H¨ older continuity of quotients of solutions for the Schr¨ odinger equation ∗ ¨ P. KROGER AND K.-TH. STURM∗
0
Summary
We the behaviour of local weak solutions u of the Schr¨odinger equation investigate − 12 ∆ + V u = 0 with a potential V ≥ 0 on Rd , d ≥ 2. There exist potentials V for which the solutions are discontinuous but satisfy Harnack’s inequality. Nevertheless, the quotients of any two positive solutions are H¨older continuous. We aim to give lower bounds for the H¨older exponents of quotients of positive solutions of the Schr¨odinger equation. Our main result states that the quotients are H¨older continuous of order arbitrarily close to 1 if the solutions of the original equation are continuous.
1
Statement of results
We consider local weak solutions u of the Schr¨odinger equation
1 − ∆+V u=0 2
(1.1)
on an open subset D ⊂ Rd , d ≥ 2. For the sake of simplicity, we assume throughout this d paper that V ∈ Lloc 1 (R ), and V ≥ 0. Definition odinger equation 1 A function u is called local weak solution of the Schr¨ 1 loc loc − 2 ∆ + V u = 0 on D if u ∈ L1 (D), V · u ∈ L1 (D), and −
Z 1Z u · ∆ψ dx + V · u · ψ dx = 0 2 D D
∞ ∀ ψ ∈ Ccomp (D) .
A function odinger u is called V -harmonic on D if it is a local weak solution of the Schr¨ 1 equation − 2 ∆ + V u on D and, in addition, finely continuous in D. The class of V -harmonic functions on D will be denoted by HV (D). Since we are interested in pointwise estimates, we quote the following result (see [5], Propo1 sition 2): For every local weak solution u of the Schr¨ odinger equation − 2 ∆ + V u = 0 on D there exists a (pointwise) uniquely defined V -harmonic function u˜ on D that coincides with u a.e. on D. Moreover, every V -harmonic function on D is locally bounded in D. ∗ ∗
Research supported by the Deutsche Forschungsgemeinschaft 1991 Mathematics Subject Classification 35J10, 35J15, 31B15, 60J57.
1
We say that a potential V ≥ 0 on Rd , d ≥ 3, belongs to the Dynkin class (V ∈ K∞ (Rd )) if and only if the Kato norm of V is finite, i.e., sup
Z
x∈Rd
Br (x)
V (y) dy < ∞ kx − ykd−2
(1.2)
for some (and hence all) r > 0, and that it belongs to the Kato class (V ∈ K0 (Rd )) if and only if lim sup
Z
r→0 x∈Rd
In dimension d = 2 the kernel
Br (x)
1 kx−ykd−2
V (y) dy = 0 . kx − ykd−2
1 will be replaced by ln kx−yk
(1.3) +
.
loc We write V ∈ K∞ (D) if 1C · V ∈ K∞ (Rd ) for every compact set CbD. The class K0loc (D) is defined analogously. loc It is easy to see that a non-negative function V belongs to the class K∞ (D) (resp., loc K0 (D)) if and only if every point x ∈ D has a neighborhood B on which the Green potential of V " # B
y
y 7→ U V (y) ≡ E
Z
0
τ (B)
V (Xs ) ds
(1.4)
is bounded (resp., continuous) (cf. [1], [3], and [7]). Here Ey denotes expectation w.r.t. Brownian motion starting in y ∈ Rd and τ (B) denotes the first exit time of B, i.e., τ (B) ≡ inf{t > 0 : Xt 6∈ B}. loc (D) and K0loc (D) can also be According to the following propositions, the classes K∞ loc (D) (or K0loc (D) characterized by the fact that a potential V ≥ 0 belongs to the class K∞ resp.) if and only if for every point x ∈ D there exist a neighbourhood of x and a strictly positive function which is V -harmonic in that neighbourhood and satisfies Harnack’s inloc equality (or is continuous resp.). Moreover, V ∈ K∞ (D) implies that all V -harmonic loc functions satisfy Harnack’s inequality and V ∈ K0 (D) implies that all V -harmonic functions are continuous. Proposition 2 a) If d ≥ 3 and 1BR (z) · V ∈ K∞ (Rd ) (for some R > 0 and z ∈ Rd ) then for every r ∈]0, R[ the following holds: ∀ u ∈ HV (BR (z)), u ≥ 0 ,
sup u(x) ≤ K · inf u(x) x∈Br (z)
x∈Br (z)
with K = K(r, R, d, β) ≡
R+r R−r
d
· exp β and β ≡ 3 · (d − 2) · 2d+1 · kU (1BR (z) · V )k∞ .
loc b) V ∈ K∞ (D) if and only if for every point x ∈ D there exist a neighborhood B of x and a V -harmonic function u on B with inf u(y) > 0. y∈B
Proposition 3 a) (See [1], Theorem 1.5) If V ∈ K0loc (D) then all V -harmonic functions on open subsets of D are continuous. b) V ∈ K0loc (D) if and only if for every point x ∈ D there exists at least one strictly positive continuous function which is V -harmonic in an appropriate neighbourhood of x. 2
The proofs of these results (as well as that of the following theorem) will be postponed to the end of the paper. We emphasize that in order to obtain that the V -harmonic functions are continuous, we need a more restrictive condition on V than for the validity of Harnack’s inequality. In loc (D) according to the well-known fact that there fact, K0loc (D) is a proper subset of K∞ exist functions V whose Green potentials U B V (for suitable B ⊂ D) are bounded but not continuous (recall that d ≥ 2). An explicit example of such a function V is given in [1], Example 1. For those potentials V the positive V -harmonic functions satisfy Harnack’s inequality (cf. Proposition 3), but they are not continuous in D (cf. Proposition 2). According to Proposition 2, the continuity of the V -harmonic functions on D is equivalent to V ∈ K0loc (D). Note, however, that even in the case V ∈ K0loc (D) the V -harmonic functions are not necessarily H¨older continuous of any order; cf. the example below. We recall that there is a classical argument due to J. Moser [6] in order to prove H¨older continuity of solutions using Harnack’s inequality (cf. also the proof of our Theorem). This argument does obviously not apply directly to our situation. The reason is that the constants are in general not solutions of the Schr¨odinger equation. The situation is quite different if we consider quotients of solutions. We set u | u, v ∈ HV (D), v(x) > 0 for every x ∈ D RV (D) ≡ v for every open set D ⊆ Rd . Obviously, 1 + RV (D) = RV (D) and λRV (D) = RV (D) for every real number λ with λ 6= 0. Our main result is the following theorem: Theorem 4 Suppose that 1D V ∈ K∞ (Rd ). Then there exists a positive number γ which depends only on kU (1D V )k∞ and the dimension d such that every function u from RV (D) is locally γ-H¨older continuous. older continuous If 1D V ∈ K0loc (Rd ) then every function u from RV (D) is locally γ-H¨ for every γ < 1. We notice that Moser’s original argument applied to the equation ∇(v 2 ∇ uv ) = 0 yields that u is H¨older continuous. However, it seems to be difficult to prove γ-H¨older continuity for v every γ < 1 by arguments of this type. Combining the above propositions with the theorem, we obtain the following corollary: d Corollary 5 Let V be an arbitrary non-negative potential on R , d ≥2, and let u and v be any two local weak solutions of the Schr¨ odinger equation − 12 ∆ + V u = 0 on an open set D ⊂ Rd .
a) If ess inf v(x) > 0 then x∈D
u v
coincides a.e. with a locally H¨ older continuous function in D.
b) If in addition v is continuous, then uv coincides a.e. with a function which is locally γ-H¨older continuous in D for every γ < 1. 3
Example 6 Let us consider the potential V (x) =
1 (kxk · ln kxk)2
on D ≡ B1 (0) ⊂ Rd , d ≥ 2. Elementary calculations show that V ∈ K0loc (D) (cf. [1], Theorem 4.11 and the subsequent Remark 2.2). On the other hand, it is easy to see that the (strictly positive) function u(x) =
1 d−2 − 2 ln kxk
— which is not H¨ of any order — is a local weak solution of the Schr¨o older continuous dinger equation − 12 ∆ + V u = 0 in D. However, if v is another local weak solution of that equation, then, according to our Theorem from above, the quotient uv is γ-H¨older continuous in D for every γ < 1. Remarks 7 In this article we have restricted ourselves of (quotients to the investigation 1 of) local weak solutions u of the Schr¨odinger equation − 2 ∆ + V u = 0 with non-negative d potentials V ∈ Lloc 1 (R ). Essentially the same results hold true if one replaces the Schr¨odinger operator − 12 ∆+V by a generalized Schr¨odinger operator − 21 ∆ + µ with a measure µ ≥ 0 on Rd charging no polar sets (cf. [3], [2], and [5]). loc loc A measure µ belongs to the generalized class K∞ (D) (resp., K0 (D)) if there exist open sets Bi with ∪i∈I Bi = D and bounded (resp., continuous) superharmonic functions ui on Bi with − 12 ∆ui = µ. loc loc In the case d = 1, we obviously have K0 (D) = K∞ (D). For dimension d ≥ 2, however, it is easy to construct superharmonic functions u which are bounded but not continuous. loc loc Any such function u leads to a measure µ = − 21 ∆u ∈ K∞ (D) \ K0 (D), and finally to a generalized Schr¨odinger equation − 12 ∆ + µ v = 0 with discontinuous solutions v whose quotients are (H¨older) continuous. Similar results also hold true for Schr¨odinger operators − 21 ∆ + V with arbitrary po1 d tentials V ∈ Lloc 1 (R ) (or, more generally, for operators − 2 ∆ + µ with signed measures µ charging no polar sets). The main difference to our situation occurs in the proof of Harnack’s inequality for the V -harmonic functions (cf. Proposition 2). In order to estimate the conditional gauge " !#
Ex,z exp −
τ (B)
Z
0
V (Xs )ds
one uses Jensens’s inequality together with Khas’minskii’s Lemma. This leads to u(x) R+r ≤ R−r x,y∈Br (z) u(y)
sup
d
·
eβ 1−β
provided that β ≡ 3 · (d − 2) · 2d+1 · kU (1BR (z) · |V |)k∞ < 1 (cf. also [8], [4]). 4
2
Proofs
Proof of Proposition 2. a) Let B ≡ BR0 (z) for an arbitrary R0 with r < R0 < R. Then, according to [5], every u ∈ HV (BR (z)) admits the representation " x
exp −
u(x) = E
!
τ (B)
Z
#
V (Xs )ds · u(Xτ (B) )
0
for x ∈ B. Since V is non-negative, we obviously have h(x) ≡ Ex [u(Xτ (B) ]] ≥ u(x) on B; here h is harmonic (w.r.t. the Laplacian) in B. On the other hand, for x ∈ B, " x
exp −
u(x) = E
"
" x,Xτ (B)
= E
E
exp −
"
≥
Z
inf E x∈B
exp −
!#
V (Xs )ds
#
· u(Xτ (B) )
V (Xs )ds
τ (B)
Z 0
y∈∂B
!#
τ (B)
0 x,y
#
V (Xs )ds · u(Xτ (B) )
0
x
!
τ (B)
Z
h
i
· Ex u(Xτ (B) ) ;
here Ex,y denotes expectation with respect to the Brownian motion restricted to B, starting in x, and conditioned to converge to y. By Jensen’s inequality "
inf Ex,y exp −
x∈B y∈∂B
0
!#
τ (B)
Z
V (Xs )ds
x,y ≥ exp − sup E x∈B y∈∂B
"Z
#
τ (B)
0
V (Xs )ds
and by the 3G-Theorem (see [8] or [4]) sup Ex,y
x∈B y∈∂B
"Z 0
τ (B)
#
V (Xs )ds ≤ 3(d − 2)2d+1 · kU (1BR (z) V )k∞ .
The right-hand side of the last inequality is by definition equal to β. We have shown that u ≤ h and u ≥ e−β · h in B. Therefore u(x1 ) h(x1 ) β sup ≤ sup ·e ≤ x1 ,x2 ∈Br (z) u(x2 ) x1 ,x2 ∈Br (z) h(x2 )
R0 + r R0 − r
!d
· eβ ;
here we have applied the classical Harnack’s inequality to the harmonic function h. Now we let R0 tend to R. This establishes part a) of the assertion. loc b) Let V ∈ K∞ (D) and let B be any relatively compact open subset of D. Then the function " !# Z τ (B) y 7→ Ey exp − V (Xs )ds 0
5
is V -harmonic in B and bounded below by y
exp − sup E y∈B
"Z
#!
τ (B)
V (Xs )ds
0
> 0.
Conversely, let u ∈ HV (B) with ≡ inf u(y) > 0. y∈B
For every relatively compact open subset B 0 ⊂ B the function 0
h ≡ u + U B (V · u)
(2.5)
is harmonic (w.r.t. the Laplacian) in B 0 . In particular, h is locally bounded in B 0 . According to 1 1 0 0 U B V ≤ · U B (V · u) ≤ · h loc this implies V ∈ K∞ (B 0 ). 2 Proof of Proposition 3.b If V ∈ K0loc (D), then the existence of suitable strictly positive V -harmonic functions follows from Proposition 2.b. According to Proposition 3.a, those functions are continuous. Let us conversely assume that there exists a strictly positive continuous V -harmonic function u on a suitable neighborhood B 0 of x. For any bounded open set BbB 0 we obtain ≡ inf u(y) > 0. Since u is V -harmonic on a neighborhood of the closure of B, the y∈B
function h ≡ u + U B (V · u) is harmonic on B and hence continuous on B. The continuity of u therefore implies the continuity of U B (V · u) which in turn implies that V · u ∈ K0loc (B) . But V ≤
1
· V · u on B, hence we have also V ∈ K0loc (B).
2
Proof of the Theorem. Our proof is based on the explicit version of Harnack’s inequality given in Proposition 2.a and a simple iteration argument due to Moser (see [6], Section 2). The proof of the Theorem for d = 2 can be reduced to the case d = 3 by consideration of functions which are constant with respect to the variable x3 . We therefore restrict ourselves to the case d ≥ 3. Let z ∈ D be given. We introduce the notation osc w ≡ sup w − inf w M
M
M
for every set M and every bounded real function w on M . We consider positive numbers r and R with r < R and BR (z)bD. If we apply Harnack’s inequality (Prop. 2.a) to pairs of non-negative functions in HV (BR (z)), then we obtain a Harnack inequality for 6
non-negative functions in RV (BR (z)). More specifically, for every non-negative function v ∈ RV (BR (z)) the following holds: sup v ≤ K(r, R, d, β)2 · inf v Br (z)
Br (z)
where K(r, R, d, β) =
R+r R−r
d
· exp β and β = 3(d − 2)2d+1 · kU (1BR (z) V )k∞ . Hence,
osc v ≤ (K(r, R, d, β)2 − 1) · inf v .
Br (z)
Br (z)
Now we set v ≡ sup u − u. Thus, BR (z)
osc u =
Br (z)
osc v
Br (z)
≤ (K(r, R, d, β)2 − 1) · inf v Br (z)
2
≤ (K(r, R, d, β) − 1) · ( osc u − osc u) . BR (z)
Br (z)
We can conclude that 1 osc u ≤ 1 − Br (z) K(r, R, d, β)2
!
· osc u . BR (z)
Let δ be a positive number with r = δR. Since K(r, R, d, β) = by iteration that u is γ-H¨older continuous at z for
ln 1 − γ≡
1−δ 1+δ
2d
1+δ 1−δ
d
· exp β, we obtain
exp(−2β)
ln δ
.
This establishes the first part of the assertion since δ can be chosen arbitrarily in the interval (0, 1). In order to prove the second part, notice that there exists a positive constant Cd such that 1−β 1− 1+β
!2d
· exp(−2β) ≤ Cd β
for every positive β with β < 1. We choose δ ≡ β. Then, γ ≥1+
ln Cd . ln β
Since the right-hand side of the last inequality tends to 1 if kU (1BR (z) V )k∞ tends to 0, the proof of the theorem is now complete. Acknowledgement. The authors are very grateful to Professor E. B. Davies for helpful comments. 7
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