Advanced Diff Theory - JAN MALY

7/23/2019 Advanced Diff Theory - JAN MALY

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ADVANCED THEORY OF DIFFERENTIATION LORENTZ SPACES

JAN MALY

Contents

1. Hardys inequality 12. Rearrangement 13. Lorentz spaces 34. Rearrangement and Lorentz spaces in Rn 65. Embedding to continuous functions 86. Embedding into the Lorentz space 10

7. The critical case 138. Sobolev inequalities 149. Compact embedding 1610. Notes 18References 18

1. Hardys inequality

1.1. Theorem (Hardys inequality). Let1 p p 1.So, ifF(0) =F() = 0, then the Hardy inequality on (0 , ) for Fholds for all =p 1.

2. Rearrangement

2.1. Distribution and rearrangement of a function. Let (X, S, ) be a -finite measure space andf :X Rbe a measurable function. We define the distribution function off as

f(s) =({|f|> s}), s > 0,

and the nonincreassing rearrangement off as

f(t) = inf{s > 0, f(s) t}, t >0.

2.2. Observations.

(a) s < f(t) t < f(s).

(b) (f) = f.

Date: March 2003.

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(c) f = (f).(d) f,f

are nonincreassing, right continuous (and thus lower semicontinuous).(e)

0

f(t) dt=0

f(s) ds=X

|f| d.

Proof. The facts that the functionf,f are nonincreassing and that f is right continuous are obvious.

Fort >0 denoteMt = { >0, f() t}. Ifs < f(t) = infMt, thens does not belong to Mt and thusf(s)> t. Ift < f(s), then sdoes not belong to Mt. ButMt is an interval from somewhere to infinityandMt is left closed unless infMt = 0. It follows that s s}= (0, f(s)) and thus (f

)(s) =((0, f(s))) =f(s). This proves (b). Similarly,{f > t} = (0, f

(t)) and thus (f)(t) = f(t), which is (c). Then the right continuity of f follows

from (c) (but also easily seen directly). The right equality of (e) will follow from Proposition 2.3 (justfollowing) and the left equality uses the same and, in addition, takes into account (b).

2.3. Proposition (Distribution integration).X

|f| d=

0

f(s) ds

Proof. By the Fubini theorem, both integrals are equal to the product measure of the set

{(x, s) X (0, ) : s < |f(x)|}.

2.4. Theorem (Hardy-Littlewood-Polya inequality).X

|f g| d

0

f(t)g(t) dt.

Proof. We use the Fubini theorem to estimateX

|f g| d=

X

0 r}

drds

=

(0,)2

{f > s} {g > r}

drds

=

0

0


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Proof. Suppose that g Gt. By Observations 2.2(c), the integral ofg is 1. Using the monotonicity of

f andg we obtain t

f(s)g(s) ds f(t)

t

g(s) ds= f(t)

(1

t0

g(s) ds

=f(t)

t

0

1t

g(s)

ds t

0

f(s)

1t

g(s)

ds,

which shows, with the help of the Hardy-Littlewood-Polya inequality (Theorem 2.4) thatX

|f g| d

0

f(s)g(s) ds1

t

t0

f(s) ds= f(t).

To prove converse inequality, we denote

t =({f > f(t)}), t ={ff(t)}

and set

g=

1t on{f > f

(t),tt

t(tt) on{f=f(t)}.

A simple calculation shows thatg Gt andX

|f|g d= f(t).

2.7. Proposition (Subadditivity). Let : (0, ) Rbe right continuous, nonnegative and nonincreas-ing. Then the operator

f

0

(t)f(t) dt

is subadditive. In particular, the double-star operator is subadditive.

Proof. We will prove the statements in the converse order. The double-star operator is subadditive by thecharacterization proved in Proposition 2.6. Now, each right continuous, nonnegative and nonincreasing

function can be represented as(t) =c + ((t, )),

whereis a Lebesgue-Stieltjes measure on R. There is no problem with the constant part of. Now,by the Fubini theorem,

0

f(t) ((t, )) dt=

0

s0

f(t) dt

d(s) =

0

sf(s) d(s),

which is subadditive by the special statement.

3. Lorentz spaces

3.1. Lorentz norm. Let 1 m , 1 q . We define

fLm,q =tt 1m f(t)Lq(dtt ) .

We observe

fqLm,q =

0

(t1/mf(t))qdt

t , q 0

t1/mf(t).

We define

Lm,q =

f : f :X R measurable, fLm,q


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Proof. First, we observe that

{f+ g > s} {f > s/2} {g > s/2}

and thus

(3) (f+ g)(s) f(s/2) + g(s/2).Iff(t/2) + g(t/2) s/2, then f(t/2) s/2 and thus f(s/2)t/2, similarly g(s/2) t/2. Using(3) we obtain

(f+ g)(s) t/2 + t/2 =t.

It follows that

(f+ g)(t) 2

f(t/2) + g(t/2)

.

From this we easily deduce that all the Lorentz spaces are linear.

3.3. Proposition (Triangle inequality). Lm,q is a norm if1 q m.

Proof. Let f , g be nonnegative measurable functions on X. Then tqm1[(f+g)]q1(t) is nonincreasing

and thus by Proposition 2.7

h 0

tqm1[(f+ g)(t)]q1h(t) dt

is subadditive. Hence, as in the standard proof of the Minkowski inequality, 0

t qm1[(f+ g)]q(t) dt

0

tqm1[(f+ g)]q1(t)f(t) dt +

0

t qm1[(f+ g)]q1(t)g(t) dt

0

tqm1[(f+ g)]q(t) dt

11/q 0

tqm1[f]q(t) dt

1/q+

0

t qm1[g]q(t) dt

1/qand a cancellation yields the result. The cancellation is legitimate by Proposition 3.2.

3.4. Theorem (Equivalent norm). If 1 < m < , then L(m,q) is a genuine norm equivalent to

Lm,q .

Proof. The fact that this is a genuine norm follows from the subadditivity of the double-star operator,Proposition 2.7. Since f f, the inequality

fLm,q fL(m,q)

is trivial. The converse inequality follows from the Hardy inequality (Theorem 1.1) ifq 0.

3.5.Remark. We will not prove here that Lm,q is not norm form < qand L1,q is not equivalentto any norm forq >1. Instead of this, we will demonstrate this phenomenon on a particular caseq= .Example of failure of the triangle inequality in Lm,((0, 2)) is the following: f(x) = min{1, x1/m},g(x) =f(2 x). Then f =g =f, (f+ g)(2) = 1 + 21/m. We have

fLm, =gLm, = 1, f+ gLm, 21/m(f+ g)(2) = 21/m(1 + 21/m) = 21/m + 1> 2.

For the space L1,, consider the function

fn(x) = 1

2n

n

i=1

1

|x kn |.

Thenfn belongs to the convex hull of the unit ball, but fn Clog n.

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3.6. Proposition (Lorentz norm via distribution).

fLm,q =m1/qss f

1/m (s)Lq( dss ) .

Proof. We consider the set

(4)m(f) :=

[r, s] (0, )2 :f(rm)> s}

={[r, s] (0, )2 : f(s)> rm

;

the equality follows from Observation 2.2a. Forq 0

r[f(rm)] = supt>0

t1/mf(t)

or as

sups>0

s[f(s)]1/m.

3.7. Theorem (Holder type inequality).

X |f g| d fLm,qgLm,q .(Recall thatp denotes the conjugated exponent to p, so that 1p +

1p = 1).

Proof. By the Hardy-Littlewood-Polya inequality (Theorem 2.4),X

|f g| d

0

t1/mf(t)t1/m

g(t)dt

t tt1/mf(t)Lq(dtt )tt

1/mg(t)Lq ( dtt )

=fLm,qgLm,q .

3.8. Theorem (Inclusions). Suppose that1 m,q, M,Q .

(a) Ifq < Q, thenfLm,Q CfLm,q .

(b) Ifm < M then (X) 1m fLm,q C(X) 1MfLM,Q .Proof. (a) We have

t 1m f(t)

q qm [f

(t)]q t0

sqm1 ds qm

t0

s qm1[f(s)]q ds qm

0

sqm1[f(s)]q ds,

which proves (a) for Q = . In the general case we refer to the previous estimate and obtain 0

t1m f(t)

Q dtt

=

0

t 1m f(t)

Qqt1m f(t)

q dtt

sups>0

s1m f(s)

Qq 0

t 1m f(t)

q dtt

qm

0

sqm1[f(s)]q ds

Qqq

0

s qm1[f(s)]q ds

=qm

Qq 1

0

sqm1[f(s)]q ds

Q/q.

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(b) In view of (a), it is enough to consider the case q= 1,Q = . We have 0

t1m1f(t) dt=

(X)0

t 1m

1M1t

1Mf(t) dt sup

s>0

s 1Mf(s)

(X)0

t1m

1M1 dt

(X) 1m 1M

1m

1M

sups>0

s 1Mf(s)

.

3.9. Remark. The inclusion LM,Q Lm,q fails if the measure of the space is infinite.

3.10. Lemma. Suppose that Ej are pairwise disjoint measurable subsets of X and f Lm,q(X), 1 q m. Then

j

f Ej

mLm,q fmLm,q .

Proof. We may assumeq < m. Denotefj =fEj . Then

j

(fj) f.

LetS= inf{s > 0 : f(s) = 0}. (S= iff is strictly positive everywhere.) Holders inequality yields S0

sq1(fj)q/m (s) ds

m/q S

0

sq1(fj)(s)fqm1 (s) ds

S0

sq1fqm (s) ds

mq 1

for each j N. Summing over j we obtain

mm/qj

f Ej

mLm,q =j

S0

sq1(fj)q/m (s) ds

m/q

S

0

sq1fqm (s) ds

mq 1

j S

0

sq1(fj)(s)fqm1 (s) ds

S

0

sq1fq/m (s) ds

m/q=mm/qfmLm,q .

3.11. Lemma (Absolute ontinuity of the norm). Let{Ek} be a sequence of measurable sets,

E1 E2, . . . , k

Ek

= 0.

Letu Lm,q(X), q s}

|f(x)| dx, s >0,

whereC depends only onn.

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Proof. Without loss of generality we may assume f0. We set C= 5n + 1. Then for each z {M f >Cs} there exists a ball Bz =B(z, rz) such that

Cs|Bz|

Bz

f(x) dx

Bz{f>s}

f(x) dx + s|Bz|,

so that(C 1)s|Bz|

Bz{f>s}

f(x) dx.

By a standard Vitali type covering argument we find a sequence {Bj} of pairwise disjoint balls Bj =B(zj , rj) {Bz : z {Mf > Cs}}such that {Mf > Cs}

jB(zj , 5rj). We obtain

s(M f)(Cs) =s|{Mf > Cs}| j

|B(zj , 5rj)| 5nj

|Bj| 5n

(C 1)

j

Bj

f(x) dx

{f>s}

f(x) dx.

4.2. Lemma. Suppose thatfL1

loc(Rn

). Then{Mf>s}

|f(x)| dx C s|{Mf > s}| s > 0,

whereC depends only onn.

Proof. Without loss of generality we may assume that f 0. The set G= {Mf > s} is open, we mayassume that |G|< . For each z G there exists a ball Bz =B(z, rz) such that 5rz = dist(z, R

n \ G).This means that there exists a point y = yz Gsuch that |z y|= 5rz, which implies that

B(z, 5rz) B(y, 10rz).

Hence B(z,5rz)

f(x) dx

B(y,10rz)

f(x) dx M f(y)|B(y, 10rz)| 10n s|B(z, rz)|.

By a standard Vitali type covering argument we find a sequence {Bj} of pairwise disjoint balls Bj =B(zj , rj) {Bz : z G}such that G

jB(zj , 5rj). We obtain

G

f(x) dxj

B(zj ,5rj)

f(x) dx C sj

B(zj , rj) C s|G|.

4.3. Theorem (HerzRiesz inequality). Suppose thatfL1loc(Rn). Then

Cf(t) (Mf)(t) C f(t), t >0,

whereC, C >0 depends only onn.

Proof. Without loss of generality we may assume that f 0. We will use Lemma 4.1 with the sameconstantC. We have

f(t) (Mf)

Cf(t)

{f>f(t)}

f(x) dx

{f>f(t)}

f(x) dx

t0

f() d=t f(t).

ThusCf(t) {s 0 : (Mf)(s) t},

which yields(Mf)(t) C f(t).

To prove the converse inequality, we use Lemma 4.2 with s = (Mf)(t) and denote G = {M f > s},F = Rn \ G= {M fs}. Then obviously |G|= |{Mf >(Mf)(t)}| t and fF M fF s a.e. onRn. (Thus also (fF)

s and (fF) s.) We have

(5) t(fG)(t) =

t0

(fG)() d =

G

f(x) dx C s|G| C st

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and

(6) t(fF)(t) ts.

Since the double star operator is subadditive (Proposition 2.7), adding together (5) and (6) we obtain

tf

(t) C ts= C t(Mf)

(t),which concludes the proof.

4.4. Theorem (Boundedness of the maximal operator). IffLm,q(Rn), 1< m < , then

M fLm,q CfLm,q .

Proof. By Theorem 4.3 and Theorem 3.4

M fLm,q CfL(m,q) CfLm,q .

4.5. Proposition (Regularization). Let1< m


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which is the desired estimate. Now, in the non-smooth case, we aproximateu by standard mollificationsand observe that the estimate shows that the j uis a uniform Cauchy sequence. It follows that thereis a continuous representative for which the estimate (8) holds.

5.3. Corollary. Suppose that u Ln,1() and B = B(y, r) be a ball. Then for the continuousrepresentative ofu

oscBu CuBLn,1 ,

where we denote

oscEu = diam(u(E)).

Proof. Forz, z B we infer from Theorem 5.2

|u(z) u(z)| u(z)

B

u+ u(z)

B

u CuBLn,1 .

5.4. Theorem. Ifu Ln,1(), then the continuous representative ofu is a.e. differentiable.

Proof. We denote

lipf(x) = lim supyx

|f(y) f(x)|

|x y| .

We will estimate the measure of the set

E = {x : lipf(x)> }.

Ifx E, by the oscillation estimate we obtain that there exists a radius rx (0, 1) such that

(9) CoscB(x,rx) fr

Cr

uB(x,rx)Ln,1

By the Vitali type covering theorem, we find a disjointed sequence B(xj , rj) of balls from {B(x, rx) : x E} such that

E j=1

B(xj , 5rj).

Using (9) and Lemma 3.10 we obtain

|E| Cj=1

rnj Cj=1

nuB(xj ,rj)nLn,1

C

nunLn,1

It follows that

|{x : lipf(x) = +}|= 0.

By the Rademacher-Stepanov theorem, the functionu is differentiable a.e.

5.5.Theorem. Ifn >1 andu Ln,1(Rn), then there existsc R such thatu c C0(Rn). Moreover,

u c CuLn,1 .

Proof. Ifz, z /B(0, 2R), then there is a chain of 3 balls in Rn \ B(0, R) which connects z and z, andthus

oscRn\B(0,2R)

u CuLn,1(

Rn

\B(0,R))

and the expression on the right tends to zero as R by Lemma 3.11.

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6. Embedding into the Lorentz space

6.1.Median. We say that a valuec is a median of a function u in a measurable set M(with 0< (M) c}| 1

2|M|, |M {u < c}|

1

2|M|.

6.2. Proposition (Potential estimate). Let u be a Sobolev function with a zero median in a ball Bcontainingz. Suppose thatz is a Lebesgue point foru. Then

(10) u(z) C

B

|x z|1n|u(x)| dx

withC depending only onn.

Proof. The proof follows the standard proofs of similar potential estimates.

6.3. Proposition (Weak type estimate). Suppose thatu W1,1(B) and0 is a median of u inB. Lets > 0. Then

(11) s|{u > s}|11n C

B

|u(x)| dx,


Proof. We denote by G the set {x B ; |u(x)| s}. Then, by Proposition 6.2, Lemma 5.1 and Fubinistheorem

s |G|

G

|u(x)| dx C

G

B

|u(y)|

|x y|n1dy

dx

C

B

|u(y)|

G

dx

|x y|n1

dy C|G|

1n

B

|u(y)| dy.

Thus

s |G|11n C

B

|u(y)| dy.

6.4.Lemma. Suppose thatu W1,1(B)and0 is a median ofu inB . Let0 a b


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Proof. Case q p. We settk = 2k|B|, use Lemma 6.4 with

ak =u(tk), bk = u

(tk+1), k= 0, 1, . . . .

Write

Ek ={ak < u < bk}.Since

(13) tk 2tk+1 2|{u bk}|, |{u > ak}| tk,

we obtain

(14)

u(tk+1) u(tk)

qtq qnk C t

q1k

Ek

|u|q dx.

Let

g=k

tqp1

k

Ek.

Ifq < p, then

g(t

q

p1

k ) =|{g > t

q

p1

k }| |{u > ak}| tk

and thus

g(tk) tqp1

k , k= 0, 1, 2, ,

which implies

(15) g(t) C tqp1, t >0.

If q = p, then g 1 and thus (15) holds as well. Applying (15) and the Hardy-Littlewood-Polyainequality (Theorem 2.4) to f=|u|q andg we obtain

(16)

k=0

tqp1

k

Ek

|u|q dx=

B

f(x)g(x) dx C

0

f(t)g(t) dt

C 0

tqp1[|u|(t)]q dt

=uqLp,q .

Now, we have

(17)

uqp,q =

0

tqp

qn1[u(t)]q dt

Ck

tktk+1

tqp

qn1[u(t)]q dt

C

ktqp

qn

k [u(tk)]

q.

Given


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Casep < q p, measurable f and disjoint measurable sets Ek we havek

fEkqLp,q

fqLp,q .

Caseq= . Lets > p. Then we find t0> 0 such that

uLp, = supt>0

t1/p

u(t) 21/st1/p

0 u(t0).

Setb = u(t0), a= u(2t0). Then

(2t0)1/pa uLp, 2

1/st1/p0 b

and thus

a 21s

1p b, which meansb C(b a).

Then we have

bt1 1n0 C(b a)t

1 1n0 C

{a


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By Lemma 6.4 (with q= 1), for each i 1, 2 . . . we have

ii

|v(s)| ds=

ii

s1/n

(u)(s) ds (2i)1/n

ii

(u)(s) ds

(2i)1/n

[u

(i) u

(i)]

C

Ei

|u(x)| dx.

Summing over i and applying the HardyLittlewoodPolya inequality and (20) we obtain

F

|v(s)| dsi

ii

|v(s)| ds

C

E

|u(x)| dx

C

0

(E)(s)|u|(s) ds

C

t0

|u|(s) ds= t|u|(t).

Passing to supremum with respect to all measurable sets F (0, ) with |F|< t we obtain

tv(t) C t|u|(t).

7. The critical case

In this section we assume n 2.

7.1. Theorem (Mazya, Hansson, Brezis & Wainger). LetB Rn be a ball of measure 1. Suppose thatu W1,1(B) and0 is a median ofu inB. Then

(21)

10

ulog et

n dtt

Cunn


Proof. We may assume that u is bounded. We set tk = 2k+1, use Lemma 6.4 with

ak =u(tk), bk =u

(tk+1), k= 1, 2 . . . .

Sincetk 2tk+1 2|{u bk}|, |{u > ak}| tk,

we obtain

(22)

u(tk+1) u(tk)

nC

{ak


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Hence (passing to limit form )

(23)

k=1

ankkn

Ck=1

ank

1kn1

1

(k+ 1)n1

C

k=1

ank+1 ank

(k+ 1)n1

Ck=1

an1

k+1(ak+1 ak)(k+ 1)n1

C k=1

ak+1 ak

n 1n k=1

ank+1(k+ 1)n

1 1n.

Recalling thata1= 0, we infer from (22) and (23) thatk=1

u(tk)n

kn =

k=1

ankkn

Ck=1

ak+1 ak

nC

B

|u|n dx,

a discrete version of (21).

7.2. Theorem (Trudinger inequality). Let B Rn be a ball of measure 1. Suppose thatu W1,1(B)

and0 is a median ofu inB. Then there exists >0 depending only onn such that

(24)

B

e( |u|un

)n

dx 2.

Proof. We use (21) to estimate

(u(t))n

logn1(e/t)(n 1)(u(t))n

t0

ds

slogn(e/s)(n 1)

t0

(u(s))nds

slogn(e/s)

Cunn.

Thus there exists >0 such that

u(t)

un

n log(e/t),

where (0, 1) is so small that e

/(1 )< 2. We obtainB

e( |u|un

)n

dx=

10

e|u(t)|un

ndt

10

et

dt 2.

8. Sobolev inequalities

8.1.Theorem(Global inequality on the entire space). Suppose thatu W1,1loc(Rn)andu Lp,q, where

1 p < n and1 q . Then there exists a constantc R such thatu c Lp,q(Rn). Moreover

u cLp,q CuLp,q

withC depending only onp, q andn.

Proof. Letck be medians ofu on Bk :=B(0, 2k). Set

uk =

u ck inBk,

0 outside Bk

Then by Theorems 6.5 and 3.8

(25) ukqLp,q

CuqLp,q(Bk) CuqLp,q(Rn)

.

For the constants we have

|ck1 ck| C

Bk

|u ck| dx C

Bk

|u ck|p dx

1/pC2kn/p

ukLp,q

C2kn/p

uLp,q(Rn).

Hence the sequence ck has a limit c. We may assume that c = 0. Then uk u a.e. and thusu lim inf

k(uk).

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By the Fatou lemma, from (25) it easily follows

uqLp,q CuqLp,q

8.2. Theorem (Sobolev-Poincare inequality). Suppose that B = B(z, r) is a ball in Rn, p, q 1 areexponents andu W1,1(B). If

(26) 1

q

1

p

1

n,

then

(27)

B

|u uB|q1/q

C r

B

|u|p1/p

,

whereuB is the average ofu onB andC depends only onn, p andq.

Proof. We may assume that u has zero median in B. Suppose first that there is an equality in (26).Then by Theorem 6.5 and Theorem 3.8

B

|u|q dx1/q

=rn/q

B

|u|q dx1/q

C rn/q

B

|u|p dx1/p

=C rnq+

np

B

|u|p dx1/p

=C r

B

|u|p dx1/p

If there is not equality in (26), we find an exponent s [1, n) (the case n = 1 is left to the reader) suchthats < p and s > q. Then using Holder inequality we obtain

B

|u|q dx1/q

B

|u|s

dx1/s

C r

B

|u|s dx1/s

C r

B

|u|p dx1/p

We have

|uB|=B

u dx

B

|u|q dx1/q

and thus

B

|u uB|q dx

1/q2

B

|u|q dx1/q

which allows us to obtain the desired left hand side.

8.3. Remarks on Sobolev-Poincare inequalities. 1. The inequalities are stated in the scale ofLebesgue spaces, the Lorentz space versions are of course also available.

2. Instead of balls we may consider scaled copies of a fixed convex domain D1. Then the inequalitiesholds for integrating over sets of type

D= {x: (x z)/r D1}

and the constant Cdepends in addition on the parameter (diam D1)n/|D1|. The place where the shape

of the domain plays a role is the estimate (10), where the convexity ofD is essential for the method ofthe proof.

3. We say that is a q , p-Poincare domain, if (27) holds with in place ofB . There are much morePoincare domains than convex ones. The problem which domains are Poincare domains is not simple.

8.4. Theorem (Embedding W1,p(Rn) Lq()). Suppose that u Lp(Rn) W1,ploc(Rn) and u

Lp(Rn). Suppose that ||< and1

q

1

p

1

n.

Then

||1/q

|u|q dx1/q

C||1/p

Rn

|u|p dx1/p

+ C||1n

1p

Rn

|u|p dx1/p

.

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16/18

Proof. We may assume that||= 1, otherwise we make the rescaling v(y) =r u(y/r), wherer is such anumber thatrn =||. Further we may asume that

Rn

|u|p dx

1/p

+

Rn

|u|p dx

1/p

1.

We may assume that q > p, otherwise the inequality follows directly from the Holder inequality on without using the term with the gradient. We find an exponents [1, n) (the case n= 1 is left to thereader) such that s p and s q. We may assume that u 0. Denote

v= min{u, 1}, w= u v.

Thenw 0 and

|{w >0}|= |{u >1}|

Rn

up dx 1.

Then by Theorem 8.1 there exist constants C >0 andc R; Cdepending only on n and s,c dependingalso on u, such that

1

CRn

|w c|s

dx

1/s

Rn

|w|s dx

1/s

=

{w>0}

|w|s dx

1/s

{w>0}

|w|p dx

1/p

Rn

|u|p dx1/p

1.

and sincew c= ceverywhere except on the set {w >0}of measure 1, the convergence of the integralon the left implies thatc = 0. It follows

(28)

wq dx1/q

ws

dx1/s

C.

On the other hand, since 0 v 1 and q > p,

(29)

vq dx

Rn

vq dx

Rn

vp dx

Rn

up dx 1.

From (28) and (29) we obtain the assertion.

8.5. Remarks on Sobolev embedding. 1. The embedding of typeuq C||

1q

1p+

1n up

can be obtained for u W1,p0 () by the chain

||1q+

1p

1n uq C(TE u)Lp(Rn) CuLp(),

where TE: W1,p0 ()W1,p(Rn) is the trivial extension operator by zero.

2. We say that is a W1,p-extension domain if there exists a W1,p-extension operator, this means abounded linear operator E: W1,p()W1,p(Rn) such that E u= u on for each u W1,p(). If E isaW1,p-extension operator, then the embedding

uq Cu1,p

can be obtained by the chain

uq C E uW1,p(Rn) CuW1,p().

9. Compact embedding

9.1. Lemma. Suppose that ||< and1 q < a be exponents. Let{uk} be a sequence of measurablefunctions bounded inLa(). Suppose that there is a functionu such thatuk u a.e. Thenukuq 0.

Proof. By the Fatou lemma, u Lq(). Now we may assume that u = 0. Let Ek = {|uk| > 1}. Wesplituk =vk+ wk, wherevk = uk\Ek is the good part and wk =uk

Ek

is the bad part. We have

vk 1 and vk 0 a.e., hence by the Lebesgue theorem

(30)

|vk|q dx 0.

For the bad part, we denote Sk =jk

Ej ,

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and observe thatS1 S2 . . . and that

kSk is contained in the null set where {uk} diverges. Hence|Ek| |Sk| 0 and

(31)

|wk|q dx=

Ek

|uk|q dx

|uk|a dx

qa

|Ek|1 qa 0.

From (30) and (31) we obtain the assertion.

9.2.Lemma. Suppose that{uk}is a bounded sequence inW1,p(Rn). Then there exists a subsequence of

uk which converges a.e.

Proof. Since the matter is local, we may assume that that {uk} is bounded in W1,10 (B) for a fixed ball

B (otherwise we multiply with a cut-off function). Let 1 be a standard mollification kernel and >0.Recall the notation

(x) =n1(x/).

Then for each k N

(32)

Rn

|( uk)(x) uk(x)| dx

Rn

Rn

(x y)|uk(y) uk(x)| dy

dx

C nRn

B(x,)

|uk(y) uk(x)| dy

dx

C

Rn

B(x,)

|uk(y)|

|y x|n1dy

dx

=C

Rn

B(y,)

|uk(y)|

|y x|n1dx

dy

C

Rn

|uk(y)| dy.

There exists q (1, ) so that {uk} is locally bounded in Lq(B). This implies that there exists a

subsequence (labelled again asuk) such thatuk converges to some functionu weakly in Lq, in particular

uk

u pointwise.

Now, by Lemma 9.1, uk uk in L1(B). Using (32) we obtainB

|uk uj | dx

B

|uk uk| dx +

B

| uk uj| dx +

B

| uj uj| dx

B

| uk uj | dx + 2Csupi

Rn

|ui| dx

It follows that uk u in L1(B), and, passing if necessaty to a subsequence, uk u a.e. in B .

9.3. Examples. 1. Let be a nontrivial nonnegative smooth function with support in B(0, 1) andrk 0. Let p, q [1, ) be exponents. Set

uk(x) =r1npk (x/rk).

Then Rn

(|uk|p + |uk|

p) dx=

Rn

(rpk||p + |p) dy

and thus the sequence{uk} is bounded in W1,p(Rn). Since

Rn

(|uk|q dx= r

n+qnqpk

Rn

||q dy,

the only chance for a sequence ukj to converge to 0 in Lq is if

1

q >

1

p

1

n.

2. Let , p, q be as above. Suppose that contains an infinite sequence{Bk} of mutually disjointcopies of the unit ball, Bk = B(zk, 1). Then the sequence {vk}, vk(x) = (zk +x), is bounded inW1,p(Rn) and converges to 0 a.e. Nevertheless, none of its subsequences can converge to 0. Therefore,for compactness we assume that the target domain is of finite measure.

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9.4. Theorem (Rellich, Kondrashev). Suppose that ||< and that1 p, q 1p

1n . Then

the embeddingW1,p(Rn)Lq() is compact.

Proof. Let uk be a sequence bounded in W1,p(Rn). We find an exponent s [1, n) (the case n = 1 is

left to the reader) such that s p and s > q. Then, by the embedding theorem (Theorem 8.4, the

sequenceuk is bounded in Ls

(). By Lemma 9.2, a subsequence ofuk (labelled again as uk) convergesto a function u pointwise in . By Lemma 9.1, uk u in L

q().

10. Notes

The Hardy inequality is from 1920 and 1926, see [6]. The method of rearrangement has been sys-tematically developed by Hardy, Littlewood and Polya [6]. For Theorem 2.4 see Hardy, Littlewood andPolya [6]. The Lorentz spaces were introduced in [10], [11]. The reader can find more information onLorentz spaces and their history e.g. in books [1], [16], [21], [24].

The inequalities of Theorem 4.3 are due to F. Riesz [18] (upper estimate) and Herz [7] (lower estimate).The borderline case of the embedding into continuous functions and differentiability a.e. goes back toCalderon [3] and Stein [20], see also [4], [8]. The Sobolev type embedding in the setting of Lorentz spacesis due to ONeil [14] and Peetre [15], see also Tartar [22]. Our proof follows [12]. Theorem 6.7 is from [4].

Theorem 7.1 as it is stated here has been discovered indendently by Hansson [5] and Brezis and Wainger[2], the idea goes back to Mazja [13]. Our proof follows [12]. The Trudinger inequality is from [23]. Theoriginal reference to Sobolev inequalities is [19]. The compact embedding theorem is due to Rellich [17]and Kondrachov [9].

References

[1] C. Bennett and R. Sharpley.Interpolation of operators, volume 129 ofPure and Applied Mathematics. Academic PressInc., Boston, MA, 1988.

[2] H. Brezis and S. Wainger. A note on limiting cases of Sobolev embeddings and convolution inequalities.Comm. PartialDifferential Equations, 5(7):773789, 1980.

[3] A. P. Calderon. On the differentiability of absolutely continuous functions. Rivista Mat. Univ. Parma, 2:203213,1951.

[4] A. Cianchi and L. Pick. Sobolev embeddings into BMO, VMO, and L. Ark. Mat., 36(2):317340, 1998.[5] K. Hansson. Imbedding theorems of Sobolev type in potential theory. Math. Scand., 45(1):77102, 1979.

[6] G. H. Hardy, J. E. Littlewood, and G. Polya.Inequalities. Cambridge, at the University Press, 1952. 2d ed.[7] C. Herz. The HardyLittlewood maximal theorem. InSymposium on Harmonic Analysis. University of Warwick, 1968.[8] J. Kauhanen, P. Koskela, and J. Maly. On functions with derivatives in a Lorentz space.Manuscripta Math., 100(1):87

101, 1999.[9] W. Kondrachov. Sur certaines proprietes des fonctions dans lespace. C. R. (Doklady) Acad. Sci. URSS (N. S.),

48:535538, 1945.[10] G. G. Lorentz. Some new functional spaces.Ann. of Math. (2), 51:3755, 1950.[11] G. G. Lorentz. On the theory of spaces .Pacific J. Math., 1:411429, 1951.[12] J. Maly and L. Pick. An elementary proof of sharp Sobolev embeddings. Proc. Amer. Math. Soc., 130(2):555563

(electronic), 2002.[13] V. G. Mazja. Sobolev spaces. Springer Series in Soviet Mathematics. Springer-Verlag, Berlin, 1985. Translated from

the Russian by T. O. Shaposhnikova.[14] R. ONeil. Convolution operators and L(p, q) spaces. Duke Math. J., 30:129142, 1963.[15] J. Peetre. Espaces dinterpolation et theoreme de Soboleff. Ann. Inst. Fourier (Grenoble), 16(fasc. 1):279317, 1966.[16] L. Pick, A. Kufner, O. John, and S. Fuck. Function spaces. Vol. 1, volume 14 ofDe Gruyter Series in Nonlinear

Analysis and Applications. Walter de Gruyter & Co., Berlin, extended edition, 2013.[17] F. Rellich. Ein Satz uber mittlere Konvergenz. Nachr. Akad. Wiss. Gottingen Math.Phys., K1:3035, 1930.[18] F. Riesz. Sur un theoreme de maximum de MM. Hardy and Littlewood. J. London Math. Soc., 7:1013, 1932.[19] S. L. Sobolev. Applications of functional analysis in mathematical physics. Translated from the Russian by F. E.

Browder. Translations of Mathematical Monographs, Vol. 7. American Mathematical Society, Providence, R.I., 1963.[20] E. M. Stein. Editors note: the differentiability of functions in Rn. Ann. of Math. (2), 113(2):383385, 1981.[21] E. M. Stein and G. Weiss.Introduction to Fourier analysis on Euclidean spaces . Princeton University Press, Princeton,

N.J., 1971. Princeton Mathematical Series, No. 32.[22] L. Tartar. Imbedding theorems of Sobolev spaces into Lorentz spaces.Boll. Unione Mat. Ital. Sez. B Artic. Ric. Mat.

(8), 1(3):479500, 1998.[23] N. S. Trudinger. On imbeddings into Orlicz spaces and some applications.J. Math. Mech., 17:473483, 1967.[24] W. P. Ziemer.Weakly differentiable functions, volume 120 of Graduate Texts in Mathematics. Springer-Verlag, New

York, 1989. Sobolev spaces and functions of bounded variation.

Charles University, Faculty of Mathematics and Physics KMA, Sokolovska 83, 18675 Praha 8, Czech

Republic

E-mail address: [email protected]

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Advanced Diff Theory - JAN MALY