Volume 8 (2007), Issue 3, Article 80, 8 pp.
ON Lp -ESTIMATES FOR THE TIME DEPENDENT SCHRÖDINGER OPERATOR
ON L2
MOHAMMED HICHEM MORTAD
D ÉPARTEMENT DE M ATHÉMATIQUES
U NIVERSITÉ D ’O RAN (E S - SENIA )
B.P. 1524, E L M ENOUAR , O RAN . A LGERIA .
mortad@univ-oran.dz
Received 07 July, 2007; accepted 29 August, 2007
Communicated by S.S. Dragomir
A BSTRACT. Let L denote the time-dependent Schrödinger operator in n space variables. We
consider a variety of Lebesgue norms for functions u on Rn+1 , and prove or disprove estimates
for such norms of u in terms of the L2 norms of u and Lu. The results have implications for
self-adjointness of operators of the form L + V where V is a multiplication operator. The proofs
are based mainly on Strichartz-type inequalities.
Key words and phrases: Schrödinger Equation, Strichartz Estimates and Self-adjointness.
2000 Mathematics Subject Classification. Primary 35B45; Secondary 35L10.
1. I NTRODUCTION
Let (x, t) ∈ Rn+1 where n ≥ 1. The Schrödinger equation ∂u
= i4x u has been much
∂t
studied using spectral properties of the self-adjoint operator 4x . When a multiplication operator
(potential) V is added, it becomes important to determine whether 4x + V is a self-adjoint
operator, and there is a vast literature on this question (see e.g. [9]).
∂
− 4x as a self-adjoint operator on
One can also, however, regard the operator L = −i ∂t
2
n+1
L (R ), and that is the point of view taken in this paper. We ask what can be said about the
domain of L, more specifically, we ask which Lq spaces, and more generally mixed Lqt (Lrx )
space, a function u must belong to, given that u is in the domain of L (i.e. u and Lu both belong
to L2 (Rn+1 )). We answer this question and, using the Kato-Rellich theorem, deduce sufficient
conditions on V for L + V to be self-adjoint.
Our approach is based on the fact that any sufficiently well-behaved function u on Rn+1 can
be regarded as a solution of the initial value problem (IVP)
(
−iut − 4x u = g(x, t),
(1.1)
u(x, α) = f (x)
I would like to express all my gratitude to my ex-Ph.D.-supervisor Professor Alexander M. Davie for his helpful comments, especially in
the counterexamples section.
227-07
2
M. H. M ORTAD
where α ∈ R, f (x) = u(x, α) and g = Lu.
To apply this, we will use estimates for u based on given bounds for f and g. A number of
such estimates are known and generally called Strichartz inequalities, after [12] which obtained
such an Lq bound for u. This has since been generalized to give inequalities for mixed norms
[13, 4]. The specific inequalities we use concern the case g = 0 of (1.1) and give bounds for
u in terms of kf kL2 (Rn ) - see (3.2) below. The precise range of mixed Lqt (Lrx ) norms for which
the bound (3.2) holds is known as a result of [13, 4] and the counterexample in [6].
2
In Section 2 we prove a special case of our main theorem, namely a bound for u in L∞
t (Lx ),
which does not require Strichartz estimates, only elementary arguments using the Fourier transform. The main theorem, giving Lqt (Lrx ) bounds for the largest possible set of (q, r) pairs, is
proved in Section 3. In fact, we prove a somewhat stronger bound, in a smaller space L2,q,r defined below. The fact that the set of pairs (q, r) covered by Theorem 3.1 is the largest possible
is shown in Section 4.
Some results on a similar question for the wave operator can be found in [7]. For Strichartztype inequalities for the wave operator, see e.g. [11, 12, 2, 3, 4].
We assume notions and definitions about the Fourier Transform and unbounded operators
and for a reference one may consult [8], [5] or [10]. We also use on several occasions the wellknown Duhamel principle for the Schrödinger equation (see e.g. [1]).
Notation. The symbol û stands for the Fourier transform of u in the space (x) variable while
the inverse Fourier transform will be denoted either by F −1 u or ǔ.
We denote by C0∞ (Rn+1 ) the space of infinitely differentiable functions with compact support.
We denote by R+ the set of all positive real numbers together with +∞.
For 1 ≤ p ≤ ∞, k · kp is the usual Lp -norm whereas k · kLpt (Lqx ) stands for the mixed spacetime
Lebesgue norm defined as follows
1q
Z
ku(t)kqLrx dt
kukLqt (Lrx ) =
.
R
We also define some modified mixed norms. First we define, for any integer k,
Z
kukLqt,k (Lrx ) =
1q
k+1
k
ku(t)kqLrx dt
,
and then
! p1
kukLp,q,r =
X
kukpLq
r
t,k (Lx )
.
k∈Z
We note that kukLp,q1 ,r ≥ kukLp,q2 ,r if q1 ≥ q2 , and that kukLqt (Lrx ) ≤ kukLp,q,r if q ≥ p.
Finally we define
MLn = {f ∈ L2 (Rn+1 ) : Lf ∈ L2 (Rn+1 )},
where L is defined as in the abstract and where the derivative is taken in the distributional sense.
We note that MLn = D(L), the domain of L, and also that C0∞ (Rn+1 ) is dense in MLn in the graph
norm kukL2 (Rn+1 ) + kLukL2 (Rn+1 ) .
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 80, 8 pp.
http://jipam.vu.edu.au/
Lp -E STIMATES FOR THE S CHRÖDINGER O PERATOR
3
2
2. L∞
t (Lx ) E STIMATES .
Before stating the first result, we are going to prepare the ground for it. Take the Fourier
transform of the IVP (1.1) in the space variable to get
(
−iût + η 2 û = ĝ(η, t),
û(η, α) = fˆ(η)
which has the following solution (valid for all t ∈ R):
Z t
2
−iη 2 t
ˆ
(2.1)
û(η, t) = f (η)e
+i
e−iη (t−s) ĝ(η, s)ds,
α
where η ∈ Rn .
Duhamel’s principle gives an alternative way of writing the part of the solution depending on
g. Taking the case f = 0, the solution of (1.1) can be written as
Z t
(2.2)
u(x, t) = i
us (x, t)ds,
α
where us is the solution of
(
Lus = 0,
t > s,
us (x, s) = g(x, s).
Now we state a result which we can prove using (2.1). In the next section we prove a more
general result using Strichartz inequalities and Duhamel’s principle (2.2).
Proposition 2.1. For all a > 0, there exists b > 0 such that
kukL2,∞,2 ≤ akLuk2L2 (Rn+1 ) + bkuk2L2 (Rn+1 )
for all u ∈ MLn .
Proof. We prove the result for u ∈ C0∞ (Rn+1 ) and a density argument allows us to deduce it
for u ∈ MLn .
We use the fact that any such u is, for any α ∈ R, the unique solution of (1.1), where
f (x) = u(x, α) and g = Lu, and therefore satisfies (2.1).
Let k ∈ Z and let t and α be such that k ≤ t ≤ k + 1 and k ≤ α ≤ k + 1. Squaring (2.1),
integrating with respect to η in Rn , and using Cauchy-Schwarz (and the fact that |t − α| ≤ 1),
we obtain
Z
Z Z t
2
2
(2.3)
kû(·, t)kL2 (Rn ) ≤ 2
|û(η, α)| dη + 2
|ĝ(η, s)|2 dsdη.
Rn
Rn
α
Now integrating against α in [k, k + 1] allows us to say that
Z k+1 Z
Z
2
2
ku(·, t)kL2 (Rn ) ≤ 2
|û(η, α)| dηdα + 2
k
Rn
k+1
k
Z
|ĝ(η, s)|2 dηds.
Rn
Now take the essential supremum of both sides in t over [k, k + 1], then sum in k over Z to get
(recalling that g = Lu)
∞
X
ess sup ku(·, t)k2L2 (Rn ) ≤ 2kLuk2L2 (Rn+1 ) + 2kuk2L2 (Rn+1 ) .
k=−∞
k≤t≤k+1
Finally to get an arbitrarily small constant in the Lu term we use a scaling argument: let m
be a positive integer and let v(x, t) = u(mx, m2 t). Then we find
kvkL2 (Rn+1 ) = m−1−n/2 kukL2 (Rn+1 )
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 80, 8 pp.
http://jipam.vu.edu.au/
4
M. H. M ORTAD
and
kLvkL2 (Rn+1 ) = m1−n/2 kLukL2 (Rn+1 ) .
Also,
kv(·, t)kL2 (Rn ) = m−n/2 ku(·, m2 t)kL2 (Rn )
and so
sup kv(·, t)k2L2 (Rn ) = m−n
k≤t≤k+1
sup
m2 k≤t≤m2 (k+1)
ku(·, t)k2L2 (Rn )
m2 (k+1)−1
≤ m−n
X
j=m2 k
sup ku(·, t)k2L2 (Rn ) .
j≤t≤j+1
Summing over k gives
kvk2L2,∞,2 ≤ m−n kuk2L2,∞,2
≤ m−n 2kLuk2L2 (Rn+1 ) + 2kuk2L2 (Rn+1 )
≤ 2m−2 kLvk2L2 (Rn+1 ) + 2m2 kvk2L2 (Rn+1 )
and choosing m so that 2m−2 < a completes the proof.
Now we recall the Kato-Rellich theorem which states that if L is a self-adjoint operator on a
Hilbert space and V is a symmetric operator defined on D(L), and if there are positive constants
a < 1 and b such that kV uk ≤ akLuk + bkuk for all u ∈ D(L), then L + V is self-adjoint on
D(L) (see [9]).
Corollary 2.2. Let V be a real-valued function in L∞,2,∞ . Then L + V is self-adjoint on
D(L) = MLn .
Proof. One can easily check that
kV ukL2 (Rn+1 ) ≤ kV kL∞,2,∞ kukL2,∞,2 .
Choose a < kV k−1
L∞,2,∞ and then Proposition 2.1 shows that L + V satisfies the hypothesis of
the Kato-Rellich theorem.
In particular, it follows that L + V is self-adjoint whenever V ∈ L2t (L∞
x ).
3. Lqt (Lrx ) E STIMATES .
Now we come to the main theorem in this paper, which depends on the following Strichartztype inequality. Suppose n ≥ 1 and q and r are positive real numbers (possibly infinite) such
that q ≥ 2 and
2 n
n
(3.1)
+ = .
q
r
2
When n = 2 we exclude the case q = 2, r = ∞. Then there is a constant C such that if
f ∈ L2 (Rn ) and g = 0, the solution u of (1.1) satisfies
(3.2)
kukLqt (Lrx ) ≤ Ckf kL2 (Rn ) .
This result can be found in [13] for q > 2; the more difficult ‘end-point’ case where q = 2,
n ≥ 3 is treated in [4]. That (3.2) fails in the exceptional case n = 2, q = 2, r = ∞ is shown in
[6].
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 80, 8 pp.
http://jipam.vu.edu.au/
Lp -E STIMATES FOR THE S CHRÖDINGER O PERATOR
5
For n ≥ 1 we define a region Ωn ∈ R+ × R+ as follows: for n 6= 2,
2 n
n
+
+
(3.3)
Ωn = (q, r) ∈ R × R : + ≥ , q ≥ 2, r ≥ 2
q
r
2
and for n = 2, Ω2 is defined by the same expression, with the omission of the point (2, ∞).
The sets Ωn are probably most easily visualized in the ( 1q , 1r )-plane. Then Ω1 is a quadrilateral with vertices ( 14 , 0), ( 12 , 0), (0, 12 ), ( 12 , 12 ) and for n ≥ 2, Ωn is a triangle with vertices
( 12 , n−2
), (0, 12 ), ( 12 , 12 ), the point ( 12 , 0) being excluded in the case n = 2.
2n
Theorem 3.1. Let n ≥ 1, and let (q, r) ∈ Ωn . Then for all a > 0, there exists b > 0 such that
kukL2,q,r ≤ akLukL2 (Rn+1 ) + bkukL2 (Rn+1 )
(3.4)
for all u ∈ MLn .
Proof. By the inclusion L2,q1 ,r ⊆ L2,q2 ,r , when q1 ≥ q2 it suffices to treat the case where
2
+ nr = n2 , for which (3.2) holds.
q
Let k ∈ Z and let α ∈ [k, k + 1]. As in the proof of Proposition 2.1 we use the fact that u is
the solution of (1.1) with f = u(·, α) and g = Lu. Now we split u into two parts u = u1 + u2 ,
where u1 , u2 are the solutions of
(
(
Lu1 = g,
Lu2 = 0,
u1 (x, α) = 0,
u2 (x, α) = f.
The estimate for u2 is deduced from (3.2):
ku2 kLqt (Lrx ) ≤ Ckf kL2 (Rn ) ≤ Cku(·, α)kL2 (Rn ) .
(3.5)
For u1 we apply (2.2) to obtain
Z
t
u1 (x, t) = i
(3.6)
us (x, t)ds,
α
from which we deduce
Z
k+1
ku1 (·, t)kLr (Rn ) ≤
kus (·, t)kLr (Rn ) ds
k
for t ∈ [k, k + 1], and hence
Z
ku1 k
Lqt,k (Lrx )
k+1
≤
kus kLqt (Lrx ) ds
k
Z
≤C
k+1
kg(·, s)kL2 (Rn ) ds
k
≤ CkgkL2 (Rn ×[k,k+1]) .
Combining this with (3.5) we have
kuk2Lq
r
t,k (Lx )
≤ 2C 2 ku(·, α)k2L2 (Rn ) + 2C 2 kLuk2L2 (Rn ×[k,k+1]) .
Integrating w.r.t. α from k to k + 1 gives
kuk2Lq
r
t,k (Lx )
≤ 2C 2 kuk2L2 (Rn ×[k,k+1]) + 2C 2 kLuk2L2 (Rn ×[k,k+1]) .
Summing over k, we obtain
kuk2L2,q,r ≤ 2C 2 kukL2 (Rn+1 ) + 2C 2 kLukL2 (Rn+1 ) ,
and the proof is completed by a similar scaling argument to that used in Proposition 2.1.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 80, 8 pp.
http://jipam.vu.edu.au/
6
M. H. M ORTAD
Using the inclusion L2,q,r ⊆ Lqt (Lrx ) for q ≥ 2 we deduce
Corollary 3.2. Let n ≥ 1, and let (q, r) ∈ Ωn . Then for all a > 0, there exists b > 0 such that
kukLqt (Lrx ) ≤ akLukL2 (Rn+1 ) + bkukL2 (Rn+1 )
(3.7)
for all u ∈ MLn .
In particular, we get such a bound for kukLq (Rn+1 ) whenever 2 ≤ q ≤ (2n + 4)/n.
By applying the Kato-Rellich theorem we can deduce a generalization of Corollary 2.2 from
Theorem 3.1. We first define
2 n
∗
+
+
(3.8)
Ωn = (p, s) ∈ R × R : + ≤ 1, p ≥ 2, s ≥ 2
p s
for n 6= 2, and for n = 2, Ω2 is defined by the same expression, with the omission of the point
(2, ∞).
Corollary 3.3. Let n ≥ 1 and let (p, s) ∈ Ω∗n . Let V be a real-valued function belonging to
L∞,p,s . Then L + V is self-adjoint on MLn .
2p
2s
Proof. Let q = p−2
and r = s−2
. Then (q, r) ∈ Ωn and the conclusion (3.4) of Theorem 3.1
applies. Now we have
Z k+1
Z k+1
2
kV u(·, t)kL2 (Rn ) ≤
ku(·, t)k2Lr (Rn ) kV (·, t)k2Ls (Rn )
k
k
≤ kuk2Lq
r
t,k (Lx )
kV k2Lp
s
t,k (Lx )
and summation over k gives
kV ukL2 (Rn+1 ) ≤ kukL2,q,r kV kL∞,p,s .
Then, using (3.4), the result follows in the same way as Corollary 2.2.
It follows from Corollary 3.3 that L+V is self-adjoint whenever V ∈ Lpt (Lsx ) for (p, s) ∈ Ω∗n .
Taking the case s = p, we find that L + V is self-adjoint if V ∈ Lp (Rn+1 ) for some p ≥ n + 2.
4. C OUNTEREXAMPLES
Now we show that Theorem 3.1 is sharp, as far as the allowed set of q, r is concerned.
Proposition 4.1. Let n ≥ 1 and let q and r be positive real numbers, possibly infinite, such that
(q, r) ∈
/ Ωn . Then there are no constants a and b such that (3.7) holds for all u ∈ MLn .
Proof. For (q, r) to fail to be in Ωn one of the following three possibilities must occur: (i) q < 2
or r < 2; (ii) 2q + nr < n2 ; (iii) n = 2, q = 2 and r = ∞. We consider these cases in turn.
(i) If q < 2, choose a sequence (βk )k∈Z which is in l2 but not in lq . Let φ(x, t) be a smooth
function of compact support on Rn+1 which vanishes for t outside [0, 1], and let u(x, t) =
P
n
/ Lqt (Lrx ) for any r.
k∈Z βk φ(x, t − k). Then u ∈ ML , but u ∈
2
r
The case r < 2 can be treated similarly. We chose a sequence βk which
P is in l but not l ,
and a smooth φ which vanishes for x1 outside [0, 1], then set u(x, t) = k∈Z βk φ(x − ke1 , t),
where e1 is the unit vector (1, 0, . . . , 0) in Rn . Then u ∈ MLn , but u ∈
/ Lqt (Lrx ) for any q.
(ii) In this case we use the scaling argument which shows that the Strichartz estimates fail,
together with a cutoff to ensure u and Lu are in L2 .
We start with a non-zero f ∈ L2 (Rn ), and let u be the solution of (1.1) with α = 0 and g = 0.
2
2
(An explicit example would be f (x) = e−|x| and then u(x, t) = (1 + 4it)−n/2 e−|x| /(1+4it) ).
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 80, 8 pp.
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Lp -E STIMATES FOR THE S CHRÖDINGER O PERATOR
7
Choose a smooth function φ on R such that φ(0) 6= 0 and such that φ and φ0 are in L2 . Then for
λ > 0 define
vλ (x, t) = λn/2 u(λx, λ2 t)φ(t).
Then (using Lu = 0) we find Lv(x, t) = −iλn/2 u(λx, λ2 t)φ0 (t). We calculate kvλ kL2 (Rn+1 ) =
kf kL2 (Rn ) kφkL2 and kLvλ kL2 (Rn+1 ) = kf kL2 (Rn ) kφ0 kL2 . Also
kvλ kLqt (Lrx ) = λ
β
1q
Z
ku(·, t)kqLr (Rn ) |φ(λ−2 t)|q dt
,
R
where β = n2 − nr − 2q > 0. So λ−β kvλ kLqt (Lrx ) → |φ(0)|kukLqt (Lrx ) (note that the norm on the
right may be infinite) and hence kvλ kLqt (Lrx ) tends to ∞ as λ → ∞, completing the proof.
(iii) This exceptional case we treat in a similar fashion to (ii), but we need the result from [6],
that the Strichartz inequality fails in this case. We start by fixing a smooth function φ on R such
that φ = 1 on [−1, 1] and φ and φ0 are in L2 .
Now let M > 0 be given and we use [6] to find f ∈ L2 (R2 ) with kf kL2 (R2 ) = 1 such that the
solution u of (1.1) with α = 0 and g = 0 satisfies kukL2t (L∞
> M . Then we can find R > 0 so
x )
RR
2
2
1/2
that −R ku(·, t)kL∞ (R2 ) dt > M . Let λ = R and define v(x, t) = λn/2 u(λx, λ2 t)φ(t). Then
kvkL2 (R3 ) = kφkL2 , kLvkL2 (R3 ) = kφ0 kL2 and
kvk2L2t (L∞
x )
Z
1
≥
kv(·, t)k2L∞ (R2 ) dt > M 2 ,
−1
which completes the proof, since M is arbitrary.
We remark that [6] also gives an example of f ∈ L2 (R2 ) such that u ∈
/ L2t (BM Ox ) and the
argument of part (iii) can then be applied to show that no inequality
kukL2t (BM Ox ) ≤ akLukL2 (R3 ) + bkukL2 (R3 )
can hold.
5. Q UESTION
We saw as a result of Corollary 3.3 that if (p, s) ∈ Ω∗ , then L + V is self-adjoint on MLn
whenever V ∈ Lpt (Lsx ). One can ask whether this can be extended to a larger range of (p, s)
with p, s ≥ 2. If one asks whether L + V is defined on MLn , then we would require a bound
kV ukL2 (Rn+1 ) ≤ akLukL2 (Rn+1 ) + bkuk to hold for all u ∈ MLn . If such a bound is to hold for
2p
2s
all V ∈ Lpt (Lsx ), then, in fact, we require (3.7) to hold for q = p−2
and r = s−2
, which we know
∗
cannot hold unless (p, s) ∈ Ω .
One can instead ask for L + V , defined on say C0∞ (Rn+1 ), to be essentially self-adjoint. This
is equivalent to saying that the only (distribution) solution in L2 (Rn+1 ) of the PDE
−iut − 4x u + V u = ±iu
is u = 0 (see e.g. [8]).
We do not know if there are any values of (p, s) not in Ω∗n such that this holds for all V ∈
p
Lt (Lsx ). The analogous question for the Laplacian is extensively discussed in [9].
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 80, 8 pp.
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8
M. H. M ORTAD
R EFERENCES
[1] L.C. EVANS, Partial Differential Equations, American Mathematical Society (G.S.M. Vol. 19),
1998.
[2] J. GINIBRE AND G. VELO, Generalized Strichartz inequalities for the wave equation, J. of Functional Analysis, 133 (1995), 50–68.
[3] J. HARMSE, On Lebesgue space estimates for the wave equation, Indiana Univ. Math. J., 39(1)
(1990) 229–248.
[4] M. KEEL AND T. TAO, Endpoint Strichartz estimates, Amer. J. of Math., 120 (1998), 955–980.
[5] E.H. LIEB AND M. LOSS, Analysis, American Mathematical Society (G.S.M. Vol. 14), Second
Edition 2001.
[6] S.J. MONTGOMERY-SMITH, Time decay for the bounded mean oscillation of solutions of the
Schrödinger and wave equations, Duke Math. J., 91(2) (1998), 393–408.
[7] M.H. MORTAD, Self-adjointness of the perturbed wave operator on L2 (Rn ), Proc. Amer. Math.
Soc., 133(2) (2005), 455–464.
[8] M. REED AND B. SIMON, Methods of Modern Mathematical Physics, Vol. 1, “Functional Analysis,” Academic Press, 1972.
[9] M. REED AND B. SIMON, Methods of Modern Mathematical Physics, Vol.2, “Fourier Analysis,
Self-Adjointness,” Academic Press, 1975.
[10] E.M. STEIN AND G. WEISS, Introduction to Fourier Analysis on Euclidean Spaces, (PMS; 32)
Princeton University Press, 1971.
[11] R.S. STRICHARTZ, A priori estimates for the wave equation and some applications, J. of Functional Analysis, 5 (1970), 218–235.
[12] R.S. STRICHARTZ, Restrictions of Fourier transforms to quadratic surfaces and decay of solutions
of wave equations, Duke Mathematical Journal, 44(3) (1977), 705–714.
[13] K. YAJIMA, Existence of solutions for Schrödinger evolution equations, Comm. Math. Phys., 110
(1987), 415–426.
J. Inequal. Pure and Appl. Math., 8(3) (2007), Art. 80, 8 pp.
http://jipam.vu.edu.au/