Notes on

Lecture (0157600) - Fourier analysis and itsapplications to PDEs

Xian LiaoInstitute for Analysis, Karlsruhe Institute for Technology

Summer Term 2019

These are short incomplete notes, only for participants of the course Lecture(0157600) at the Karlsruhe Institute for Technology, Summer Term 2019.Corrections are welcome to be sent to

xian.liao(at)kit.edu.The following textbooks/notes are recommended:

• H. Bahouri, J.-Y. Chemin and R. Danchin: Fourier analysis and non-linear partial differential equations. Springer, 2011.

• H. Koch: Lecture notes on PDE and modelling. http://www.math.

uni-bonn.de/ag/ana/SoSe2017/V3B2_SS_17_PDEM/pdem.pdf

If you have more questions about the lecture, you are welcome to comedirectly to

• Zihui He, Room 3.030, Math. Building 20.30, zihui.he(at)kit.edu,

• Xian Liao, Room 3.027, Math. Building 20.30, xian.liao(at)kit.edu.

1 [June 4, 2019]

Contents

1 Fourier transform 31.1 Definition on L1pRdq . . . . . . . . . . . . . . . . . . . . . . . 31.2 Schwartz space SpRdq . . . . . . . . . . . . . . . . . . . . . . . 41.3 Inverse Fourier transform . . . . . . . . . . . . . . . . . . . . . 71.4 Tempered distribution space . . . . . . . . . . . . . . . . . . . 81.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.6 Functions with compactly supported Fourier transforms . . . . 12

2 Littlewood-Paley theory 142.1 Homogeneous Besov spaces . . . . . . . . . . . . . . . . . . . . 152.2 Homogeneous paradifferential calculus . . . . . . . . . . . . . 222.3 Nonhomogeneous Besov spaces . . . . . . . . . . . . . . . . . . 252.4 Commutator estimates . . . . . . . . . . . . . . . . . . . . . . 26

3 Applications to PDEs 293.1 Transport equation . . . . . . . . . . . . . . . . . . . . . . . . 29

3.1.1 The Cauchy-Lipschitz theorem . . . . . . . . . . . . . . 303.1.2 Transport equation . . . . . . . . . . . . . . . . . . . . 31

3.2 Navier-Stokes equation . . . . . . . . . . . . . . . . . . . . . . 343.2.1 Weak solutions . . . . . . . . . . . . . . . . . . . . . . 353.2.2 Strong solutions . . . . . . . . . . . . . . . . . . . . . . 36

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[23.04.2019]

1 Fourier transform

In this section we will introduce the Fourier transform in the whole spacesetting Rd, d ¥ 1.

1.1 Definition on L1pRdq

Definition 1.1. Let f P L1pRd;Cq, d ¥ 1. We define its Fourier transformas a function f P L8pRd;Cq below

fpξq : Fpfqpξq 1

p2πq d2ˆRdeixξfpxq dx , @ξ P Rd .

Proposition 1.1 (Riemann-Lebesgue). Let f P L1pRd;Cq, then its Fouriertransform f is continuous and satisfies

lim|ξ|Ñ8

|fpξq| 0.

Proof. Exercise. We make use of the continuity of the function in ξ: eixξ

and the Lebesgue convergence theorem to show the continuity of fpξq. Thedecay at infinity of fpξq follows from the density argument and integrationby parts.

By Lemma 1.1, the Fourier transform

F : L1pRdq ÞÑ CbpRdq C0pRdq

is a continuous map and

FpfqL8pRdq ¤1

p2πq d2fL1pRdq.

In the above, C0pRdq is the space of continuous functions which decay to 0at infinity and CbpRdq is the space of bounded continuous functions endowedwith the supremum norm.

We next investigate the application of the Fourier transform on τapfq,eiaxf , f A, f g.

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Lemma 1.1. Let a P Rd, f, g P L1pRdq and A be a real invertible d dmatrix. Then

Fpτapfqq eiaξFpfq, τapfq fpx aq,Fpeiaxfq Fpfqpξ aq,Fpf Aq | detA|1Fpf AT q, and in particular Fpfpλqq λdpFpfqqpλ1q, @λ ¡ 0,

Fpf gq p2πq d2FpfqFpgq,ˆRdfFpgq dx

ˆRdFpfqg dx .

Proof. Exercise. Here the convolution is defined as

pf gqpxq ˆRdfpx yqgpyq dy

ˆRdfpyqgpx yq dy ,

and hencef gL1 ¤ fL1gL1 .

1.2 Schwartz space SpRdq

In this subsection we study the Fourier transform of Schwartz functions, i.e.smooth rapidly decaying functions.

Definition 1.2 (Schwartz space). The Schwartz space SpRdq is the set ofthe smooth functions f P C8pRdq satisfying for any k P N

fk,S : sup xPRd, |α|¤kp1 |x|kq|Bαfpxq| 8.Remark 1.1. Exercise. It is equivalent to say that

SpRdq tf P C8pRdq | sup xPRd |xαBβfpxq| 8, @ multiindices α, βu.Notice that for f P S, fk,S may depend on k P N.

It is also easy to see that

• SpRdq CkpRdq for all k P N and SpRdq LppRdq for all p P r1,8s;• If f P SpRdq, then xαf, Bαf P SpRdq for any multiindex α;

• If f P SpRdq and g P C8b pRdq, then fg P SpRdq;

• If f P SpRdq and g P D1pRdq pC80 pRdqq1 with compact support, then

f g P SpRdq;

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• If f, g P SpRdq, then f g P SpRdq;• The Gaussian function e

12|x|2 P SpRdq.

We say that fn Ñ f in S if fn fk,S Ñ 0 for all k P N. We can thenintroduce a metric dp, q on S:

dpf, gq :¸kPN

2kf gk,S

1 f gk,S ,

such that dpfn, fq Ñ 0 if and only if fn Ñ f in S.

Proposition 1.2. The space pSpRdq, dp, qq is a complete metric space andthe space DpRdq C8

0 pRdq of smooth compactly supported functions is densein it. Hence SpRdq is dense in LppRdq, @p P r1,8q.Proof. Exercise. Observe that for the smooth cutoff function χ, f χpR1qfk,S ¤ CkR

1fk1,S Ñ 0 as RÑ 8.

Since SpRdq L1pRdq, we can define the Fourier transform on SpRdq:Theorem 1.1. The Fourier transform maps continuously from SpRdq toSpRdq: For any integer k P N, there exist a constant C and an integer N P Nsuch that

Fpfqk,S ¤ CfN,S , @f P SpRdq.Furthermore, the equalities in Lemma 1.1 together with the following equali-ties hold true:

Fpp1iBxqαfq ξαFpfq, Fpxαfq piBξqαFpfq, @multiindex α.

Proof. It is straightforward (Exercise) to use Lebesgue’s theorem and inte-gration by parts to show the above two equalities. Then for any |α||β| ¤ k,there exist C and N (we can simply take N k d 1) such thatξαBβξFpfq Fp1i Bxqαp1i xqβf ¤ p2πq d

2 Bαx pxβfqL1

¤ p1 |x|qd1L1p1 |x|qd1Bαx pxβfqL8 ¤ CfN,S .

[23.04.2019][29.04.2019]

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Corollary 1.1. The Fourier transform maps the Gaussian function e12|x|2 P

SpRdq to itself: Fpe 12|x|2q e

12|ξ|2.

Proof. We first notice that

pBxj xjqe 12|x|2 0, j 1, , d,

and hence by Theorem 1.2,

ipξj BξjqFpe12|x|2q 0, j 1, , d.

Let d 1, then the function φpξq : Fpe 12|x|2qpξq P SpRq satisfies the

following first-order differential equation

φ1 ξφ 0, i.e. pe 12ξ2φq1 0,

and hence there exists a constant C P R such that

φpξq Ce12ξ2 .

In particular,

C φp0q Fpe 12x2qp0q 1?

2π

ˆRe

12x2 dx 1,

and thus

Fpe 12x2q φpξq e

12ξ2 .

For d ¥ 2, we simply notice that

Fpe 12|x|2q 1

p2πq d2ˆRdeixξe

12|x|2 dx

1

p2πq d12

ˆRd1

eix1ξ1e

12|x1|2 dx 1 1

p2πq 12

ˆReixdξde

12pxdq2 dx d

1

p2πq d12

ˆRd1

eix1ξ1e

12|x1|2 dx 1e

12pξdq2 Fpe 1

2|x1|2qe 1

2pξdq2 ,

with x1 px1, , xd1q, ξ1 pξ1, , ξd1q. An easy induction argumentimplies the result for d ¥ 2.

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1.3 Inverse Fourier transform

Definition 1.3. Let f P L1pRdq. We define

fpxq : F1pfqpxq p2πq d2

ˆRdeixξfpξq dξ .

Theorem 1.2. Let f P SpRdq, then F1pfqpxq Fpfqpxq and

F1Fpfqpxq fpxq FF1pfqpxq.The Fourier transform is an automorphism on SpRdq.Proof. As limεÑ0 e

ε2

2|x|2 1 pointwisely, we calculate straightforward

F1Fpfqpxq p2πq d2

ˆRdeixξfpξq dξ lim

εÑ0p2πq d

2

ˆRde

ε2

2|ξ|2eixξfpξq dξ

limεÑ0

p2πqdˆRde

ε2

2|ξ|2eixξ

ˆRdeiyξfpyq dy dξ

limεÑ0

p2πqdˆRd

ˆRde

ε2

2|ξ|2eipyxqξ dξ fpyq dy

limεÑ0

p2πq d2

ˆRdFpe ε2

2|ξ|2qpy xqfpyq dy .

We recall Lemma 1.1 and Corollary 1.1 to derive that

p2πq d2Fpe ε2

2|ξ|2qpzq p2πq d

2 εde1

2ε2|z|2 ,

which is a Dirac sequence. Hence F1Fpfqpxq fpxq and FF1 Idfollows simply from F1pfqpxq Fpfqpxq.Remark 1.2. Exercise. If f P SpRdq, then e

ε2

2|x|2f Ñ f in SpRdq as

εÑ 0. Hence p2πq d2 εde

12ε2

|x|2 f Ñ f in SpRdq as εÑ 0.

Corollary 1.2. Let f, g P SpRdq, then

Fpf gq p2πq d2FpfqFpgq, Fpfgq p2πq d2Fpfq Fpgq,

and ˆRdfg dx

ˆRdf ¯g dξ .

In particular, fL2 fL2.

Proof. Exercise. We show the result by the following equalities from Lemma1.1:

Fpf gq p2πq d2FpfqFpgq,ˆRdfg

ˆRdf g.

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1.4 Tempered distribution space

Definition 1.4. A tempered distribution on Rd is a continuous linear mapfrom SpRdq to C. We denote the set of tempered distributions by S 1pRdq. Wesay that Tj Ñ T in S 1pRdq if

Tjpfq Ñ T pfq for any f P SpRdq.Remark 1.3. • By the definition above, a tempered distribution T is a

distribution T P D1pRdq pC80 pRdqq1 such that there exists k P N and

C P R s.t.|T pfq| ¤ Cfk,S , @f P DpRdq.

Indeed, if T P D1pRdq such that the above inequality holds, then asDpRdq SpRdq is dense, there exists a unique continuation of T as acontinuous linear map from SpRdq.• The convergence Tj Ñ T in S 1pRdq means indeed that there exists k P N

such thatsup fk,S¤1|Tjpfq T pfq| Ñ 0

Definition 1.5. Let T P D1pRdq. We define the support of T : Supp pT q asthe complement of the following set

tx P Rd | Dr ¡ 0 s.t. T pfq 0, @f P C80 pBp0, rqqu.

Example 1.1. • Every function g P L1pRdq defines a tempered distribu-tion Tg P S 1pRdq as

Tgpfq ˆRdgf dx , @f P SpRdq.

Similarly it holds for g P Lp for any p P r1,8s.• Let us denote the set of distributions with compact support by E 1pRdq,

which is the dual space of C8pRdq endowed with the seminorms sup |α|¤kBαfL8pBp0,kqq,k P N. Then E 1pRdq S 1pRdq and in particular the Dirac functionδ0 P S 1pRdq.

If A : SpRdq ÞÑ SpRdq is a linear operator, then by duality we can definethe operator At : S 1pRdq ÞÑ S 1pRdq as follows:

pAtT qpfq T pAfq, @T P S 1pRdq, f P SpRdq.Definition 1.6. Let φ P C8 with at most polynomial growth, f P SpRdq andT P S 1pRdq. Then we define the following operators on S 1pRdq:

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• The product by φ : pφT qpfq T pφfq,• The derivative Bxj : pBxjT qpfq T pBxjfq,• The convolution with f : pT fqpxq T pfpx qq,• The Fourier transform F : FpT qpfq T pFpfqq,• The inverse Fourier transform F1pT qpfq T pF1pfqq.

Remark 1.4. We can simply check (Exercise) the above equalities directlywhen T Tg, g P SpRdq.

Hence by Theorem 1.2 the Fourier transform is an automorphism onS 1pRdq.

[29.04.2019][06.05.2019]

Proposition 1.3. SpRdq is dense in S 1pRdq.Proof. By Remark 1.2,

p2πq d2 εde

12ε2

||2 pe δ2

2|x|2fq Ñ f in SpRdq, as εÑ 0, δ Ñ 0.

By the above definition, for any T P S 1pRdq, there exist k and C such thatT gpxq T pgpx qq ¤ Cgpx qk,S Csup yPRd, |α|¤kp1 |y|kq|Bαy gpx yq|, @g P SpRdq,

which is a smooth function with at most polynomial growth, and for anyf, g P S,

pT gqpfq ˆRdT pgpx qq fpxq dx T

ˆRdgpx qfpxq dx

T pgpq fq.

Hence for any T P S 1,

Tp2πq d

2 εde1

2ε2||2 pe δ2

2|x|2fq

T p2πq d

2 εde1

2ε2||2pe δ2

2|x|2fq

e

δ2

2|x|2p2πq d

2 εde1

2ε2||2 Tpfq Ñ T pfq, as εÑ 0, δ Ñ 0.

Thus

S Q e δ2

2|x|2p2πq d

2 εde1

2ε2||2 TÑ T in S 1.

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Theorem 1.3. The Fourier transform defines a unitary operator from L2pRdqto L2pRdq.Proof. Let f P L2pRdq and let pfnqn SpRdq converge to f in L2pRdq. Thenby Corollary 1.2, fm fnL2 fm fnL2 and hence pfnqn SpRdq isa Cauchy sequence in L2. We define the unique limit of pfnqn in L2 as theFourier transform of f . This is a unitary operator from

´Rd fg

´Rd f

¯g.

Proposition 1.4. Let T P S 1pRdq. Then Supp pT q t0u ô FpT q is apolynomial.

Proof. ”ð” is straightforward (Exercise).Now let Supp pT q t0u. For any ε P p0, 1q, we take a cutoff function

χε χpε1q P C80 pBp0, εqq. Then T pp1 χεqfq 0 for any f P SpRdq and

hence T pfq T pχεfq. As T P S 1pRdq, there exist k, C such that

|T pfq| |T pχεfq| ¤ Cχεfk,S .Let f P SpRdq such that Bαfp0q 0 for any |α| ¤ k, then for any

|x| ¤ ε 1, by Taylor’s expansion formula,

|fpxq| ¤ C0sup |β|k1BβfL8 |x|k1,

|Bγfpxq| ¤ C0sup |β|k1BβfL8 |x|k1|γ|, @|γ| ¤ k.

Hence

|T pfq| ¤ Cχpε1qfk,S ¤ C1εfk1,S ,

which tends to 0 as εÑ 0. Thus T pfq 0.Let f P SpRdq. Then by Supp pT q t0u and Taylor’s formula again,

T pfq Tf

¸|α|¤k

1

α!Bαfp0qxα ¸

|α|¤k

1

α!Bαfp0qT pχxαq,

where the first summand on the righthand side vanishes by the above argu-ment. Let g F1f , then

T pgq T pgq ¸|α|¤k

1

α!pBαgqp0qT pχxαq

¸|α|¤k

1

α!

T pχxαqˆ

Rdpixqαgpxq dx

¸|α|¤k

1

α!

T pχxαqpixqαpgq.

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1.5 Examples

Example 1.2. Calculate the Fourier transform of the delta function δ p2πq d

2 , and hence yBαδ p2πq d2 piξqα. Thus by Proposition 1.4, the tempered

distribution T with Supp pT q t0u can only be a linear combination of δ andits derivatives: T °|α|¤k aαBαδ0.

Indeed, it is easy to calculate

δpfq δpfq fp0q p2πq d2

ˆRdfpξq dξ p2πq d

2

pfq.Example 1.3. When d 1, then Fpe|x|q

b2π

11ξ2 (Exercise).

Example 1.4. When σ P p0, dq, then Fp|x|σq c|ξ|σd for some constantc. (Exercise)

Let d ¥ 2. Define the operators

R d

j1

xjBxj , Zj,k xjBxk xkBxj ,

such that

|x|2Bxk xkR d

j1

xjZj,k,

and

Rp|x|σq σ|x|σ, Zj,kp|x|σq 0.

Then we show that

RpFp|x|σqq FpRp|x|σqq dFp|x|σq pσ dqFp|x|σq,Zj,kpFp|x|σqq 0,

such that

|ξ|2Bxkp|ξ|dσFp|x|σqq ξkRp|ξ|dσFp|x|σqqd

j1

ξjZj,kp|ξ|dσFp|x|σqq 0.

Therefore the distribution ∇p|ξ|dσFp|x|σqq is supported at t0u such thatT : |ξ|dσFp|x|σq c is supported at t0u for some constant c. Thus byExample 1.2, T °|α|¤k aαBαδ0 and

0 RT ¸|α|¤k

aαRBαδ0

¸|α|¤k

aαpd |α|qBαδ0,

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where we used RpBαδ0q pd|α|qBαδ0 (from the fact Bαδ0 p1ixqαp2πq d

2 ).Hence aα 0, T 0 and Fp|x|σq c|ξ|σd.

For d 1, we simply make use of the operator R x ddx

.

Remark 1.5. If f is homogeneous of degree m: fpλxq λmfpxq, then f ishomogeneous of degree pm dq: fpλ1ξq λmdfpξq. This coincides withExample 1.4.

1.6 Functions with compactly supported Fourier trans-forms

We have already seen from Proposition 1.4 that if a tempered distribution’sFourier transform is compacted supported at t0u, then it is a polynomial. Wenow investigate the relations between the Lp functions and their derivativeswhen their Fourier transforms are compactly supported on a ball or on anannulus.

Lemma 1.2 (Bernstein). Let B tξ P Rd | |ξ| ¤ Ru be a ball centered at 0with radius R ¡ 0 and C tξ P Rd | 0 r1 ¤ |ξ| ¤ r2u be an annulus. Thenthere exists a constant C such that the following facts hold for any k P N,λ ¡ 0, p, q P r1,8s with p ¤ q and u P LppRdq:

Supp puq λB ñ sup |α|kBαuLq ¤ Ck1λkdp dq uLp ,

Supp puq λC ñ Ck1λkuLp ¤ sup |α|kBαuLp ¤ Ck1λkuLp .

Proof. By scaling argument (Exercise) we can restrict ourselves to the casewith λ 1 and by density argument it remains to consider u P SpRdq.

Let χ P DpRdq be a cutoff function with value 1 on B. If Supp puq B,then

u χuñ u p2πq d2 pχ uq ñ Bαu p2πq d

2 pBαχq u,and hence by Young’s inequality,

BαuLq ¤ p2πq d2 BαχLruLp , 1 1

q 1

p 1

r.

Obviously χ P S and Bαχ P S Lr, while we can also estimate

fLr ¤ fL1 fL8 ¤ Cp1 |x|2qdfL8 fL8 ¤ Cp1∆qdfL1 ,

we derive BαχLr ¤ Cp1 ∆qdpξαχqL1 ¤ Ck1, such that BαuLq ¤Ck1uLp follows.

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[06.05.2019][07.05.2019]

If Supp puq C, then we can take ϕ P DpRd zt0uq with ϕ 1 on C suchthat

u ϕu ϕ|ξ|2k¸|α|k

piξqαpiξqαu ¸|α|k

ϕpiξqα|ξ|2k

yBαu.Then by a similar argument as above we have

uLp ¤¸|α|k

F1ϕpiξqα|ξ|2k

BαuLp¤¸|α|k

F1ϕpiξqα|ξ|2k

L1BαuLp ,

where F1ϕpiξqα|ξ|2k

L1 ¤ C

p1 |x|2qdF1ϕpiξqα|ξ|2k

L8

¤ Cp1∆qdϕpiξqα|ξ|2kL1 ¤ Ck1.

We also have the following lemma describing the action of the heat semi-group on Lp functions whose Fourier transform are supported on an annulus.

Lemma 1.3. Let C be an annulus as in Lemma 1.2. Then there exists aconstant C such that the following fact holds true for any t ¡ 0, λ ¡ 0,p P r1,8s and u P Lp:

Supp puq λC ñ et∆uLq ¤ CeC1tλ2λ

dp dq uLp , @q P rp,8s.

Proof. By scaling argument (Exercise) we can restrict ourselves to the caseλ 1. Take ϕ P DpRd zt0uq as in the above proof, then if Supp puq C,

u ϕuñzet∆u ϕet|ξ|2

u.

It remains to show (Exercise)

F1ϕet|ξ|

2Lr ¤ Cp1∆qdpϕet|ξ|2qL1 ¤ CeC1t.

It is straightforward (Exercise) to derive from Lemma 1.2 and Lemma1.3 that

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Corollary 1.3. Let C be an annulus as above. Then there exists a constantC such that for any T ¡ 0, λ ¡ 0, 1 ¤ p ¤ q ¤ 8, 1 ¤ b ¤ a ¤ 8 there hold

Supp pu0q λCpBtµ∆qu0ùñu|t0u0

uLapr0,T s;Lqq ¤ Cpµλ2q 1aλ

dp dq u0Lp ,

and

Supp pfpt, qq λC, @t P r0, T spBtµ∆qufùñu|t00

uLapr0,T s;Lqq ¤ Cpµλ2q1 1b 1aλ

dp dq fLbpr0,T s;Lpq.

2 Littlewood-Paley theory

We have seen from Berstein’s inequalities in Lemma 1.2 that if the Fouriertransform u of a function u P LppRdq is compactly supported on an annulus,then the application of the derivatives ∇ on u works as a multiplication ofλ on u: ∇uLp λuLp , with λ denoting the size of Supp puq. We henceintroduce the following dyadic partition of unity

1 χpξq ¸j¥0

ϕjpξq, ϕjpξq ϕp2jξq @ξ P Rd, (2.1)

where the radial functions

χ P DpBq, B tξ P Rd | |ξ| ¤ 4

3u,

ϕ χp2q χ P DpCq, C tξ P Rd | 1 ¤ |ξ| ¤ 8

3u,

take the values in the interval r0, 1s as follows:

We can then do the Littlewood-Paley decomposition (formally) for u PLppRdq as follows:

u ∆1u¸j¥0

∆ju, with z∆1u χpξqupξq, y∆ju ϕjpξqupξq, (2.2)

14 [June 4, 2019]

such that by Bernstein’s inequalities

∆juLq ¤ C2jpdp dqq∆juLp , @j ¥ 1, @q ¥ p,

sup |α|kBα∆juLp ¥ C12jk∆juLp , @j ¥ 0.

We also introduce the low-frequency cut-off operator Sj as follows:

Sju ¸

j1¤j1

∆j1u, i.e. ySju χp2jqu, j ¥ 0, (2.3)

and we have

SjuLq ¤ C2jpdp dqquLp , @j ¥ 0, @q ¥ p,

and Sj Ñ Id , j Ñ 8 on S 1pRdq:Proposition 2.1. Let u P S 1pRdq, then

SjuÑ u in S 1pRdq, as j Ñ 8.

Proof. Exercise. By duality and Theorem 1.2 it suffices to show

ySjf χp2jqf Ñ f in SpRdq.

Hence the Littlewood-Paley decomposition (2.2): Id °j¥1 ∆j is well-

defined on S 1pRdq.

2.1 Homogeneous Besov spaces

It is easy to notice from the dyadic partition of unity (2.1) that

1 ¸jPZ

ϕjpξq, @ξ 0.

We now restrict ourselves to the following tempered distributions whoseFourier transforms vanish at the origin in the following sense:

Definition 2.1. We denote by S 1hpRdq the space of tempered distributionsu P SpRdq such that

limλÑ8

θpλDquL8 0 with θpλDqu θpλξqupξq, @θ P DpRdq.

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Remark 2.1. Any tempered distribution u P S 1 with u P L1loc belongs to S 1h.

Any Lp, p P r1,8q function u belongs to S 1h since θpλDquL8 λdθpλ1quL8 ¤ uLpλdθpλ1qLp1 Ñ 0 as λÑ 8, as long as p1 1.

The polynomial P 0 does not belong to S 1h as θpλqP θp0qP .

We then denote by 9∆j, 9Sj, j P Z the following operators (to distinguishfrom (2.2) and (2.3) where j P N):

9∆ju ϕp2jDqu with ϕp2jDqu ϕp2jξqupξq, j P Z,9Sju χp2jDqu with χp2jDqu χp2jξqupξq, j P Z,

(2.4)

such thatu

¸jPZ

9∆ju and limjÑ8

9Sju 0 in S 1, if u P S 1h. (2.5)

Definition 2.2 (Homogeneous Besov spaces). Let s P R, pp, rq P r1,8s2.We denote by 9Bs

p,rpRdq the set of tempered distributions u P S 1h such that

u9Bsp,r

2js 9∆juLppRdq

jPZ

`r 8. (2.6)

Example 2.1. The function |x|σ, σ P p0, dq belongs to 9Bdpσ

p,8 for all p Pr1,8s, but not to 9B

dpσ

p,r for any r P r1,8q.Let us take p 1 and show |x|σ P 9Bdσ

1,8 for σ P p0, dq (Exercise).Indeed, we take a smooth cutoff function ρ to decompose the function f |x|σ into

f f1 f2, f1 ρf P L1, f2 p1 ρqf P Lq, @q ¡ d

σ,

such that f P S 1h by Remark 2.1. Let h F1ϕ, then we calculate

9∆jf p2πq d2 2jdhp2jq f 2jσp 9∆0fqp2jq,

such that 2jpdσq 9∆jfL1 9∆0fL1 . It remains to show 9∆0f P L1 by Bern-stein’s inequality:

9∆0f1L1 p2πq d2h f1L1 ¤ hL1f1L1 ¤ Cf1L1 8,

9∆0f2L1 ¤ Ck 9∆0pDkf2qL1 ¤ CkDkpp1 ρqfqL1 8 if k ¡ d σ.

16 [June 4, 2019]

However, as 9∆0fL1 0 (otherwise f 0), p2jpdσq 9∆jfL1qqj R `r for any

r 8. By Proposition 2.2 (see below), f P 9Bdσ1,8 9B

dpσ

p,8 for any p P r1,8s.On the other hand, as

2jpdpσq 9∆jfLp 2j

dp p 9∆0fqp2jqLp 9∆0fLp ,

p2jp dpσq 9∆jfLpqqj R `r for any r 8.[07.05.2019][13.05.2019]

Theorem 2.1. The homogeneous Besov space 9Bsp,r S 1h, s P R, pp, rq P

r1,8s2 is a normed space. If pp, rq P r1,8q2, then the space S0pRdq : tf PSpRdq | Supp pfq X t0u Hu is dense in 9Bs

p,rpRdq.The definition of the homogeneous Besov space 9Bs

p,r is independent of thechoice of the function ϕ in the dyadic partition. Let C 1 be an annulus andpujqjPZ be a sequence of functions such that

Supp pujq 2jC 1,2jsujLp

jPZ P `rpZq,

¸jPZ

uj Ñ u in S 1 with u P S 1h,

then u P 9Bsp,r and

u9Bsp,r

¤ Cs

2jsujLpjPZ`r .If furthermore s ¡ 0, then for the sequence of functions pvjqjPZ satisfying(with some ball B)

Supp pvjq 2jB,2jsvjLp

jPZ P `rpZq,

¸jPZ

vj Ñ v in S 1 with v P S 1h,

we have v P 9Bsp,r and

v9Bsp,r

¤ Cs

2jsvjLpjPZ`r .Proof. By view of (2.5), the seminorm

9Bsp,ris indeed a norm. In order to

show the density result, for any u P 9Bsp,r, p, r 8, for any ε ¡ 0, we take

the approximated function

uRM,N pId 9SMqθp RquN, uN :

¸|j|¤N

9∆ju,

17 [June 4, 2019]

where N is chosen such that u uN

9Bsp,r ε

2,

M ¡ N , θ is a smooth cutoff function and R will be determined later. Then(Exercise) uRM,N P S0 (while uN P W8,ppRdq is not necessarily in S), and asM ¡ N , we calculate

uRM,N uN 9Bsp,rpId 9SMq

θp Rq 1

uN

9Bsp,r

¸j¥M1

2jsr 9∆j

θp Rq 1

uNrLp 1

r.

We decompose the summation°j¥M1 into high and low frequency parts

separately again, such that (Exercise: Check the following calculation)¸j¥0

2js 9∆j

θp Rq 1

uNLp ¤ Csup j¥02jprss2q 9∆j

θp Rq 1

uNLp

¤ C∇rss2θp Rq 1

uNLp

1

jM1

2js 9∆j

θp Rq 1

uNLp ¤ CpMqθp

Rq 1

uNLp ,

and therefore for fixed M , N , as p 8, we can choose R large enough suchthat

uRM,N uN 9Bsp,r ε

2,

and hence uRM,N u9Bsp,r

ε.

We now show the independence of the choice of the function ϕ in thedyadic partition in the definition of the homogeneous Besov space 9Bs

p,r. In-deed, if we take another function ϕ in the dyadic partition, then there existsJ P N such that Suppϕp2jq X Supp ϕp2j1 q H if |j j1| ¥ J and hence(Exercise) the norms u

9Bsp,rand u

9Bsp,r

are equivalent.

Similarly, if Supp pujq 2jC 1, then there exists J P N such that

uj ¸

|kj|¤J9∆kuj and hence 9∆ku

¸|jk|¤J

9∆kuj.

Thusp2ks 9∆kuLpq`r¤ p ¸

|jk|¤J2pkjqs2jsujLpqk

`r¤ CpJ, sqp2jsujLpqj`r .

18 [June 4, 2019]

If Supp pvjq 2jB, then there exists J P N such that 9∆kv °j¥kJ 9∆kvj

and hencep2ks 9∆kvLpq`r¤ p ¸

j¥kJ2pkjqs2jsvjLpqk

`r¤ CpJ, sqp2jsvjLpqj`r ,

where the last inequality is ensured by s ¡ 0.

Remark 2.2. As S0 S 9Bsp,r, S is also dense in 9Bs

p,r if p 8, r 8. If

s dp

or if ps, p, rq pdp, p, 1q, then 9Bs

p,rpRdq can be a Banach space, while inother cases, due to the possible infrared divergence in the low frequency part(keeping in mind the polynomials), 9Bs

p,rpRdq is not a Banach space. Indeed,in general, if

s d

por ps, p, rq pd

p, p, 1q, (2.7)

then by Lemma 1.2,

limjÑ8

¸j1 j

9∆j1uL8 ¤ limjÑ8

2jpdpsq ¸

j1 j

2js 9∆j1uLp

0,

and hence u P S 1h if u P S 1 and u9Bsp,r

8.

Proposition 2.2. We have the following basic properties for the homoge-neous Besov spaces:

• Homogeneity: upλq9Bsp,r

λsdp u

9Bsp,r, @λ ¡ 0;

• Embedding: 9Bsp,r ãÑ 9B

sdp 1p 1p1q

p1,r1 , if p ¤ p1, r ¤ r1 and in particular9Bsp,1 ãÑ 9Bs

p,r ãÑ 9Bsp,8;

• Interpolation: 9Bs1p,rX 9Bs2

p,r ãÑ 9Bsp,r, @s P rs1, s2s and 9Bs1

p,8X 9Bs2p,8 ãÑ 9Bs

p,1,@s P ps1, s2q. Furthermore, we have for any u P S 1h,

u9Bθs1p1θqs2p,r

¤ uθ9Bs1p,ru1θ

9Bs2p,r, @θ P r0, 1s;

u9Bθs1p1θqs2p,1

¤ Cs2s1 p1

θ 1

1θ quθ9Bs1p,8u1θ9Bs2p,8, @θ P p0, 1q. (2.8)

Proof. The homogeneity property follows easily from the following calcula-tion of change of variables (Exercise):

2js 9∆j

upλxqLp λps

dpqp2jλ1qsp2jλ1qdhp2jλ1q uLp .

where h p2πq d2 ϕ.

19 [June 4, 2019]

We derive from Lemma 1.2 that if p ¤ p1, then

9∆juLp1 ¤ C2jdp 1

p 1p1q 9∆juLp .

This, together with `r1 ¤ `r if r ¤ r1, yields the embedding property

9Bsp,r ãÑ 9B

sdp 1p 1p1q

p1,r1 .The first inequality in (2.8) follows easily from

2jpθs1p1θqs2q 9∆juLp 2js1 9∆juLp

θ2js2 9∆juLp

1θand Holder’s inequality. In order to show the second inequality in (2.8), weseparate the high frequency and low frequency part in the definition of Besovnorms:

u9Bθs1p1θqs2p,1

¸j¤N

2jpθs1p1θqs2q 9∆juLp ¸j¡N

2jpθs1p1θqs2q 9∆juLp ,

with the frequency threshold 2N to be determined later. We use the Besovnorm 9Bs1

p,8 to control the low frequency part:¸j¤N

2jpθs1p1θqs2q 9∆juLp ¤¸j¤N

2jpθs1p1θqs2q2js1u

9Bs1p,8

¸j¤N

2jp1θqps2s1qu9Bs1p,8

2Np1θqps2s1q

2p1θqps2s1q 1u

9Bs1p,8,

and similarly we use the Besov norm 9Bs2p,8 to control the high frequency part:¸

j¡N2jpθs1p1θqs2q 9∆juLp ¤

¸j¡N

2jpθs1p1θqs2q2js2u

9Bs2p,8

¸j¡N

2jθps2s1qu9Bs2p,8

¤ 2Nθps2s1q

1 2θps2s1qu

9Bs2p,8.

We choose N such that

2Nps2s1q u

9Bs2p,8

u9Bs1p,8

such that the second inequality in (2.8) holds.

[13.05.2019][20.05.2019]

We define the homogeneous Sobolev spaces 9Hs as follows:

20 [June 4, 2019]

Definition 2.3. Let s P R. We denote by 9HspRdq the set of tempered distri-butions u P S 1 such that u P L1

loc pRdq and

u9Hs

ˆRd|ξ|2s|upξq|2 dξ

12 8.

Then 9Hs S 1h. Since 9Hs

9Bs2,2, 9Hs 9Bs

2,2 for any s P R. If s d2

(i.e. (2.7) holds), then 9Hs 9Bs2,2.

We also have the following relations between the Besov spaces and theLebesgue spaces:

Proposition 2.3. Let pp, qq P r1,8s2. Then

• 9B0p,1 ãÑ Lp ãÑ 9B0

p,8 and more generally 9Bdp dq

p,1 ãÑ Lq whenever p ¤ q;

• 9B0p,2 ãÑ Lp ãÑ 9B0

p,p whenever p P r2,8q and 9B0p,p ãÑ Lp ãÑ 9B0

p,2 when-

ever p P p1, 2s, and in particular L2 9B02,2;

• 9Bdp

p,1 ãÑ C0 whenever p P r1,8q;

• The space of bounded measures on Rd is continuously embedded in 9B01,8;

• uLq ¤ Cu1θ9Bα8,8

uθ9Bβp,p

whenever 1 ¤ p q 8, α P R with θ pq

and β αp qp 1q.

We may also define the homogeneous Triebel-Lizorkin space 9F sp,rpRdq,

1 p 8 as

9F sp,rpRdq tu P S 1h | u 9F sp,r

2js| 9∆jupxq|

jPZ`r

Lp 8u,

and the Lorentz space Lp,rpRdq, 1 ¤ p 8 as

Lp,rpRdq tu : Rd ÞÑ C | uLp,r s 1

rupsqLrpR; ds

sq 8u,

where u is the rearrangement function of u:

upsq inftλ |mptx P Rd | |upxq| ¥ λuq ¤ su.

There are relations among the above mentioned functional spaces which wedo not go to details here.

21 [June 4, 2019]

2.2 Homogeneous paradifferential calculus

Let u, v P S 1h and we would like to consider their product uv (which is ingeneral not well defined). Recall the Littlewood-Paley decomposition (2.4)-(2.5):

u ¸jPZ

9∆ju and limjÑ8

9Sju 0 in S 1,

v ¸jPZ

9∆jv and limjÑ8

9Sjv 0 in S 1,

where

9∆ju ϕp2jDqu with ϕp2jDqu ϕp2jξqupξq, j P Z,9Sju χp2jDqu with χp2jDqu χp2jξqupξq, j P Z .

Then formally we can write the product uv as

uv ¸j,kPZ

9∆ju 9∆kv,

and we can split the above sum into three parts:

uv ¸jPZ

9Sj1u 9∆jv ¸jPZ

9∆ju 9Sj1v ¸

|jk|¤1

9∆ju 9∆kv.

Notice that the Fourier transform of 9Sj1u χp2pj1qDqu is compactly

supported on the ball of size 432pj1q and the Fourier transform of 9∆ju

ϕp2jDqu is compactly supported on the annulus tξ P Rd | 2j ¤ |ξ| ¤ 832ju,

such that 9Sj1u 9∆jv p2πq d

29Sj1u y9∆jv, j P Z

is compactly supported on the annulus tξ P Rd | 132j ¤ |ξ| ¤ 10

32ju, while

9∆ju 9∆kv p2πq d

2y9∆ju y9∆kv, |j k| ¤ 1

is compactly supported on the ball of size 8 2j.Definition 2.4. Let u, v P S 1h. We denote by 9Tuv the homogeneous paraprod-uct of v by u as follows:

9Tuv ¸jPZ

9Sj1u 9∆jv.

22 [June 4, 2019]

We denote by 9Rpu, vq the homogeneous remainder of u and v as follows:

9Rpu, vq ¸

|jk|¤1

9∆ju 9∆kv.

We call the above product decomposition of uv the Bony decomposition:

uv 9Tuv 9Tvu 9Rpu, vq.

The summations in the above definitions of the bilinear operators 9T, 9Rare formal and we are going to make it rigorous in the setting of homogeneousBesov spaces, by use of the above consideration of the supports of the Fouriertransforms and Theorem 2.1.

Theorem 2.2. Let ps, p, rq P Rr1,8s2 satisfy (2.7): s dp

or ps, p, rq pdp, p, 1q. Then there exists C (depending only on s) such that 9Tuv P S 1h and

9Tuv 9Bsp,r¤ CuL8v 9Bsp,r

.

If furthermore σ 0, then there exists C (depending only on |s σ|) suchthat

9Tuv 9Bsσp,r¤ Cu

9Bσ8,r1v

9Bsp,r2, with

1

r mint1, 1

r1

1

r2

u.

Proof. Since 9Tuv °jPZ 9Sj1u 9∆jv with

9Sj1u 9∆jv compactly supported on

the annulus tξ P Rd | 132j ¤ |ξ| ¤ 10

32ju, we apply Theorem 2.1 and Bern-

stein’s inequality to derive

9Tuv 9Bsp,r¤ C

2js 9Sj1u 9∆jvLp`r¤ C

2js 9Sj1uL8 9∆jvLp`r

¤ Csup jPZ 9Sj1uL82js 9∆jvLp

`r¤ CuL8v 9Bsp,r

.

We claim (Exercise) that if σ 0, then2jσ 9SjuLp`r¤ Cu

9Bσp,r, (2.9)

and hence the same argument as above yields

9Tuv 9Bsσp,r¤ C

2jσ 9Sj1uL8q`r1

2js 9∆jvLp`r2

Cu9Bσ8,r1

v9Bsp,r2

.

23 [June 4, 2019]

Theorem 2.3. Let ps, σq P R2 with sσ ¡ 0. Let psσ, p, rq P Rr1,8s2satisfy (2.7), pp1, p2q P r1,8s2 satisfy 1

p 1

p1 1

p2and pr1, r2q satisfy 1

r

minp1, 1r1 1

r2q. Then there exists C (depending only on s σ) such that

9Rpu, vq9Bsσp,r

¤ Cu9Bsp1,r1

v9Bσp2,r2

.

Proof. Since sσ ¡ 0 and 9Rpu, vq °jPZp°

|jk|¤19∆ju 9∆kvq with

°|jk|¤1

9∆ju 9∆kv

compactly supported on the ball of size 8 2j, we derive from Theorem 2.1that

9Rpu, vq9Bsσp,r

¤ C2jpsσq ¸

|jk|¤1

9∆ju 9∆kvLp`r

¤ C2js 9∆juLp1

`r1

2jσ ¸|jk|¤1

9∆kvLp2`r2

¤ Cu9Bsp1,r1

v9Bσp2,r2

.

Remark 2.3. If s σ 0 and r 1, then from the above proof we have

9Rpu, vq9B0p,8

¤ Cu9Bsp1,r1

v9Bsp2,r2

.

Corollary 2.1. Let ps, p, rq P Rr1,8s2 satisfy (2.7). Then L8 X 9Bsp,r is

an algebra and

uv9Bsp,r

¤ CpuL8v 9Bsp,r u

9Bsp,rvL8q.

Proof. We simply derive from Theorem 2.2 and Theorem 2.3 that

uv9Bsp,r

¤ 9Tuv 9Bsp,r 9Tvu 9Bsp,r

9Rpu, vq9Bsp,r

¤ CuL8v 9Bsp,r CvL8u 9Bsp,r

Cu9B08,8

v9Bsp,r,

which, together with the fact u9B08,8

sup jPZ 9∆juL8 ¤ CuL8 , implies

the result.

[20.05.2019][21.05.2019]

Corollary 2.2. Let ps, σq P pd2, d

2q2 such that s σ ¡ 0, then

uv9Bsσ d22,1

¤ Cu9Hsv 9Hσ .

24 [June 4, 2019]

Proof. Since 9Hs 9Bs2,2 ãÑ 9B

s d2

8,2 , we derive from Theorem 2.2 that

9Tuv9Bsσ d22,1

¤ Cu9Bs d28,2

v9Bσ2,2

¤ Cu9Hsv 9Hσ ,

and similarly we have the above inequality for 9Tvu. We now apply Theorem

2.3 and the embedding 9Bsσ1,1 ãÑ 9B

sσ d2

2,1 to 9Rpu, vq to arrive at

9Rpu, vq9Bsσ d22,1

¤ 9Rpu, vq9Bsσ1,1

¤ Cu9Bs2,2v

9Bσ2,2.

Corollary 2.3 (Hardy’s inequality). Let s P r0, d2q, then there exists C such

that ˆRd

|fpxq|2|x|2s dx ¤ Cf2

9Hs .

Proof. Exercise. Notice that

ˆRd

|fpxq|2|x|2s dx ¤

¸|jk|¤1

9∆jp|f |2qL2 9∆kp|x|2sqL2 ¤ C|f |29B2s d22,1

|x|2s9Bd22s

2,8

.

Remark 2.4. If d ¥ 3 and s 1, then we simply have

ˆRd

|fpxq|2|x|2 dx ¤ C∇f2

L2 .

2.3 Nonhomogeneous Besov spaces

In this subsection we give the definition of the nonhomogeneous Besov space.Most of the results in Subsection 2.1 still hold true in the nonhomogeneousframework, and the proofs are even simpler as we take the Littlewood-Paleydecomposition (2.2): u ∆1u

°j¥0 ∆ju and we do not have to worry

about the low frequencies 9∆ju, j ¤ 1.

Definition 2.5. Let s P R, pp, rq P r1,8s2. The nonhomogeneous Besovspace Bs

p,r consists of all tempered distributions u P S 1pRdq such that

uBsp,r :2js∆juLp

j¥1

`r 8.

25 [June 4, 2019]

Remark 2.5. When p r 2, then

Bs2,2 Hs : tu P S 1 | u2

Hs :ˆRdp1 |ξ|2qs|u|2 dξ 8u.

When s P p0, 1q, p r 8, then

Bs8,8 : Cs tu P CbpRdq | russ : sup xy

|upxq upyq||x y|s 8u.

We just list the properties that still hold true in the nonhomogeneoussetting and the proofs are left to the interested readers:

• Theorem 2.1 holds in the nonhomogeneous setting except that we nowhave the fact that DpRdq is dense in Bs

p,rpRdq when p 8 and r 8.

Furthermore, Bsp,rpRdq is a Banach space.

• The embedding property and interpolation property in Proposition 2.2hold true in the nonhomogeneous setting.

• Theorem 2.2 and Theorem 2.3 hold in the nonhomogeneous setting,where the nonhomogeneous paraproduct is defined by

Tuv ¸j

Sj1u∆jv, Sj1u ¸

1¤k¤j2

∆ku,

and the nonhomgeneous remainder is defined by

Rpu, vq ¸

|kj|¤1

∆ju∆kv.

Correspondingly Corollary 2.1 and Corollary 2.2 hold true in the non-homogeneous setting.

2.4 Commutator estimates

Let us consider the transport equation" Btρ v ∇ρ f, t P R, x P Rd,ρ|t0 ρ0,

(2.10)

where the unknown function ρ ρpt, xq P R is transported by the velocityvector field v vpt, xq P Rd and f fpt, xq P R is the source term. Weapply the operator ∆j to it to arrive at the transport equation for ∆jρ:" Btp∆jρq v ∇p∆jρq ∆jf rv ∇,∆jsρ,

p∆jρq|t0 ∆jρ0,(2.11)

26 [June 4, 2019]

where rv ∇,∆jsρ denotes the commutator v ∇p∆jρq∆jpv ∇ρq. In order totransport the Bs

p,r-regularity of the unknown function ρ, we have to estimatethe commutator as follows:2jsrv ∇,∆jsρLp

j¥1

`r.

Theorem 2.4. Let s P R, pp, rq P r1,8s2 and v : Rd ÞÑ Rd be a vector field.Assume further that

s ¡ max tdp, d

p1u with

1

p1 1 1

p,

or s ¡ 1max tdp, d

p1u if div v :

d

j1

Bxjvj 0. (2.12)

Then there exists a constant C such that2jsrv ∇,∆jsρLpj¥1

`r¤ C∇v

Bdpp,8XL8

ρBsp,r if s 1 d

p. (2.13)

For general s ¡ 0 or s ¡ 1 when div v 0, then2jsrv ∇,∆jsρLpj¥1

`r¤ C

∇vL8ρBsp,r∇vBs1p,r∇ρL8

. (2.14)

Proof. We decompose v into the low frequency and high frequency parts:

v ∆1v v, v ¸j¥0

∆jv.

To warm up, we have (Exercise) for the low frequency part that2jsr∆1v ∇,∆jsρLpj¥1

`r¤ C∇vL8∇ρBs1

p,r, (2.15)

by calculating straightforward

r∆1v ∇,∆jsρ p2πq d2

ˆRdϕjpx yqp∆1vqpxq p∆1vqpyq

∇ρpyq dy

p2πq d2

ˆRdϕjpx yq

px yq

ˆ 1

0

p∇∆1vqpx τpx yqqdτ∇ρpyq dy

and noticing

r∆1v ∇,∆jsρ ¸

|j1j|¤1

r∆1v ∇,∆jsp∆j1ρq

and ϕjpx yq px yq 2jdϕp2jpx yqq px yq 2j

2jdzϕpzq|z2jpxyq

.

27 [June 4, 2019]

Now we consider the commutator

rv ∇,∆jsρ d

k1

vkBk∆jρ∆jpvkBkρq

d

k1

TvkBkp∆jρq TBk∆jρvk Rpvk, Bk∆jρq ∆jpTvkBkρ TBkρv

k Rpvk, Bkρqq

d

k1

rTvk ,∆jsBkρ TBk∆jρvk ∆jpTBkρvkq Rpvk, Bk∆jρq ∆jRpvk, Bkρq,

and we are going to estimate the terms one by one.Similarly as above (Exercise), rTvk ,∆jsBkρ

°|jj1|¤4rSj11v

k,∆jsBk∆j1ρ

satisfies (2.15).[21.05.2019][27.05.2019]

As

TBk∆jρvkLp

¸j1¥j3

Sj11Bk∆jρ∆j1 vkLp

¤ CBk∆jρLp¸

j1¥j3

∆j1 vkL8

¤ C∆jpBkρqLp¸

j1¥j3

2j1∆j1p∇vkqL8

¤ C∆jpBkρqLp2j∇vL8 ,(2.15) follows for the term TBk∆jρv

k.Now we come to∆jpTBkρvkq

Lp ¸

|jj1|¤4

∆jpSj11Bkρ∆j1 vkq

Lp¤

¸|jj1|¤4

Sj11BkρL8∆j1 vkLp .

Recall the characterisation (2.9) of Bs1 d

p8,r when s 1 d

p, such that2js∆jpTBkρvkqLp

j¥1

`r

¤ C2jps1 d

pqSj1∇ρL8

j¥1

`r

2jp1 dpq∆j v

kLpj¥1

`8

¤ C∇ρBs1 dp8,r

vB

1 dpp,8

¤ C∇ρBs1p,r∇v

Bdpp,8

.

We can also simply estimate the above as (without any restriction on s):2js∆jpTBkρvkqLpj¥1

`r¤ C∇ρL8

2js∆j vkLpj¥1

`r

¤ C∇ρL8∇vBs1p,r.

28 [June 4, 2019]

Next it is straightforward to estimate2jsRpvk, Bk∆jρqLpj¥1

`r2js ¸

|jj1|¤2

∆j1 vkBk∆jp

¸|j2j1|¤1

∆j2ρqLpj¥1

`r

¤ C∇vL8∇ρBs1p,r.

Finally we can follow the idea in the proof of Theorem 2.3 to control(Exercise)

2js∆j

d

k1

Rpvk, BkρqLpj¥1

`r

2js d

k1

∆jBkRpvk, ρq ∆jRpdiv v, ρqLpj¥1

`r.

Indeed, in the general case where we do not know div v 0, we simply rewrite

2js∆jRpvk, BkρqLp ¤ C

#2jps

dpq∆jRpvk, BkρqL p2 if s ¡ d

p¥ d

p1, i.e. p ¥ 2,

2jps d

p1q∆jRpvk, BkρqL1 if s ¡ d

p1¥ d

p, i.e. p ¤ 2,

then we make use of ∆jRpvk, Bkρq °j1¥j4

°|j1j2|¤1 ∆j

p∆j1 vkqp∆j2Bkρq

to derive (2.13). If div v 0 such that div v div ∆1v, then underthe assumption s ¡ 1 max td

p, d

p1u we still have (2.13). Under the

assumption s ¡ 0 or s ¡ 1 if div v 0, the inequality (2.14) followssimilarly.

3 Applications to PDEs

3.1 Transport equation

In this section we will consider the transport equation (2.10):" Btρ v ∇ρ f, t P R, x P Rd,ρ|t0 ρ0,

(T)

where the function ρ ρpt, xq P R is transported by the velocity vector fieldv vpt, xq P Rd and f fpt, xq P R is the source term.

We will first briefly revisit the classical Cauchy-Lipschitz theorem in theODE theory, and then we will solve (T) by use of the commutator estimateestablished in Subsection 2.4.

29 [June 4, 2019]

3.1.1 The Cauchy-Lipschitz theorem

Theorem 3.1. Let E be a Banach space, Ω E an open set, I Q 0 an opentime interval and X0 P Ω.

Let v : I Ω ÞÑ E be L1loc pI; Lip pΩ;Eqq in the following sense:

ˆK

suptpX1ptq,X2ptqqPΩ2 |X1ptqX2ptqu

vpt,X1ptqq vpt,X2ptqqEX1ptq X2ptqE dt 8, @K compact set in I.

Then there exists an open time interval J Q 0 (with J I) such that theequation

Xptq X0 ˆ t

0

vpt1, Xpt1qq dt1 , (ODE)

i.e. ddtXptq vpt,Xptqq with Xp0q X0 has a unique continuous solution

X Xptq : J ÞÑ Ω.

Proof. (Exercise) The theorem follows from the Picard iteration scheme:

Xk1ptq X0 ˆ t

0

vpt1, Xkpt1qq dt1 , @k ¥ 0.

Indeed, since

Xk1 XkE ¤ˆ t

0

vpt1, Xkpt1qq vpt1, Xk1pt1qqE dt1

¤ˆ t

0

Xkpt1q Xk1pt1qEγpt1q dt1 ,

where the function γ γptq characterizes the Lipschitz dependence of thefunction v vpt, q:

γptq : suptpX1ptq,X2ptqqPΩ2 |X1ptqX2ptqu

vpt,X1ptqq vpt,X2ptqqEX1ptq X2ptqE ,

then there exists some time interval J Q 0 such that the sequence X0 °k¥0pXk1ptqXkptqq converges to Xptq in Ω E uniformly on J by virtue

of

suptPJ

Xk1ptq XkptqE ¤ suptPJ

Xkptq Xk1ptqEˆJ

γpt1q dt1

and γ P L1loc pIq. Hence Xptq X0

°k¥0pXk1 Xkq P CpJ ; Ωq is the

solution of (ODE) and is unique by a similar argument as above.

30 [June 4, 2019]

Proposition 3.1 (Property of the flow). Assume the hypotheses in Theorem3.1 with E CpI;Rdq, x P Rd and v P L1

loc pI; Lip pRd;Rdqq. Then the flowψt ψpt, q : Rd ÞÑ Rd of the vector field v:

ψtpxq xˆ t

0

vpt1, ψt1pxqq dt1 (3.16)

is a C1 diffeomorphism on Rd on the entire time interval I and satisfy

∇ψ1t L8 ¤ e

´ t0 ∇vL8 ,

∇ψ1t Id L8 ¤ e

´ t0 ∇vL8 1. (3.17)

Proof. (Exercise) By Theorem 3.1, for any fixed x P Rd, there exists aunique solution Xpt;xq P CpJ ;Rdq of (ODE) on the time interval J I with´Jvpt1, qLip dt1 ¤ 1

2and a unique continuation argument ensures the unique

existence of the solution Xpt;xq P CpI;Rdq. We define ψtpxq Xpt;xq.Differentiate (3.16) with respect to x yields

Bxjpψtqk δj,k ˆ t

0

d

l1

pBlvkqpt1, ψt1qBxjpψt1ql dt1 .

Then the first inequality in (3.17) for ψt follows from Gronwall’s lemma.Correspondingly we derive the second inequality in (3.17) for ψt as follows:

∇ψt Id L8 ¤ˆ t

0

∇vL8e´ t10 ∇vL8 dt1

ˆ t

0

de´ t0 ∇vL8 e

´ t0 ∇vL8 1.

The inequalities for ψ1t follow immediately since

ψ1t pxq x

ˆ 0

t

vpt1, ψt1pψ1t pxqqq dt1 .

[27.05.2019][03.06.2019]

3.1.2 Transport equation

Theorem 3.2. Let s, p, r satisfy (2.12). Assume the initial data ρ0 P Bsp,r,

the source term f P L1pr0, T s;Bsp,rq, and the vector field v P L1pr0, T s; Lip qX

Lqpr0, T s;BM8,8q for some q ¡ 1 and M ¡ 1 s such that V pT q 8 with

V ptq $&%´ t

0∇v

Bdpp,8XL8

, if s 1 dp,

´ t0∇vBs1

p,r, if s ¡ 1 d

por ps, rq p1 d

p, 1q.

31 [June 4, 2019]

Then the equation (T) has a unique solution ρ P L8pr0, T s;Bsp,rq such

that for a.e. t P r0, T s,

ρptqBsp,r ¤ eCV ptqρ0Bsp,r

ˆ t

0

eCV pt1qfpt1qBsp,r dt1

. (3.18)

Furthermore, if r 8 then ρ P Cpr0, T s;Bsp,rq and if r 8 then ρ P

Cpr0, T s;Bs1

p,rq for any s1 s.

Proof. Step 1 A priori estimateLet ρ P L8pr0, T s;Bs

p,rq be a solution of (T). Let us apply the operator∆j to the transport equation (T) to arrive at the equation (2.11). Then bymultiplying both sides of the equation by sgnp∆jρq|∆jρ|p1 and integratingwith respect to x P Rd yields

∆jρptqLp ¤ ∆jρ0Lp ˆ t

0

∆jfLp rv ∇,∆jsρLp

1

pdiv vL8∆jρLp

dt1 . (3.19)

We hence derive from the commutator estimate in Theorem 2.4 that

ρptqBsp,r ¤ ρ0Bsp,r ˆ t

0

fBsp,r Cp ddtV qρBsp,r

dt1 ,

and the estimate (3.18) follows from the Gronwall’s lemma.Step 2 Approximate solution sequence

Let us regularize the data for the transport equation as follows:

ρn,0 Snρ0 P B8p,r, fn Φn t pSnfq P Cpr0, T s;B8

p,rq,vn Φn t pSnvq P Cbpr0, T s Rdq such that ∇vn P Cpr0, T s;B8

p,rqwhere Φn Φptq is a mollifier sequence with respect to the time variable.Then the regularized equation

Btρn vn ∇ρn fn, pρnq|t0 ρ0,n,

has a unique solution ρn P Cpr0, T s;Bsp,rq:

ρnpt, xq ρ0,npψ1n,tpxqq

ˆ t

0

fnpt1, ψn,t1pψ1n,tpxqqq dt1 ,

where ψn,t : Rd ÞÑ Rd is the flow of the vector field vn. Then by the estimate(3.18), we have the following estimate for ρn:

ρnptqBsp,r ¤ eCVnptqρn,0Bsp,r

ˆ t

0

eCVnpt1qfnpt1qBsp,r dt1

,

32 [June 4, 2019]

and hence the uniform estimate for tρnu:

ρnL8pr0,T s;Bsp,rq ¤ eCV pT qρ0Bsp,r

ˆ T

0

fpt1qBsp,r dt1.

Step 3: Convergence of the approximate solution sequenceIn order to show the convergence of the approximate solution sequence,

we prove its compactness with respect to the time variable. To this end,we will show Btρn is uniformly bounded in Lqpr0, T s;Bm

p,8q for some m smallenough, q ¡ 1 and T 8. However, as f P L1pr0, T s;Bs

p,rq, we have to first

consider rρn : ρn ´ t

0fn instead.

By the above uniform estimate on ρnL8pr0,T s;Bsp,rq and the assumption

v P Lqpr0, T s;BM8,8q, M ¡ 1s, we derive from the estimates for paraproduct

and remainder the uniform bound on vn ∇ρnLqpr0,T s;Bmp,8q, m mints1M, s 1,Mu:

vn ∇ρnLqpr0,T s;Bmp,8q ¤ CvnLqpr0,T s;BM8,8q∇ρnL8pr0,T s;Bs1p,r q.

Therefore rρn ρn´ t

0fn

´ t0vn∇ρn is uniformly bounded in L8pr0, T s;Bs

p,rqandW 1,qpr0, T s;Bm

p,8q. By the compact embeddingW 1,qpr0, T sq ãÑãÑ Cpr0, T sqfor some q ¡ 1 and T 8 and the compactness of the multiplication operatorMϕ : Bs

p,r ÞÑ Bmp,8 via g ÞÑ ϕg, @ϕ P C8

0 pRdq, the sequence tϕ rρnu is com-pact in Cpr0, T s;Bm

p,8q and hence there exists a subsequence (still denoted bypρn, fnq) such that rρn Ñ rρ in Cpr0, T s;S 1q and ϕ rρn Ñ ϕrρ in Cpr0, T s;Bm

p,8qand hence in Cpr0, T s;Bs1

p,rq, @s1 s. Thus @s1 s,

ϕρn ϕ rρnˆ t

0

ϕfn Ñ ϕrρˆ t

0

ϕf : ϕρ in Cpr0, T s;Bs1

p,rq, @ϕ P C80 pRdq.

Hence the limit ρ satisfies Btρ v ∇ρ f at least in the distribution sense.Thus ρ P L8pr0, T s;Bs

p,rq X Cpr0, T s;Bs1

p,rq satisfies the estimate (3.18).In order to show ρ P Cpr0, T s;Bs

p,rq if r 8, we simply use Sjρ PCpr0, T s;B8

p,rq to approximate ρ in L8pr0, T s;Bsp,rq: We estimate the dif-

ference ρ Sjρ °k¥j ∆kρ by use of the (3.19) as

∆kρLp ¤ eV ptq∆kρ0Lp

ˆ t

0

∆kfLp CρL8pr0,T s;Bsp,rqˆ t

0

ckpt1q dt1,

for some ckptq P L1pr0, T s; `rq. Therefore if r 8 then2ks∆kρ0Lp

ˆ t

0

∆kfLp

k¥j

`r,2ks

ˆ t

0

ckpt1qk¥j

`rÑ 0,

as j Ñ 8. This implies ρ SjρL8pr0,T s;Bsp,rq Ñ 0 as j Ñ 8 and henceρ P Cpr0, T s;Bs

p,rq.

33 [June 4, 2019]

Remark 3.1. In particular if p r 2, then the Hs-regularity of ρ canbe transported by the velocity vector field v with ∇v P L1pr0, T s;Hσ X L8q,provided with

d2 s 1 d

2and σ d

2,

or s σ 1 ¡ 1 d2.

[03.06.2019][04.06.2019]

3.2 Navier-Stokes equation

In this subsection we consider the initial value problem for the Navier-Stokesequation $&% Btu u ∇u∆u∇p 0,

div u 0,u|t0 u0,

(NS)

where the time variable t ¥ 0, the space variable x P Rd, the unknownvelocity vector field u upt, xq : r0,8qRd ÞÑ Rd and the unknown pressureterm p ppt, xq : r0,8q Rd ÞÑ R. The Navier-Stokes equation can modelthe evolution of the incompressible fluid (including liquid and gas), where

• The material derivative Btu ∇ Bt°dk1 u

kBxk corresponds to thetransport of the fluid along the velocity vector field u;

• The second order derivative term ∆u °dk1 Bxkxku describes the

viscosity effect in the fluid;

• The divergence free condition div u °dk1 Bxkuk 0 corresponds to

the incompressibility of the fluid:

d

dtpdetp∇ϕqq tr

adjp∇ϕq d

dt∇ϕ tr

adjp∇ϕq∇ϕ∇upt, ϕpt, xqq

div upt, ϕpt, xqq 0,

where ψ ψpt, xq : r0,8q Rd ÞÑ Rd is the flow of the velocity vectorfield u: d

dtψpt, xq upt, ψpt, xqq;

• The unknown pressure term ∇p pBxjpqj (not necessarily the physicalpressure in the fluid) can be simply viewed as a Lagrangian multiplierassociated to the divergence free constraint.

34 [June 4, 2019]

If the solution u is not smooth enough, the term u ∇u in (NS) will alwaysbe understood as div pub uq with the matrix pub uqjk : pujukqjk. Indeed,if u is smooth, then thanks to div u 0,

pu ∇uq d

k1

ukpBxkujq,

div pub uq d

k1

Bxkpujukq d

k1

pujBkuk Bkujukq ujdiv u u ∇uj pu ∇uq.

Notice that we need more regularity assumption on u in oder to make senseof the term u ∇u, than of the term ub u.

3.2.1 Weak solutions

It is straightforward to deduce the energy equality for (NS) if the solution isregular enough (say u P C1pr0,8q; pH8pRdqqdq, ∇p P Cpr0,8q, pH8pRdqqdq).Let us take L2pRdq inner product between the equation (NS) and u itself,and we calculate the resulting terms one by one:

•´Rd Btu u

´Rd

12Btp|u|2q 1

2ddt

´Rd |u|2,

•´Rd u ∇u u

´Rd

12u ∇p|u|2q ´Rd 1

2pdiv uq|u|2 0,

•´Rd ∆u u ´Rd |∇u|2;

•´Rd∇p u ´Rd p div u 0.

Thus we arrive at

1

2

d

dt

ˆRd|u|2

ˆRd|∇u|2 0,

which implies immediately the energy equality by integration in time:

1

2uptq2

L2pRdq ˆ t

0

∇upt1q2L2pRdq dt1 1

2u02

L2pRdq. (3.20)

Thanks to the above energy estimate, J. Leray proved the global-in-timeexistence of the weak solution to (NS) in 1934:

Theorem 3.3. Let u0 be a divergence-free vector field in pL2pRdqqd. Thenthere exists a weak solution u P L8pR; pL2pRdqqdq X L2pR; p 9H1pRdqqdq of(NS) satisfying the energy inequality:

1

2uptq2

L2pRdq ˆ t

0

∇upt1q2L2pRdq dt1 ¤ 1

2u02

L2pRdq. (3.21)

35 [June 4, 2019]

Here a weak solution u P L2loc pRRdq of (NS) means that the following

equality holds

ˆRdupt, xq ϕpt, xq dx

ˆ t

0

ˆRdpu Btϕ ub u : ∇ϕ u ∆ϕq dx dt

ˆRdu0pxq ϕp0, xq dx ,

for all ϕ P C80 pr0,8q; pC8

0 pRdqqdq with divϕ 0.

The proof could be similar as the proof of Theorem 3.2. We construct asequence of approximated solutions which satisfy the energy estimate (3.21)uniformly. Then the compactness in the time variable as well as a cut-offprocedure in the space variable will imply the compactness of the approxi-mated solution sequence in a weaker topology and the limit will satisfy (NS)in the weak sense. However, only the energy inequality (3.21) can hold forthe weak solution, while the energy equality (3.20) holds only under moreregularity assumptions in three dimensional case. In particular if d 2, thenthe above weak solution is unique and satisfies the energy equality (3.20),which was proved by O. Ladyzhenskaya in 1959.

3.2.2 Strong solutions

It is also convenient to rewrite the equation (NS) by eliminating the pressureterm ∇p. Indeed, as div u 0, we introduce the projection operator P :

P pId ∇p∆q1div q, i.e. xPv pvj d

k1

ξjξk|ξ|2

pvk d

k1

pδj,k ξjξk|ξ|2 q

pvk,such that

Pu u, P p∇pq 0.

Hence we apply the operator P to the equation (NS) to arrive at" Btu∆u Qpu, uq,u|t0 u0,

(PNS)

where the bilinear operator Q reads as

Qpv, wq 1

2P pdiv pv b wq div pw b vqq,

i.e. pQpv, wqqj 1

2

d

k1

d

l1

pδj,k 1qiξjξkξl|ξ|2pvkwl vlwkq.

36 [June 4, 2019]

We can show the local-in-time well-posedness result of (PNS) in differentfunctional frameworks and here we will follow Kato’s Lp approach to showthe well-posedness result of (PNS) in L3pR3q in three dimensional case.

Theorem 3.4. Let u0 P pL3pR3qq3. Then there exists a positive time T anda unique solution u P Cpr0, T s; pL3pR3qq3q of the initial value problem (PNS).

There exists a positive constant c such that if u0L3 ¤ c then T can bechosen as 8.

Proof. Step 1 A priori estimateLet us take the Fourier transform of the semilinear heat equation (PNS)

and then apply Duhamel’s formula to arrive at

upt, ξq et|ξ|2 pu0pξq i

d

k,l1

ˆ t

0

eptt1q|ξ|2pδj,k 1qξjξkξl|ξ|2

ukpt1qulpt1q dt1 .

Denote

Γjklpt, q p2πq 32F1

iet|ξ|2pδj,k 1qξjξkξl|ξ|2

,

then the solution u reads as

upt, xq et∆u0 d

k,l1

ˆ t

0

Γklpt t1, q pukulqpt1, q dt1 . (3.22)

As

et∆u0 F1et|ξ|

2 pu0pξq p4πtq 3

2 e||2

4t u0,

we apply Young’s inequality to derive for β ¥ 3

et∆u0LβpR3q ¤ p4πtq 32 e |x|2

4t LαpR3qu0L3pR3q, with 1 1

β 1

α 1

3,

¤ Ct32 3

2α u0L3pR3q Ct12p1 3

βqu0L3pR3q.

For any p P r1,8s, T P p0,8q, we define the norm

uKppT q sup tPp0,T st12p1 3

pquptqLppR3q, (3.23)

thenet∆u0KβpT q ¤ Cu0L3pR3q, @β ¥ 3. (3.24)

37 [June 4, 2019]

Now we focus on Γjkl. We can rewrite Γjkl as

Γjkl p2πq 32 pδj,k 1qBjBkBlF1pet|ξ|2 |ξ|2q

p2πq 32 pδj,k 1qBjBkBlF1

ˆ 8

t

et|ξ|2

dt1

p2πq 32 pδj,k 1qBjBkBl

ˆ 8

t

p2tq 32 e

|x|2

4t dt1

p2πq 32 pδj,k 1q

ˆ 8

t

p2tq 32 p4tq 3

2

BjBkBle||2p x?4tq dt1

π 32 pδj,k 1q|x|4

ˆ |x|2

4t

0

sBjBkBle||2p x|x|?sqds, with s |x|2

4t,

which implies the pointwise bound for Γ:

|Γjkl| ¤ C mint|x|4, t2u ¤ Cp|x| ?tq4.

Hence

Γjklpt, qLαpR3q ¤ Cˆ ?

t

0

pt2qαr2 dr ˆ 8

?t

pr4qαr2 dr 1α ¤ Ct2 3

2α , @α P r1,8s.

Therefore Young’s inequality ensures for any pp, qq P r1,8s2 with 1p 1

q¤ 1,ˆ t

0

Γklpt t1, q pukulqpt1, q dt1LβpR3q

¤ C

ˆ t

0

pt t1q2 32p1 1

β 1p 1qqupt1qLppR3qupt1qLqpR3q dt1

¤ C

ˆ t

0

pt t1q 12 3

2p 1β 1p 1qqpt1q1 3

2p 1p 1qq dt1

sup t1¥0pt1q12p1 3

pqupt1qLppR3q

sup t1¥0pt1q

12p1 3

qqupt1qLqpR3q

.

If 13 1

β¡ 1

p 1

q, we can control the above time integral by

Ct12 3

2p 1β 1p 1qqˆ t2

0

pt1q1 32p 1p 1qq dt1 Ct1 3

2p 1p 1qqˆ t

t2pt t1q 1

2 3

2p 1β 1p 1qq dt1

¤ Ct12 3

2β Ct12p1 3

βq.

Then we have arrived at´ t0 Γklpt t1, q pukulqpt1q dt1KβpT q

¤ CuKppT quKqpT q,if 1

β¤ 1

p 1

q 1

3 1

β, 1p 1

q¤ 1.

(3.25)

38 [June 4, 2019]

Step 2 Existence & Uniqueness of the solution in K6pT qWe have established the a priori estimates (3.24)-(3.25) for the solution

(3.22) to (PNS) in Step 1. We would like to use the contraction mappingargument to show the existence and uniqueness of the solution in the Banachspace

K6pT q tu P Cpp0, T s;L6pR3qq | uK6pT q 8u.Indeed, we first rewrite (3.22) into the following form

u aBpu, uq, a : et∆u0, Bpu, uq :ˆ t

0

Γklpt t1, q pukulqpt1, q dt1 .

It is easy to see that if u0 P L3pR3q, then et∆u0 P K6pT q for any T P p0,8q.As Γ P Cpp0,8q;LαpR3qq, @α P r1,8q, the bilinear map

B : K6pT q K6pT q ÞÑ K6pT q, with Bpu, vqK6pT q ¤ CuK6pT qvK6pT q.

For any u0 P L3pR3q, for any ε ¡ 0, there exists ϕ P SpR3q such thatu0 ϕL3pR3q ε. On the other hand, et∆ϕL8pr0,T s;L6q ¤ CϕL6 . Thus

et∆u0K6pT q ¤ et∆pu0 ϕqK6pT q et∆ϕK6pT q

¤ Cu0 ϕL3 CT12p1 3

6qϕL6 ¤ Cε CT

14 ϕL6 .

We can choose T sufficiently small (depending on u0, ε) such that

et∆u0K6pT q ¤ Cε. (3.26)

Therefore for ε ¡ 0, T ¡ 0 sufficiently small, we derive from the contractionmapping argument that there exists a unique fixed point u of the map u ÞÑaBpu, uq in the Banach space K6pT q, with

uK6pT q ¤ 2et∆u0K6pT q. (3.27)

If u0L3pR3q c, then

et∆u0K6pT q ¤ Cu0L3 ¤ Cc, @T P p0,8q.Hence in the small data case that c ¡ 0 is sufficiently small, there ex-ists a unique fixed point u P K6pT q for any T P p0,8q, with uK6pT q ¤2et∆u0K6pT q.Step 3 Continuity with values in L3pR3q

Although we have showed in Step 2 the existence and the uniqueness ofthe solution u P K6pT q, we have to prove further u P Cpr0, T s;L3q and theuniqueness of the solution therein.

39 [June 4, 2019]

Now u P K6pT q with uK6pT q ¤ 2et∆u0K6pT q is the known function andwe would like to show

u aBpu, uq P Cpr0, T s;L3q.Obviously a et∆u0 P Cpr0, T s;L3q. As u P K6pT q, we infer from thederivation of the estimate (3.25) (with β 3) that Bpu, uq P Cpp0, T s;L3qand for any t P p0, T q,

Bpu, uqL8pr0,ts;L3q ¤ Cu2K6ptq ¤ 4Cet∆u02

K6ptq,

where the righthand side tends to zero as tÑ 0. This implies the continuityof Bpu, uq at time zero and hence Bpu, uq P Cpr0, T s;L3q.Step 4 Uniqueness in Cpr0, T s;L3pR3qq

Let u, v P Cpr0, T s;L3q be two solutions to (PNS) and we would like toshow u v. To this end, we will use energy estimates in the L2 functionalframework for their difference

w uv pet∆u0Bpu, uqqpet∆u0Bpv, vqq Bpu, uqBpv, vq P Cpr0, T s;L3q.Indeed, by the equation (PNS), w satisfies the equation$&% Btw ∆w Qpu, uq Qpv, vq

: f Qpet∆u0, wq Qpw, et∆u0q QpBpu, uq, wq Qpw,Bpv, vqq,w|t0 0.

(w)By (3.25) with p q 3, β 2 we have

wK2pT q ¤ Bpu, uqK2pT q Bpv, vqK2pT q ¤ Cu2K3pT q Cv2

K3pT q,

with wK2pT q sup tPp0,T st14 wptqL2 , and hence w P Cpr0, T s;L2q.

By Sobolev embedding 9H12 pR3q 9B

122,2pR3q ãÑ 9B0

3,2pR3q ãÑ L3pR3q (by

Proposition 2.2 and Proposition 2.3) and L32 pR3q ãÑ 9H 1

2 pR3q (by dual-ity), we have the following estimate for the bilinear operator Qpa, bq 1

2P pdiv pa b bq div pb b aqq (noticing the zero-order projection operator

P : 9Hs ÞÑ 9Hs):

Qpa, bq9H 3

2¤ Cab b

9H 12¤ Cab b

L32

¤ C mintaL3bL3 , aL6bL2 , aL2bL6u.Thus f P Cpr0, T s; 9H 3

2 q. We take 9H 12 inner product between the w-

equation and w itself to arrive at

1

2

d

dtw2

9H 12 ∇w2

9H 12 xf, wy

9H 32 , 9H

12,

40 [June 4, 2019]

and hence

1

2wptq2

9H 12 ∇w2

L2pr0,ts; 9H 12 q

ˆ t

0

xf, wy9H 3

2 , 9H12¤ 1

2w2

L2pr0,ts; 9H12 q

1

2f2

L2pr0,ts; 9H 32 q,

which implies

wL8pr0,ts; 9H 1

2 qXL2pr0,ts; 9H12 q ¤ f

L2pr0,ts; 9H 32 q, @t P r0, T s.

Finally we would like to use Gronwall’s inequality to deduce w 0 inL8pr0, ts; 9H 1

2 q, at least in small time interval r0, ts. To this end, we decom-pose f into two parts

f f1 f2, f1 Qpet∆u0,1, wq Qpw, et∆u0,1q QpBpu, uq, wq Qpw,Bpv, vqq,f2 Qpet∆u0,2, wq Qpw, et∆u0,2q, with u0 u0,1 u0,2,

and we expect that

f1L2pr0,ts; 9H 32 q

1

2w

L2pr0,ts; 9H12 q, for small time t ¡ 0,

f22

L2pr0,ts; 9H 32 q involves low regularity of w in the form Cpu0q

ˆ t

0

w29H 1

2dt1 .

Indeed, we decompose u0 into u0,1 with small L3 value: u0,1L3 ¤ c and u0,2

with regular L6 value: u0,2 P L6 (e.g. we can simply take u0,2 Sju0 withsufficiently large j). Then

f1L2pr0,ts; 9H 32 q ¤ C

et∆u0,1L3 Bpu, uqK3ptq Bpv, vqK3ptqwL3

L2pr0,tsq

¤ Cet∆u0,1L8pr0,ts;L3q Bpu, uqK3ptq Bpv, vqK3ptq

wL2pr0,ts; 9H

12 q

1

2w

L2pr0,ts; 9H12 q,

if c and the time t is chosen small enough and

f22

L2pr0,ts; 9H 32 q ¤

ˆ t

0

et∆u0,22L6w2

L2 ¤ Cu0,22L6

ˆ t

0

w9H 1

2w

9H12

¤ 1

8w2

L2pr0,ts; 9H12 q Cu0,24

L6

ˆ t

0

w29H 1

2.

41 [June 4, 2019]