λ(y?t) 1 ? e?6λt = e ≤ C e?2λ(y?t) , 3 1 ? e?2λt

which implies immediately (c). To establish (b), let δ > 0 be any number su?ciently small and to be chosen suitably later. Write

1 ?1

dρ χλ (ρ, t) (φ(y ? ρt) ? φ(y ? t)) =

1?δ ?1

+

1 1?δ

≡ Iδ + IIδ .

(3.25)

The second term on the right hand side can be estimated by, |IIδ | ≤ δ t ||φ||Λ1 (1 + y)m (m)

1 1?δ

dρ χλ (ρ, t) ≤ δ t ||φ||Λ1 (1 + y)m, (m)

(3.26)

while the ?rst term can be estimated using Lemma 2, |Iδ | ≤ 2 ||φ||Λ1 (1 + y)m (m)

δ λt dρ e?λt(1?ρ) ?2λt ?1 1?e ?λt(2?δ) m ?λδt 1 ? e 1 = 2 ||φ||Λ(m) (1 + y) e ≤ C sup[0,2y] |φ| e?λδt . ?2λt 1?e

(3.27)

The proof of Lemma 4 is complete. Q.E.D. We can now carry out the integral in t. The precise estimates are given in the next lemma: Lemma 5 For any 0 < y, we have the following estimates,

y

dt

0

λt eλ(y?ρt) ? e?λ(y?ρt) (eλy + e?λy ) λ(y?ρt) φ(y ? ρt) ? eλt ? e?λt (e + e?λ(y?ρt) )2 ?1 y y + 1 ). ≤ C ||φ||Λ1 (1 + y)m ( (m) 1 + λy λ 2

1

y 0

dρ

dt φ(y ? t) (3.28)

with a constant C independent of y and of λ. 9

Proof. In view of the de?ning formula (3.11) for the function χλ (ρ, t) and the break up (3.18), the left hand side of the desired inequality is bounded by the integral in t of the three inequalities in Lemma 4. This gives the following upper bound, ||φ||Λ1 (1 + y)m (m)

y 0

dt (e?λy + 2e?2λ(y?t) + δt + e?δλt ).

(3.29)

The integral can be evaluated explicitly, and we ?nd 1 1 1 ye?λy + (1 ? e?2λy ) + δy 2 + (1 ? e?δλ ). λ 2 δλ (3.30)

We consider the sum of the ?rst two terms: when λy < 1, it is bounded by C y, where C is an absolute constant. When λy ≥ 1, it is bounded by C λ?1 . Thus we have y 1 . ye?λy + (1 ? e?2λy ) ? λ 1 + λy (3.31)

Next, we consider the optimal choice of δ so as to minimize the size of the sum of the remaining two terms in the above integral. We note that we may assume that δλ > 1, since otherwise the term (δλ)?1 (1 ? e?δλ ) is of size 1, and we do not even get convergence to 0. Thus we should take δλ > 1, in which case the sum of the two remaining terms is of size 1 δy 2 + (3.32) δλ which attains its lowest size yλ? 2 if we set δ = y ?1λ? 2 . This gives the estimate stated in the lemma. Q.E.D. We return now to the estimate of the contribution to Iλ (φ)(y) of the integral in t from the region |t| > y. Lemma 6 For any 0 < y, we have the following estimate 1 eλy + e?λy φ(y ? t)dt eλt ? e?λt eλ(y?t) + e?λ(y?t) 1 1 ≤ Cm ||φ||Λ0 (1 + log (1 + ) (m) λ λy

1 1

|t|>y

Proof. Consider ?rst the contribution from the region t > y. In this region, we have eλy + e?λy eλy 1 λ(2y?t) e ≤ λ(y?t) ≤ 2 λ(t?y) = 2 eλ(2y?t) 2 e + e?λ(y?t) e (3.33)

Thus the contribution from the region t > y to the integral on the left hand side of the desired inequality can be bounded by e?2λ(t?y) |φ(y ? t)| dt = 1 ? e?2λt e?2λs |φ(?s)| ds 1 ? e?2λ(s+y) 0 ∞ e?2λs (1 + |s|m ) ds. ≤ ||φ||Λ0 (m) 0 1 ? e?2λ(s+y)

∞

t>y

(3.34)

10

We claim that for all m ∈ N, we have

∞ 0

1 1 e?2λs (1 + |s|m ) ds ≤ Cm (1 + log (1 + ). ?2λ(s+y) 1?e λ λy

(3.35)

In fact, setting ? = e?2λy and making the change of variables s → u, e?2λu = s, this integral can be rewritten as 1 2λ

1 0

du 1 1 (1 + log )m . 1 ? u? 2λ u

1 2

(3.36) ≤ u ≤ 1. In the ?rst region,

We break it into two regions of integration 0 < u < the integral is of size

1 2 1

and

1 2

0

du 1 1 1 1 2 (1 + (1 + log )m ≤ 2 log )m 1 ? u? 2λ u 2λ u 0 e?λ 2 1 ≤ ( log )m du + 2m+1 m 0 λ u

1 2

e?λ

du ≤ Cm . (3.37)

In the second region, we have

1

1 2

1 1 du (1 + log )m ≤ Cm 1 ? u? 2λ u

1

1 2

du 1 ? u?

(3.38)

This last integral can be evaluated explicitly, and we ?nd that it is bounded by (1+ log (1+ 1 ). This is the desired estimate. λy Next, consider the contribution of the region t < ?y. In this region, we have instead 1 λt eλy + e?λy eλy e ≤ λ(y?t) ≤ 2 λ(y?t) ≤ 2eλt 2 e + e?λ(y?t) e (3.39)

The contribution from t < ?y to the left hand side of the desired inequality can then be bounded by ||φ||Λ0 (1 + y)m (m)

?y ?∞

1 1 e2λt (1 + |t|)m dt ≤ Cm (1 + log (1 + ), 2λt 1?e λ λy

(3.40)

as was to be shown. Q.E.D. The bound provided by Lemma 6 involves a log (λy)?1 term, and is not adequate for y close to 0. This is because the integral is only a principal value integral when |t ? y| is small, and the estimates we have just derived for the contribution of the region t > y do not take into account the cancellations inherent to principal value integrals. This issue is addressed in the next lemma: 11

Lemma 7 Assume that 0 < λy < 1. Then 1 eλy + e?λy 1 0 φ(y ? t)dt ≤ C ||φ||C[0,2] (y + ) λt ? e?λt eλ(y?t) + e?λ(y?t) λ y<|t|<1 e λy ?λy 1 1 e +e φ(y ? t)dt ≤ Cm ||φ||Λ0 . λt ? e?λt eλ(y?t) + e?λ(y?t) (m) λ |t|>1 e (3.41) (3.42)

Proof. Since we can assume that λ is large, the condition that λy < 1 implies that y < 1, say. We can exploit the cancellation by writing the integral over the region y < |t| < 1 in the form, =

1 y

y<|t|<1

1 e?λy + eλy e?λy + eλy { ?λ(y?t) φ(y ? t) ? ?λ(y+t) φ(y + t)}dt (3.43) eλt ? e?λt e + eλ(y?t) e + eλ(y+t)

Next, the expression within brackets is written as, e?λy + eλy e?λy + eλy φ(y ? t) ? ?λ(y+t) e?λ(y?t) + eλ(y?t) e + eλ(y+t) e?λy + eλy ( φ(y ? t) ? φ(y + t) ) = ?λ(y?t) e + eλ(y?t) e?λy + eλy e?λy + eλy +φ(y + t){ ?λ(y?t) ? } e + eλ(y?t) e?λ(y+t) + eλ(y+t)

(3.44)

The contribution of the ?rst term on the right hand side can be estimated as follows,

1 y

e?λy + eλy 1 1 ( φ(y ? t) ? φ(y + t) ) ≤ ||φ||C[?2,2] eλt ? e?λt e?λ(y?t) + eλ(y?t)

1 y

eλt

t eλ(2y?t) dt ? e?λt

Since λy < 1, we can estimate this last term crudely by

1 y

t e2 eλ(2y?t) dt ≤ eλt ? e?λt λ

1 y

e2λt

C tλ dt ≤ , ?1 λ

(3.45)

since the function u(e2u ? 1)?1 is a smooth and bounded function for u ≥ 0. Next, to estimate the other contribution, we also exhibit the cancellation more clearly, e?λ(y?t) 1 1 1 1 1 ? ?λ(y+t) = ( ? λ(t+y) ) λ(y?t) λ(y+t) ?2λ(t+y) eλ(t?y) +e e +e 1+e e 1 1 1 + λ(t+y) ( ? ) e 1 + e?2λ(t?y) 1 + e?2λ(y+t)

The ?rst resulting group of terms can be estimated by 1 1+ e?2λ(t+y) ( 1 eλ(t?y) ? 1 eλ(t+y) ) ≤ e?λt (eλy ? e?λy ) ≤ C λye?λt , 12 (3.46)

and the corresponding integral in turn by,

1

dt

y

1 1 1 1 0 |φ(y + t)| · ( ? λ(t+y) ) ≤ ||φ||C[0,2] λy λt ? e?λt ?2λ(t+y) eλ(t?y) e 1+e e

1 y

e2λt

dt . ?1 (3.47)

To determine the size of this expression, we break it up as follows,

1 y

dt = 2λt ? 1 e

1 dt dt + 1 2λt ≤C( 2λt ? 1 e ?1 y e λ 1 1 ≤ C ( log + 1), λ λy

1 λ

1 λ

y

dt + λt

1

1 λ

dt ) e2λt (3.48)

and hence, since λy < 1,

1

dt

y

eλt

1 1 1 1 1 0 |φ(y + t)| · ( ? λ(t+y) ) ≤ ||φ||C[0,2] ( + y). (3.49) ?λt ?2λ(t+y) eλ(t?y) ?e 1+e e λ

The remaining group of terms in (3.46) can be estimated in a similar way, 1 eλ(t+y) and hence 1 1 1 1 · |φ(y + t)| ? ?λt eλ(t+y) 1 + e?2λ(t?y) ?2λ(y+t) ?e 1+e y 1 e?3λt 0 ≤ ||φ||C[0,2] λy dt, y eλt ? e?λt

1

1 1+ e?2λ(t?y)

?

1 1+ e?2λ(y+t)

≤ (eλy ? e?λy )e?λ(3t+y) ≤ C λye?3λt ,

(3.50)

dt

eλt

(3.51)

which is even smaller than the previous integral. Finally, to estimate the integral from the region |t| > 1, we have the simple estimate, since λy < 1, say for t > 0, 1 eλy + e?λy 1 1 |φ(y ? t)| ≤ C λt λ(t?y) ||φ||Λ0 (1 + |t|)m λt ? e?λt e?λ(t?y) + eλ(t?y) (m) e e e ≤ C ||φ||Λ0 (1 + |t|)m e?2λt (m)

(3.52)

which implies readily the desired inequality upon integration in t. The proof of the lemma is complete. Q.E.D. Proof of Theorem 2. It su?ces to combine all estimates from Lemmas 4,5, and 6: when λy ≥ 1, we apply Lemmas 4 and 5, and when λy < 1, we apply Lemma 4 and 6. Q.E.D.

13

References

[1] Birkho?, G.D., “The generalized Riemann problem for linear di?erential equations and allied problems for linear di?erence and q-di?erence equations”, Proc. Amer. Acad. of Arts and Sciences 49 (1913) 521-568. [2] Borodin, A., “Isomonodromy transformations of linear systems of di?erence equations”, Ann. of Math. 160 (2004) 1141-1182. [3] Harnad, J. and A.R. Its, “Integrable Fredholm operators and dual isomonodromic deformations”, Comm. Math. Phys. 226 (2002) 497-530. [4] Krichever, I.M., “Analytic theory of di?erence equations with rational and elliptic coe?cients and the Riemann-Hilbert problem”, Russian Math. Surveys 59 (2004) 1117-1154. [5] Ladyzhenskaya, O. and N. Uraltseva, “Linear and quasilinear elliptic equations”, Academic Press, 1968. [6] Stein, E.M., “Harmonic Analysis: real-variable methods, orthogonality, and oscillatory integrals”, Princeton University Press, 1993. [7] Tracy, C.A. and H. Widom, “Fredholm determinants, di?erential equations, and matrix models”, Comm. Math. Phys. 163 (1994) 33-72.

14

- The Boundedness of the Singular Integral Operator with Variable Calder
- On the Large N Limit of the Itzykson-Zuber Integral
- Path integral representation of the evolution operator for the Dirac equation
- Weighted Estimates for the Maximal Commutator of Singular Integral Operator on Spaces of Homoge
- Double Scaling Limit of Scalar Theories on the Lattice
- The solution of mildly singular integral equation of the first kind on a disk
- ON THE EIGENVALUES OF THE VOLUME INTEGRAL OPERATOR OF ELECTROMAGNETIC SCATTERING
- On the singular limit of the quantum-classical molecular dynamics model
- On the evaluation of the norm of an integral operator associated with the stability of one-
- On the scaling limit of planar self-avoiding walk

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By comparing (1) with mined by iDφ

In particular

η ? r ? L, L being

transport equation, under

In

We study

eld

To do it, we introduce an arbitrary

Eigenfunctions