THE LINEAR SCHRÖDINGER EQUATION
This chapter will mostly be dedicated to the study of the linear Schrödinger equation
i∂t u − ∆u = 0,
(LS)
u(t = 0) = u0
where u(t, x) ∈ C is a function of t ∈ R and x ∈ Rd .
The most important phenomemon in this chapter will be dispersion, which can be summarized as follows: different frequencies travel at different speeds. To be more explicit: we will
see that, if the data of (LS) is localized in space around 0, and in frequency around ξ, then u
will travel, roughly speaking, at velocity ξ. In other words: u will be localized in space-time
around {x ∼ ξt}. Of course, general data can be broken into smaller pieces, each of which
with a given frequency localization.
This seemingly simple phenomenon has far reaching, and often delicate to understand,
consequences. The aim of the following is to review and quantify them.
When the equation is set on the torus, the picture becomes much more complicated, since
geodesics tend to wrap up the torus instead of escaping to infinity. We will say a few words
about this case.
1. Plane waves and wave packets
We establish some elementary properties of (LS) in this introductory section; for the moment, u0 is assumed to be smooth so that it is easy to make sense of all manipulations.
1.1. Plane waves. Plane waves are the simplest solutions of (LS) set on Rd : they read
2
u(t, x) = ei(xξ+t|ξ| ) ,
for ξ ∈ Rd .
By linearity of (LS), any solution of (LS) can be decomposed into plane waves. This is
achieved by taking the Fourier transform (in space) of (LS); the equation becomes
i∂t u
b(t, ξ) + |ξ|2 u
b(t, ξ) = 0
u
b(t = 0, ξ) = u
c0 (ξ).
This is simply an ODE! Its solution reads
2
u
b(t, ξ) = eit|ξ| u
c0 (ξ).
Consistently with our notation for Fourier multipliers, we write equivalently for the above
u(t, x) = e−it∆ u0 .
Coming back to u through the inverse Fourier transform gives
Z
1
2
u(t, x) =
ei(xξ+t|ξ| ) u
c0 (ξ) dξ.
d/2
(2π)
Rd
1
(1.1)
2
THE LINEAR SCHRÖDINGER EQUATION
1.2. The fundamental solution. The previous formula gives u as a superposition of plane
waves weighted by u
c0 (ξ). In order to substitute to this formula a physical space approach,
we need to compute the Fourier transform of complex Gaussians.
Lemma 1.1. If σ ∈ C with Reσ > 0,
2
F(eiσx ) =
1
2σ
d
π d
2
2 − |ξ|
e 4σ ,
2σ
1 d
2σ
with a positive real part.
R
2
Proof. It suffices to treat the case d = 1 and compute √12π R e−ixξ−σx dx. By first completing
the square, and then using contour integration, one obtains that
Z
Z
Z
ξ 2
1
1 − ξ2
1 − ξ2
2
−σ(x+i 2σ
−ixξ−σx2
)
4σ
4σ
√
e
e−σx dx.
e
dx = √ e
dx = √ e
2π R
2π
2π
R
√
2
Using once again contour integration and the well-known Gaussian integral R e−x dx = π,
Z
Z
ξ2
ξ2
1
1
1
2
−ixξ−σx2
− 4σ
√
√
e
dx =
e
e−σx dx = √ e− 4σ .
2π R
2πσ
2σ
where
2
is the square root of
The fondamental solution of the Schrödinger equation corresponds to the case where Reσ =
0; it can be reached by a limiting argument. Indeed, assuming for simplicity that u0 ∈ S,
2
2
u(t, x) = F −1 (eit|ξ| u
c0 (ξ)) = lim→0 F −1 (e(it−)|ξ| u
c0 (ξ))
1
2
F −1 (e(it−)|ξ| ) ) ∗ F −1 (c
u0 (ξ))
= lim→0
d/2
(2π)
d/2 Z
|x−y|2
1
1
− 4(it−)
u0 (y) dy
= lim→0
e
(2π)d/2 2(it − )
Z
|x−y|2
1
1
− 4it
=
u0 (y) dy.
e
(4πi)d/2 td/2
In other words, the fundamental solution of (LS) is given by the following kernel
u(t, x) = [Kt ∗ u0 ](x)
with Kt (x) =
1
1 i |x|2
e 4t .
(4πi)d/2 td/2
1.3. Wave packets. Wave packets are characterized by the fact that they are well localized
in space and frequency. In order to understand their behavior as t → ∞, we will view (2.1)
as an oscillatory integral and use the stationary phase lemma.
Lemma 1.2 (Stationary phase). Assume that φ is smooth and f ∈ C0∞ . Assume furthermore
that ∇φ only vanishes at x0 ∈ Rd and that Hess φ(x0 ) is non degenerate. Then
Z
C
1
iλφ(x)
iλφ(x0 )
I(λ) =
e
f (x) dx = d/2
e
f (x0 ) + O
.
λ (det Hess φ)d/2
λ(d+1)/2
Rd
In order to apply this lemma to (2.1), we write it as
Z
Z
1
1
i(xξ+t|ξ|2 )
e
u
c
(ξ)
dξ
=
eitΦX (ξ) u
c0 (ξ) dξ
u(t, x) =
0
(2π)d/2 Rd
(2π)d/2 Rd
where we set
x
X=
and ΦX (ξ) = Xξ + |ξ|2 .
t
THE LINEAR SCHRÖDINGER EQUATION
3
It is easy to see that
∇ΦX = X + 2ξ
Therefore, ∇ΦX only vanishes at ξ =
while
Hess ΦX = 2 Id .
− X2 .
The stationary phase lemma gives that
C −it |X|2
X
4 u
u(t, x) ∼ d/2 e
c0 −
2
t
or in other words that
u(t, x) ∼
C
e−i
d/2
|x|2
4t
x
u
c0 −
.
2t
(1.2)
t
This interesting asymptotic formula followed from an application of the stationary phase
lemma for X (hence xt ) fixed. It can be proved to hold with some uniformity in X, but we
are happy to remain at a heuristic level for the moment.
Let us now pick data which are wave packets, which means that they are well localized in
space and frequency. More precisely, assume that
Supp u
c0 ⊂ B(ξ0 , r).
Then we learn from (1.2) that
1
.
• As t → ∞, the decay rate of u is ∼ td/2
• As t → ∞, the solution u is mostly localized, at time t, in B(−2tξ0 , 2tr). In other
words, u is propagating at the group velocity −2ξ0 , and spreading on a scale ∼ t.
2. Elementary L2 estimates
2.1. Group property on Sobolev spaces. Recall that, for a regularity index s, the inhomogeneous and homogeneous Sobolev spaces, denoted H s (Rd ) and Ḣ s (Rd ) respectively, are
defined by their norms
kf kH s = khDis f kL2
and kf kḢ s = k|D|s f kL2 .
We saw that, for a smooth data u0 , the solution of (LS) is given by
2
\
e−it∆
u0 (ξ) = eit|ξ| u
c0 (ξ).
2
Since the Fourier transform is an isometry in L2 , and |eit|ξ| | = 1, we get that, for u0 smooth,
ke−it∆ u0 kH s = ku0 kH s
and ke−it∆ u0 kḢ s = ku0 kḢ s .
It is now easy to extend eit∆ to all of H s and Ḣ s . This can be done by adopting the
Fourier multiplier definition, which makes sense on these Sobolev spaces. Another (equivalent)
possibility is to extend the operator e−it∆ , using the following result: an linear operator from
a Banach space A to a Banach space B, which is defined on a dense subset of A, and bounded,
can be extended in a unique way to a bounded operator from A to B.
This leads to the following lemma.
Lemma 2.1. Let s ∈ R. For any t ∈ R, the operator e−it∆ is an isometry on H s :
keit∆ f kH s = kf kH s .
Furthermore, t 7→ eit∆ = S(t) is a strongly continuous group on H s , namely
• S(0) = Id.
• S(t)S(t0 ) = S(t + t0 )
• If f ∈ H s , S(t)f → S(t0 )f in H s as t → t0 .
4
THE LINEAR SCHRÖDINGER EQUATION
2.2. Virial identity. This is the most simple way of quantifying dispersion.
Lemma 2.2. If f , xf , and ∇f belong to L2 ,
Z
Z
d2
2 −it∆ 2
|x| e
f dx = 8 |∇f |2 dx.
dt2 Rd
Proof. Taking the Fourier transform of xe−it∆ f ),
2
2
F(xe−it∆ f )(ξ) = i∇ξ eit|ξ| fb(ξ) = eit|ξ| (−2tξ fb(ξ) + ∇ξ fb(ξ)).
Since the Fourier transform is an isometry in L2 ,
2
−it∆ 2
xe
f L2 = −2tξ fb(ξ) + ∂ξ fb(ξ) 2 = 4t2 kξ fb(ξ)k2L2 + {constant and linear terms in t}.
L
d2
dt2
to the above inequality gives the desired result.
R
R
2
One can think of the identity above as the fact that Rd |x|2 e−it∆ f dx ∼ 4t2 |∇f |2 dx.
The growth of this weighted norm Ris due to dispersion: at time Rt, the typical scale at which
the solution is localized is |x| ∼ 2t |∇f |2 . If one imagines that |f |2 = 1 and fb is localized
around ξ0 , this becomes |x| ∼ 2t|ξ0 |. This is consistent with the value of the groupe velocity
−2ξ0 .
Applying
2.3. Asymptotic equivalent. We aim at quantifying the asymptotic equivalent (1.2) in L2
|x|2
x
1 −i 4t b
for f ∈ L2 . Observe
e
f − 2t
topology. Let us first explain how to make sense of td/2
∞
that, for f ∈ C0 ,
Z
x
|x|2
|x|2
xy
1
1
−i 4t b
−i 4t
=
ei 2t ψ(y) dy
e
f −
e
d/2
d/2
2t
(4πi)t
(4πi)t
Rd
Z
2
|y|2
|x−y|
1
i 4t
−i 4t
e
f (y) dy
=
e
(4πi)td/2 Rd
|y|2
= eit∆ ei 4t f (y) .
The key point is that the right-hand side is well defined for f ∈ L2 . For f ∈ L2 , we denote
from now on
x
|y|2
|x|2
1
it∆
i 4t
−i 4t b
f −
=e
e
f (y)
(2.1)
e
2t
(4πi)td/2
Lemma 2.3. If f ∈ L2 ,
x −it∆
|x|2
1
−i 4t b
→0
e
f −
f−
e
d/2
2t L2
(4πi)t
Proof. By the definition (2.1),
x −it∆
|x|2
1
−i 4t b
e
f−
e
f −
d/2
2t (4πi)t
L2
as t → ∞.
−it∆
|y|2
it∆
i 4t
f −e
e
= e
f (y) 2
L
−it∆
|y|2
f (y) − ei 4t f (y) =
e
.
L2
Since
eit∆
is an isometry,
|y|2
i
··· . f (y) − e 4t f (y)
t→∞
−→ 0
L2
by dominated convergence.
THE LINEAR SCHRÖDINGER EQUATION
5
3. Decay estimates
3.1. Dispersive estimates. Dispersive estimates give decay at a polynomial rate for data
in Lp spaces.
Theorem 3.1. If f ∈ C0∞ ,
keit∆ f kLp .
(where p0 is such that
1
p
+
1
p0
1
|t|
d
− pd
2
kf kLp0
if 2 ≤ p ≤ ∞
= 1).
0
In particular, by density of C0∞ in Lp , eit∆ can be extended to an operator defined on all
0
of Lp satisfying the above estimate.
Proof. By the Riesz-Thorin interpolation theorem, it suffices, in order to prove this theorem,
to prove the endpoints p = 2 and p = ∞. We already proved that eit∆ is an isometry on L2 ;
this takes care of p = 2. In order to treat p = ∞, we rely on the fundamental solution of the
linear Schrödinger equation and the Young theorem:
1
1
1
1 i |x|2
1 i |x|2 it∆
ke f kL∞ = .
kf kL1 . d/2 kf kL1 .
e 4t ∗ f e 4t d/2
d/2
d/2
d/2
(4πi) t
(4πi) t
t
L∞
L∞
3.2. Strichartz estimates. These estimates give integrated decay for data in L2 spaces.
Theorem 3.2. A pair (q, r) of exponents is called admissible if

2 ≤ q, r ≤ ∞

 2 d
d
+ =

r
2
 q
(q, r, d) 6= (2, ∞, 2).
Assuming that f and F are C0∞ , and that (q, r) and (Q, R) are admissible,
(i) keit∆ f kLqt Lrx . kf kL2
Z
(ii) e−is∆ F (s) ds
2 . kF kLqt 0 Lrx0
L RZ t
i(t−s)∆
0
(iii) e
F (s) ds
R0
q r . kF kLQ
t Lx
0
Lt Lx
Proof. Step 1: (iii) if q = Q, r = R Bound first
Z
Z i(t−s)∆
ei(t−s)∆ F (s) ds
F (s) r ds
q r . e
q (by Minkowski’s inequality)
Lx
R
R
Lt Lx
Lt
Z
1
0 ds
.
(by the dispersive estimate)
d
d kF (s)kLr
x
|t − s| 2 − r
q
Lt
. kF kLq0 Lr0
t
(by the Hardy-Littlewood-Sobolev inequality).
x
Inequality (iii) if q = Q, r = R corresponds to replacing
this is possible thanks to the Christ-Kiselev lemma.
R
R
by
Rt
0
in the above left-hand side;
6
THE LINEAR SCHRÖDINGER EQUATION
Step 2: (ii) We resort to the T T ∗ trick:
Z
2
Z
Z Z
−is∆
−it∆
e−is∆ F (s) ds =
e
F (s) ds
e
F (t) dt dx
2
R
Rd
R
Lx
Z Z Z
i(t−s)∆
e
F (s) ds F (t) dx dt (by Fubini’s theorem)
=
R Rd
R
Z
0
0
i(t−s)∆
kF kLq0 Lr0 (by Lq Lr - Lq Lr duality)
F (s) ds
. e
Lqt Lrx
R
. kF k2 q0
Lt Lrx0
(by the dispersive estimate)
Step 3: (i) It is actually simply dual to (ii):
Z Z
it∆ it∆ e f q r = sup
e f F (t) dx dt
L L
t
x
kF k
0
q0
Lt Lr
x
Z
e
sup
kF k
0 ≤1
q0
Lt Lr
x
sup
.
kF k
0 ≤1
q0
Lt Lr
x
. kf kL2x
0
0
(by Lq Lr - Lq Lr duality)
Rd
R
Z
=
x
t
Rd
−it∆
F (t) dt f dx
(by Fubini’s theorem)
R
Z
e−it∆ F (t) dt
L2x
R
kf kL2x
(by the Cauchy-Schwarz inequality)
(by inequality (ii)
Step 4: (iii) for (q, r) 6= (Q, R) Bound first
Z
Z
it∆
−is∆
ei(t−s)∆ F (s) ds
=
e
e
F
(s)
ds
q r q r by (i)
R
R
Lt Lx
Lt Lx
Z
−is∆
.
F (s) ds
by (ii)
e
L2x
R
. kF kLQ0 LR0 .
Once again, inequality (iii) corresponds to replacing
this is possible thanks to the Christ-Kiselev lemma.
R
R
by
Rt
0
in the above left-hand side;
4. Smoothing estimate
The following smoothing estimate is due to Kato. It expresses the fact that a solution of
the linear Schrödinger equation is, upon localizing in space and averaging in time, 12 derivative
more regular than its data u0 (in the L2 Sobolev scale). This can be understood as follows:
the more singular, or less smooth parts of u0 correspond to very high frequencies ξ → ∞.
But we know that a function localized around ξ in Fourier travels at speed ∼ −2ξ; so the
higher the frequency, the faster the groop velocity. Therefore, very high frequency are sent
to ∞ almost immediately, and the solution becomes locally smoother.
Theorem 4.1. If > 0,
Z Z
R
Rd
1/2 it∆ 2
|D| e f dx dt . kf k2L2 .
hxi1+
THE LINEAR SCHRÖDINGER EQUATION
7
Proof. Step 1: switching to Fourier space First denote a(x) = hxi11+ . By Plancherel’s theorem, the above left-hand side can be written
Z Z 1/2 it∆ 2
|D| e f a(x) dx dt = h|D|1/2 eit∆ f , a(x)|D|1/2 eit∆ f iL2
d
R R
h
i
2
2
= h|ξ|1/2 eit|ξ| fb(ξ) , |ξ|1/2 eit|ξ| fb(ξ) ∗ a(ξ)iL2
ZZZ
2
2
=
eit(|ξ| −|η| ) |ξ|1/2 |η|1/2 b
a(ξ − η)fb(ξ)fb(η) dη dξ dt.
(4.1)
Step 2: a Fourier analysis formula and its consequence We claim that, if F is C0∞ and φ smooth
and non-degenerate,
Z Z
Z
F (x)
eiλφ(x) F (x) dx dλ = (2π)d
dΣ(x)
(4.2)
|∇φ(x)
d
R R
φ=0
(where Σ is the d−1-dimensional surface measure on the
R hypersurface {φ =R 0}R induced by the
Lebesgue measure on Rd . Recalling the coarea formula g(x)|∇u(x)| dx = R u(x)=t g(x) dΣ dt,
the above left-hand side becomes
Z
Z Z
Z Z
d(x)
iλφ(x)
iλy
e
F (x) dx dλ =
e
dΣ dy dλ
φ=y |∇φ(x)|
R Rd
R R
{z
}
|
F (y)
ZZ
Z
iλy
d/2
=
e F (y) dy dλ = (2π)
Fb(λ) dλ
Z
F (x)
d
d
= (2π) F (0) = (2π)
dΣ(x),
φ=0 |∇φ(x)
which is the desired formula (4.2). Setting φ(ξ, η) = |ξ|2 − |η|2 , the surface {φ = 0} becomes
{|ξ| = |η|}, while ∇φ = (ξ, η)T . We can now apply (4.2) to (4.1), so that the quantity to be
estimated becomes
Z Z Z
|ξ|1/2 |η|1/2
1/2 it∆ 2
b
a(ξ − η)fb(ξ)fb(η) p
dΣ,
|D| e f a(x) dx dt =
|ξ|2 + |η|2
R Rd
|ξ|=|η|
where Σ is the surface measure on the hypersurface {|ξ| = |η|} ⊂ R2d . The upshot of this
new way of writing the quantity we want to estimate is that no oscillations are present in the
above right-hand side; we can take absolute values and estimate directly.
Step 3: the Cauchy-Schwarz inequality Applying it to the above expression, we obtain
Z Z Z
h
i
1/2 it∆ 2
|b
a(ξ − η)| |fb(ξ)|2 + |fb(η)|2 dΣ
|D| e f a(x) dx dt ≤
R Rd
|ξ|=|η|
Z
Z
= 2 |fb(ξ)2 |
|b
a(ξ − η)|dΣ dξ.
|η|=|ξ|
|
{z
}
K(ξ)
In order to obtain the desired bound, it suffices to show that K ∈ L∞ .
8
THE LINEAR SCHRÖDINGER EQUATION
Step 4: K ∈ L∞ Observe first that b
a satisfies the following estimates
|b
a(ξ)| . hξi−N
|b
a(ξ)| . |ξ|1+−d .
Furthermore, K(ξ) can be written, after a change of variables
Z
Z
Z
K(ξ) ≤
|b
a(ζ)|dΣ ≤ |ζ−ξ|=|ξ| |ζ|1+−d dΣ + |ζ−ξ|=|ξ| |ζ|−3d dΣ,
|ζ−ξ|=|ξ|
|ζ|<1
|ζ|>1
which, after a moment of reflexion, is seen to be uniformly bounded in ξ.
5. The case of the torus
The question of linear estimates on the torus is very delicate, and has only been settled
very recently by Bourgain and Demeter. One obvious difference is that the linear flow on the
torus is 4π 2 -periodic, so that only local in time estimates are available. We do not attempt
to give the global picture, but only treat the most simple example of a Strichartz estimate.
Lemma 5.1 (Zygmund). If f ∈ L2 (T),
it∆ e f 4
L
t,x ([0,4π
2 ]×T)
. kf kL2 (T) .
Proof. First decompose f and eit∆ in Fourier series:
X
X
2 2
f (x) =
e2πikx fbk and eit∆ f (x) =
ei(2πkx+4π k t) fbk .
k∈Z
k∈Z
Therefore,
|eit∆ f (x)|2 =
X
ei(2π(k−`)x+4π
2 (k 2 −`2 )t)
fbk fb`
k,`∈Z
= kf k2L2 +
X
ei(2π(k−`)x+4π
2 (k 2 −`2 )t)
fbk fb` .
k6=`
Simply using that kf k2L4 = k|f |2 kL2 ,
keit∆ f k2L4 ≤ k|eit∆ f (x)|2 kL2
2
X
2
2
2
2
i(2π(k−`)x+4π
(k
−`
)t)
≤ kf kL2 + e
fbk fb` 2
k6=`
L
X
2
2
2
. kf kL2 +
|fk f` | . kf kL2 .
k6=`
In the prior to last inequality, we used Plancherel’s theorem (in space and time), along with
the fact that k − ` = k 0 − `0 and k 2 − `2 = k 02 − `02 imply that k = k 0 and ` = `0 .
When treating the nonlinear problem, we will need an estimate for the inhomogeneous
problem. In other words, we would like an estimate for the solution u of i∂t u − ∆u = F ,
in terms of F . Taking the space time Fourier transform (with dual variables τ and k) gives
(τ + k 2 )e
u(τ, k) = Fe(τ, k), so that (at least formally)
u
e(τ, k) =
1 e
F (τ, k).
τ + k2
THE LINEAR SCHRÖDINGER EQUATION
9
Therefore,solving the inhomogeneous problems is equivalent to applying to u
e the Fourier
1
multiplier τ +k2 .
This explains, at least partly, how the following lemma is related to the solution of the
inhomogeneous problem. This will become even more clear when we introduce X s,b spaces.
Lemma 5.2 (Bourgain). If u is a function on R × T,
XZ
2
kukL4 .
hτ + k 2 i3/4 |e
u(τ, k)|2 dτ.
t,x
k
Proof. First decompose
u=
X
uj ,
j≥0
where uej is supported on 2j < hτ + k 2 i < 2j+1 .
Step 1: the basic inequality. We claim that, if j, k ∈ N,
1
3
kuj uk+j kL2t,x . 2 4 k 2 4 j kuj kL2t,x kuj+k kL2t,x ,
which, by Plancherel’s theorem, is equivalent to
X Z
u
ej (σ, `)e
uk+j (τ − σ, k − `) dσ 2
`
1
(5.1)
3
. 2 4 k 2 4 j kuj kL2t,x kuj+k kL2t,x .
Lτ,k (R×Z)
By the Cauchy-Schwarz inequality (in ` and σ), followed by Hölder’s inequality (in k and τ ),
the above left-hand side is bounded by
!1/2 Z
X
2
2
LHS . |e
uj (σ, `)| |e
uk+j (τ − σ, k − `)| dσ
`
2
Lτ,k
!1/2 Z
X
1hσ+`2 i∼2j 1hτ −σ+(k−`)2 i∼2k+j dτ
`
∞
Lτ,k
X Z
= kuj kL2 kuj+k kL2 `
1/2
1hσ+`2 i∼2j 1hτ −σ+(k−`)2 i∼2k+j dτ .
∞
Lτ,k
Therefore, it suffices, in order to prove the claim (5.1), to show that, for any τ and k,
X Z
1
3
1hσ+`2 i∼2j 1hτ −σ+(k−`)2 i∼2k+j dτ . 2 2 k 2 2 j .
`
A necessary condition for the integral over τ to be nonzero is that τ = −`2 −(k−`)2 +O(2k+j ),
in which case it has size O(2j ). Therefore, the claim (5.1) would follow if, as soon as τ ==
−`2 − (k − `)2 + O(2k+j ),
X
1
1
1hσ+`2 i∼2j 1hτ −σ+(k−`)2 i∼2k+j . 2 2 k 2 2 j .
`
But this is easily established, giving the claim.
10
THE LINEAR SCHRÖDINGER EQUATION
Step 2: proof of the estimate. Start with the trick of writing the L4 norm squared as the L2
norm of the square:
X
X
2
2
kukL4 = k|u| kL2t,x = uj uj 0 ≤
kuj uj+k kL4 .
t,x
t,x
j,j 0
4
j,k≥0
Lt,x
Resort now to the above claim, followed by the Cauchy Schwarz inequality (in j) and the
Young inequality (in k)
X
X 1 3
kuj uj+k kL4 .
2 4 k 2 4 j kuj kL2t,x kuj+k kL2t,x
t,x
j,k≥0
j,k≥0
=
X
3
3
1
2 8 j 2 8 (j+k) 2− 8 k kuj kL2 kuj+k kL2
j,k

1/2 
#2 1/2
"
X 3
X X 1 3
.
2 4 j kuj k2 2  
2− 8 k 2 8 (k+j) kuj+k kL2 
L
j
.
X
j
j
3
j
4
2 kuj k2L2 ∼
XZ
k
hτ + k 2 i3/4 |e
u(τ, k)|2 dτ.
k
This is the desired inequality!
References