Note on gradient estimate of heat kernel for Schr\"odinger operators

Let $H=-\Delta+V$ be a Schr\"odinger operator on $\mathbb{R}^n$. We show that gradient estimates for the heat kernel of $H$ with upper Gaussian bounds imply polynomial decay for the kernels of certain smooth dyadic spectral operators. The latter decay property has been known to play an important role in the Littlewood-Paley theory for $L^p$ and Sobolev spaces. We are able to establish the result by modifying Hebisch and the author's recent proofs. We give a counterexample in one dimension to show that there exists $V$ in the Schwartz class such that the long time gradient heat kernel estimate fails.


Introduction
Consider a Schrödinger operator H = −∆ + V on R n , where V is a real-valued potential in L 1 loc (R n ).It is noted in [1,2] that for positive V , if H admits the following gradient estimates for its heat kernel p t (x, y) = e −tH (x, y): for all x, y ∈ R n and t > 0, |p t (x, y)| ≤ c n t −n/2 e −c|x−y| 2 /t , (1) then the kernel of Φ j (H) and its derivatives satisfy a polynomial decay as in (4), where Φ j is a function in certain Sobolev space with support in [−2 j , 2 j ].As is well-known, the decay estimate in (4) implies the Littlewood-Paley inequality for L p (R n ) [3,4,5,6].
For positive V , based on heat kernel estimates one can show (4) by a scaling argument [2].In this paper we will prove the general case, namely Theorem 1.1, by modifying the proofs in [7,8] and [2].
However, in general the gradient estimates (1), (2) do not hold for all t.This situation may occur when H is a Schrödinger operator with negative potential, or the sub-Laplacian on a Lie group of polynomial growth, cf.[9,10,11,12].A second part of this paper is to show such a counterexample, based on Theorem 1.1.
Recall that for a Borel measurable function φ : R → C, one can define the spectral operator φ(H) by functional calculus φ(H) = ∞ −∞ φ(λ)dE λ , where dE λ is the spectral measure of H.The kernel of φ(H) is denoted φ(H)(x, y) in the following sense.Let A be an operator on a measure space (M, dµ), dµ being a Borel measure on M. If there exists a locally integrable function for all f, g in C 0 (M) with supp f and supp g being disjoint, where C 0 (M) is the set of continuous functions on M with compact supports, then A is said to have the kernel A(x, y) := K A (x, y).Throughout this paper, c or C will denote an absolute positive constant.
The main result is the following theorem for V ∈ L 1 loc (R n ).Theorem 1.1.Suppose that the kernel of e −tH satisfies the upper Gaussian bound for α = 0, 1 Let Φ be supported in [−1, 1] and belong to H n+1 2 +N +δ (R) for some fixed N ≥ 0 and δ > 0. Then for each N ≥ 0, there exists a constant c N independent of Φ such that for all j ∈ Z where Φ j (x) = Φ(2 −j x) and H s := H s (R) denotes the usual Sobolev space with norm f H When V is positive, a result of the above type was proved and applied to the cases for the Hermite and Laguerre operators [1].The observation was that if V ≥ 0, then the constants corresponding to Lemma 2.1 do not change for in what follows), according to the Feynman-Kac path integral formula [13] here E x is the integral over the path space Ω with respect to the Wiener measure µ x , x ∈ R n and ω(t) stands for a brownian motion (generic path).
For general V the technical difficulty is that we do not have such a scaling-invariance.We are able to overcome this difficulty by establishing Lemma 2.5, a scaling version of the weighted L 1 inequality for Φ j (H)(x, y) with Φ ∈ H s , for which we directly use the scaling information indicated by the time variable appearing in Lemma 2.1.Thus this leads to the proof of the main theorem by combining methods of Hebisch and the author's in [7,2].
Note that under the condition in Theorem 1.1, ( 4) is valid for all for both homogeneous and inhomogeneous systems {ϕ j }, according to [1, Theorem 1.5], see also [2,3].

heat kernel having upper Gaussian bound implies rapid decay for spectral kernels
In this section we prove Theorem 1.1.Following [7] we begin with a simple lemma.
Lemma 2.1.Suppose that (1) holds.Then The next lemma can be easily proved by a duality argument and we omit the details.
If in addition ρ(L) is unitary, then the equality holds.
Let w be a submultiplicative weight on R n × R n , i.e., 0 ≤ w(x, y) ≤ w(x, z)w(z, y), ∀x, y, z ∈ R n .For simplicity we also assume w(x, y) = w(y, x).Define the norm for k ∈ L 1 loc (R 2n ) as follows: Then given two operators L 1 , L 2 , it holds that The following lemma is a scaling version of [8, Theorem 3.1] for V ∈ L 1 loc (R n ).Lemma 2.3.Suppose that (1) holds.Let L j = e −2 −j H . Then for each a ≥ 0, there exists a constant c = c(n, a) depending on n, a only such that for all j ∈ Z and k ∈ R, Proof.(a) First we show the case j = 0.For notational convenience write L = L 0 , then by Lemma 2.1 we have, with t = 1, In view of Lemma 2.2, setting ℓ = β −1 |k| L(x, y) φ , we have where we note that by (6) It is easy to calculate that Hence taking β = s/2 and fixing s = s 0 > 0 give that (b) Similarly we show the case for all j ∈ Z.
We also need a basic property on the weighted ℓ 2 norm of Fourier coefficients of a compactly supported function in Sobolev space, which can be proved by elementary Fourier expansions. where The inequality in (8) can be replaced by equality (however we will not use this improvement), which is a special case of the general norm characterization for periodic functions in H s ([0, T ]), see e.g.[14].
It follows from Lemma 2.3 and Lemma 2.4 the following weighted L 1 estimates for Φ j (H)(x, y), which is an improved version of [2, Lemma 3.1], where the restriction V ≥ 0 is removed.
It follows from Lemma 2.3, (10) and Lemma 2.4 that for each y ) , where δ = s − N − (n + 1)/2 and the last inequality follows from a change of variable and interpolation.

2.7.
Proof of Theorem 1.1.With (3) and Lemma 2.5 we are in a position to prove (4).The proof is similar to that of Proposition 3.3 in [2] in the case of positive V .For completeness, we present the details here.
By (3) we have Applying Lemma 2.5 with t = 2 −j , we obtain ✷ Remark 2.8.In the following section we will show that there exists V ∈ S, the Schwartz class, such that (4) does not hold for j → ±∞.By Theorem 1.1, this means that for such V the gradient upper Gaussian bound (3) does not hold for all t.

A counterexample to the gradient heat kernel estimate
Consider the solvable model We know from [5] that solving the Helmholtz equation for k ∈ R \ {0} H ν e(x, k) = k 2 e(x, k), yields the following formula for the continuum eigenfunctions: where P ν (x, k) = p ν (tanh x, ik) is defined by the recursion formula with p 0 ≡ 1.Note that e(x, −k) = e(−x, k) and the function ( 11) is real analytic on R 3 .Moreover, H ν has only absolutely continuous spectrum σ ac = [0, ∞) and point spectrum σ pp = {−1, −4, . . ., −ν 2 } .The corresponding eigenfunctions {e n } ν n=1 in L 2 are Schwartz functions that are linear combinations of sech m x tanh ℓ x, m ∈ N, ℓ ∈ N 0 .
Let H ac = H ν E ac denote the absolutely continuous part of H ν and is the kernel of φ(H ac ) = φ(H)E ac , cf. [16].Since H ν has eigenfunctions in S(R) and σ pp is finite, from now on it is essential to check the kernel φ(H ac )(x, y) = K(x, y) instead of the kernel of φ(H ν ).

Decay for the kernel of
In [5] we showed that for each N but (with α = 1) only holds for j ≥ 0 and does not hold for all j < 0. This suggests that (3) fails for α = 1 and t > 1 (or more precisely t → ∞), according to Theorem 1.1.Now consider the system {Φ j } j∈Z which satisfy (i), (ii) as in Section 1.We may assume Φ j (x) = Φ(2 −j x) for a fixed Φ in C ∞ ([−1, 1]) with Φ(x) = 1 on [− 1 2 , 1  2 ].Let f ∧ and f ∨ be the Fourier transform and its inverse of f on R. The following lemma shows that (13) does not hold for Φ j (H ac )(x, y) when j → ∞.
Lemma 3.2.Let K j (x, y) be the kernel of Φ j (H)E ac .a) For each N ∈ N 0 there exists c N such that for all j ≤ 0, b) For each N ∈ N 0 there exists c N such that for all j > 0, precisely In particular, the decay in (15) does not hold for all j > 0 with N > 1. c) There exist positive constants C and c such that for all j ∈ Z, where Ψ j (k) = Φ j (k 2 ) and it is easily to see that for each N, there exists c N such that for all j Proof.(a) Let λ = 2 −j/2 .By ( 12), (11) and integration by parts 2π(i(x − y)) N K j (x, y) which can be written as a finite sum of (17) We obtain for each N and all j ≤ 0 This proves (15) for j ≤ 0.
In order to show part (c) for j ∈ Z, using partial fractions we write K j (x, y) as a finite sum of which is bounded by (up to a constant multiple) where a ι , b ι ∈ R. The general term in the sum is estimated by (b) Finally we prove the sharp estimate in (16).For j > 0, (15) does not hold for N ≥ 2, instead we have only, with N = 0, 1, |K j (x, y)| ≤ c2 j/2 (1 + 2 j/2 |x − y|) −N , by using similar argument and noting (17), (19), (20).Indeed, let J > 0, N ≥ 2. It is easy to find {φ j } satisfying (i') and (ii') such that We have by (13) where α ι , β ι = 0 are of the form c tanh ℓ x tanh m y.This shows that the term (x − y) N Φ J (H ac )(x, y) cannot admit a decay of 2 −J/2(N −1) for all J > 0, otherwise one would have |(x − y) N 1 [0,∞) (H ν )(x, y)| 2 −J/2(N −1) , which leads to a contradiction that the sum of those functions in (21) must vanish, by letting J → ∞.  holds for all j > 0 but does not hold for all j < 0. Therefore the inequality in (4) does not hold for general V ∈ L 1 loc .

Remark 3 . 3 .
The argument in the proof of part (b) can be made rigorous by replacing 1 [0,∞] (H ν ) with Φ L (H ν ), and then let L → ∞ to get the same contradiction.