The principle of concentration-compactness and an application. - - PowerPoint PPT Presentation

The principle of concentration-compactness and an application.

Alexis Drouot September 3rd 2015

Plan.

Plan.

◮ The principle of concentration compactness.

Plan.

◮ The principle of concentration compactness. ◮ An application to the study of radial extremizing sequences for

the Radon transform.

Context.

Context.

◮ Let T bounded from X to Y : |Tf |Y ≤ A|f |X for some

minimal constant A.

Context.

◮ Let T bounded from X to Y : |Tf |Y ≤ A|f |X for some

minimal constant A.

◮ A natural question is: are there functions realizing the

equality?

Context.

◮ Let T bounded from X to Y : |Tf |Y ≤ A|f |X for some

minimal constant A.

◮ A natural question is: are there functions realizing the

equality?

◮ A natural approach is: take a sequence with |fn|X = 1 and

|Tfn|Y → A and show that fn converges in X.

Context.

◮ Let T bounded from X to Y : |Tf |Y ≤ A|f |X for some

minimal constant A.

◮ A natural question is: are there functions realizing the

equality?

◮ A natural approach is: take a sequence with |fn|X = 1 and

|Tfn|Y → A and show that fn converges in X.

◮ Problem: many interesting operators arise from physics and

have many symmetries.

Context.

◮ Let T bounded from X to Y : |Tf |Y ≤ A|f |X for some

minimal constant A.

◮ A natural question is: are there functions realizing the

equality?

◮ A natural approach is: take a sequence with |fn|X = 1 and

|Tfn|Y → A and show that fn converges in X.

◮ Problem: many interesting operators arise from physics and

have many symmetries.

◮ Non compact groups of symmetries are hard to fight.

The concentration compactness principle.

The concentration compactness principle.

Lemma

Let φn ≥ 0 on Rd with |φn|1 = 1. Then there exists a subsequence

• f φn, still noted φn with one of the following:

The concentration compactness principle.

Lemma

Let φn ≥ 0 on Rd with |φn|1 = 1. Then there exists a subsequence

• f φn, still noted φn with one of the following:

◮ (tightness) There exists yn ∈ Rd such that uniformly in n,

lim

R→∞

• B(yn,R)

φn = 1.

◮ (vanishing) For all R,

lim

n→∞ sup y∈Rd

• B(y,R)

φn = 0.

◮ (dichotomy) There exist 0 < α < 1 and φn ≥ φ1 n, φ2 n ≥ 0

with d(supp(φ1

n), supp(φ2 n)) → ∞ and

|φn − φ1

n − φ2 n|1 → 0,

|φ1

n|1 → α,

|φ2

n|1 → 1 − α.

Qualitative description.

Let φn satisfying the assumptions of the Lemma. Then one of the following happens:

Qualitative description.

Let φn satisfying the assumptions of the Lemma. Then one of the following happens:

◮ (tightness) There exists yn ∈ Rd such that uniformly in n,

lim

R→∞

• B(yn,R)

φn = 1.

Qualitative description.

Let φn satisfying the assumptions of the Lemma. Then one of the following happens:

◮ (tightness) There exists yn ∈ Rd such that uniformly in n,

lim

R→∞

• B(yn,R)

φn = 1. φn is mostly supported on a ball of radius R whose center yn moves around.

Qualitative description.

Let φn satisfying the assumptions of the Lemma. Then one of the following happens:

Qualitative description.

Let φn satisfying the assumptions of the Lemma. Then one of the following happens:

◮ (vanishing) For all R,

lim

n→∞ sup y∈Rd

• B(y,R)

φn = 0.

Qualitative description.

Let φn satisfying the assumptions of the Lemma. Then one of the following happens:

◮ (vanishing) For all R,

lim

n→∞ sup y∈Rd

• B(y,R)

φn = 0. φn is not really concentrated anywhere and somehow dissipates.

Qualitative description.

Let φn satisfying the assumptions of the Lemma. Then one of the following happens:

Qualitative description.

Let φn satisfying the assumptions of the Lemma. Then one of the following happens:

◮ (dichotomy) There exist 0 < α < 1 and φn ≥ φ1 n, φ2 n ≥ 0

with d(supp(φ1

n), supp(φ2 n)) → ∞ and

|φn − φ1

n − φ2 n|1 → 0,

|φ1

n|1 → α,

|φ2

n|1 → 1 − α.

Qualitative description.

Let φn satisfying the assumptions of the Lemma. Then one of the following happens:

◮ (dichotomy) There exist 0 < α < 1 and φn ≥ φ1 n, φ2 n ≥ 0

with d(supp(φ1

n), supp(φ2 n)) → ∞ and

|φn − φ1

n − φ2 n|1 → 0,

|φ1

n|1 → α,

|φ2

n|1 → 1 − α.

φn splits into two parts that get further and further from each other.

The Radon transform for radial functions in dimension 3

The Radon transform for radial functions in dimension 3

◮ The Radon transform for radial functions takes the form

T f (r) = ∞

r

f (u)udu.

The Radon transform for radial functions in dimension 3

◮ The Radon transform for radial functions takes the form

T f (r) = ∞

r

f (u)udu.

◮ T is continuous Lp(R+, u2du) → L4(R+, dr), p = 4/3.

The Radon transform for radial functions in dimension 3

◮ The Radon transform for radial functions takes the form

T f (r) = ∞

r

f (u)udu.

◮ T is continuous Lp(R+, u2du) → L4(R+, dr), p = 4/3. ◮ Goal: Prove that (radial) extremizing sequences for T

converge modulo the group of dilations.

The Radon transform for radial functions in dimension 3

◮ The Radon transform for radial functions takes the form

T f (r) = ∞

r

f (u)udu.

◮ T is continuous Lp(R+, u2du) → L4(R+, dr), p = 4/3. ◮ Goal: Prove that (radial) extremizing sequences for T

converge modulo the group of dilations.

◮ Remark: this is a very weak form of a result of Christ:

extremizing sequences for the Radon transform converge modulo the group of affine maps.

The Radon transform for radial functions in dimension 3

◮ The Radon transform for radial functions takes the form

T f (r) = ∞

r

f (u)udu.

◮ T is continuous Lp(R+, u2du) → L4(R+, dr), p = 4/3. ◮ Goal: Prove that (radial) extremizing sequences for T

converge modulo the group of dilations.

◮ Remark: this is a very weak form of a result of Christ:

extremizing sequences for the Radon transform converge modulo the group of affine maps. But the goal is to apply the concentration compactness principle in a simple setting.

The case of f ∗

n .

The case of f ∗

n .

◮ Fix now 0 ≤ fn with |fn|p = 1 and |T fn|4 → |T |.

The case of f ∗

n .

◮ Fix now 0 ≤ fn with |fn|p = 1 and |T fn|4 → |T |. ◮ If f ∗ is the nonincreasing rearrangement of f then

|T f ∗|4 ≥ |T f |4.

The case of f ∗

n .

◮ Fix now 0 ≤ fn with |fn|p = 1 and |T fn|4 → |T |. ◮ If f ∗ is the nonincreasing rearrangement of f then

|T f ∗|4 ≥ |T f |4.

◮ Thus f ∗ n is a nonincreasing extremizing sequence.

The case of f ∗

n .

◮ Fix now 0 ≤ fn with |fn|p = 1 and |T fn|4 → |T |. ◮ If f ∗ is the nonincreasing rearrangement of f then

|T f ∗|4 ≥ |T f |4.

◮ Thus f ∗ n is a nonincreasing extremizing sequence. ◮ Using some refined weak form inequalities there exists a

rescaling of f ∗

n (still called f ∗ n ) so that f ∗ n converges weakly to

a non-zero function.

The case of f ∗

n .

◮ Fix now 0 ≤ fn with |fn|p = 1 and |T fn|4 → |T |. ◮ If f ∗ is the nonincreasing rearrangement of f then

|T f ∗|4 ≥ |T f |4.

◮ Thus f ∗ n is a nonincreasing extremizing sequence. ◮ Using some refined weak form inequalities there exists a

rescaling of f ∗

n (still called f ∗ n ) so that f ∗ n converges weakly to

a non-zero function.

◮ Using Lieb’s lemma f ∗ n converges in Lp.

What about fn?

What about fn?

◮ Rescale f ∗ n so that it converges and use the same rescaling for

fn.

What about fn?

◮ Rescale f ∗ n so that it converges and use the same rescaling for

fn.

◮ f ∗ n and fn have the same distribution function.

What about fn?

◮ Rescale f ∗ n so that it converges and use the same rescaling for

fn.

◮ f ∗ n and fn have the same distribution function. ◮ Since f ∗ n converges there exists a set En with fn ≥ 1En and

|En| ∼ 1.

What about fn?

◮ Rescale f ∗ n so that it converges and use the same rescaling for

fn.

◮ f ∗ n and fn have the same distribution function. ◮ Since f ∗ n converges there exists a set En with fn ≥ 1En and

|En| ∼ 1.

◮ So vanishing does not occur!

The set En does not move away from 0.

The set En does not move away from 0.

◮ Assume that En has a large part Fn with d(Fn, 0) → ∞.

The set En does not move away from 0.

◮ Assume that En has a large part Fn with d(Fn, 0) → ∞. ◮ There are not so many planes that have a big intersection

with Fn.

The set En does not move away from 0.

◮ Assume that En has a large part Fn with d(Fn, 0) → ∞. ◮ There are not so many planes that have a big intersection

with Fn.

◮ Hence T fn cannot be so big.

The set En does not move away from 0.

◮ Assume that En has a large part Fn with d(Fn, 0) → ∞. ◮ There are not so many planes that have a big intersection

with Fn.

◮ Hence T fn cannot be so big. ◮ Then fn cannot be an extremizing sequence.

The set En does not move away from 0.

◮ Assume that En has a large part Fn with d(Fn, 0) → ∞. ◮ There are not so many planes that have a big intersection

with Fn.

◮ Hence T fn cannot be so big. ◮ Then fn cannot be an extremizing sequence. ◮ Thus En mainly remains close from 0.

fn converges weakly to a non-zero function.

fn converges weakly to a non-zero function.

◮ En remains in a ball B(0, R).

fn converges weakly to a non-zero function.

◮ En remains in a ball B(0, R). ◮ fn converges weakly to f . Can f be zero?

fn converges weakly to a non-zero function.

◮ En remains in a ball B(0, R). ◮ fn converges weakly to f . Can f be zero? ◮ |En| ≤

• fn1B(0,R) →
• f .

fn converges weakly to a non-zero function.

◮ En remains in a ball B(0, R). ◮ fn converges weakly to f . Can f be zero? ◮ |En| ≤

• fn1B(0,R) →
• f .

◮ Thus the weak limit of fn is non-zero.

fn does not blow up like a Dirac mass.

fn does not blow up like a Dirac mass.

◮ f ∗ n converge in Lp.

fn does not blow up like a Dirac mass.

◮ f ∗ n converge in Lp. ◮ f ∗ n and fn have the same distribution function.

fn does not blow up like a Dirac mass.

◮ f ∗ n converge in Lp. ◮ f ∗ n and fn have the same distribution function. ◮ Hence we have.

0 = lim

R→∞

• f ∗

n ≥R

|f ∗

n |p = lim R→∞

• fn≥R

|fn|p uniformly in R.

fn does not blow up like a Dirac mass.

◮ f ∗ n converge in Lp. ◮ f ∗ n and fn have the same distribution function. ◮ Hence we have.

0 = lim

R→∞

• f ∗

n ≥R

|f ∗

n |p = lim R→∞

• fn≥R

|fn|p uniformly in R.

◮ Thus fn cannot blow up like a Dirac mass.

Break.

Break.

◮ We ruled out vanishing.

Break.

◮ We ruled out vanishing. ◮ We proved that En remains close from 0. Thus if tightness

• ccurs then yn (the center of the ball) is 0.

Break.

◮ We ruled out vanishing. ◮ We proved that En remains close from 0. Thus if tightness

• ccurs then yn (the center of the ball) is 0.

◮ Blow-up like a Dirac mass cannot occur. Thus think

about fn as somehow uniformly locally in Lp.

Break.

◮ We ruled out vanishing. ◮ We proved that En remains close from 0. Thus if tightness

• ccurs then yn (the center of the ball) is 0.

◮ Blow-up like a Dirac mass cannot occur. Thus think

about fn as somehow uniformly locally in Lp.

◮ Usual process now: construct E 1 n , E 2 n , ... where fn is

• concentrated. Instead we follow a concentration

compactness argument.

Break.

◮ We ruled out vanishing. ◮ We proved that En remains close from 0. Thus if tightness

• ccurs then yn (the center of the ball) is 0.

◮ Blow-up like a Dirac mass cannot occur. Thus think

about fn as somehow uniformly locally in Lp.

◮ Usual process now: construct E 1 n , E 2 n , ... where fn is

• concentrated. Instead we follow a concentration

compactness argument.

◮ What’s the ennemy now?

Dichotomy

Dichotomy

◮ (dichotomy) There exist 0 < α < 1 and φn ≥ φ1 n, φ2 n ≥ 0

with d(supp(φ1

n), supp(φ2 n)) → ∞ and

|φn − φ1

n − φ2 n|1 → 0,

|φ1

n|1 → α,

|φ2

n|1 → 1 − α.

Dichotomy

◮ (dichotomy) There exist 0 < α < 1 and φn ≥ φ1 n, φ2 n ≥ 0

with d(supp(φ1

n), supp(φ2 n)) → ∞ and

|φn − φ1

n − φ2 n|1 → 0,

|φ1

n|1 → α,

|φ2

n|1 → 1 − α. ◮ Here φn = |fn|p.

Dichotomy

◮ (dichotomy) There exist 0 < α < 1 and φn ≥ φ1 n, φ2 n ≥ 0

with d(supp(φ1

n), supp(φ2 n)) → ∞ and

|φn − φ1

n − φ2 n|1 → 0,

|φ1

n|1 → α,

|φ2

n|1 → 1 − α. ◮ Here φn = |fn|p. ◮ Assume dichotomy happens and write fn ≡ f 1 n + f 2 n with

d(supp(f 1

n ), supp(f 2 n )) → ∞ and

||fn|p − |f 1

n |p − |f 2 n |p|1 → 0,

|f 1

n |p → α,

|f 2

n |p → 1 − α.

Dichotomy

◮ (dichotomy) There exist 0 < α < 1 and φn ≥ φ1 n, φ2 n ≥ 0

with d(supp(φ1

n), supp(φ2 n)) → ∞ and

|φn − φ1

n − φ2 n|1 → 0,

|φ1

n|1 → α,

|φ2

n|1 → 1 − α. ◮ Here φn = |fn|p. ◮ Assume dichotomy happens and write fn ≡ f 1 n + f 2 n with

d(supp(f 1

n ), supp(f 2 n )) → ∞ and

||fn|p − |f 1

n |p − |f 2 n |p|1 → 0,

|f 1

n |p → α,

|f 2

n |p → 1 − α. ◮ We can assume the support of f 1 n remains close from 0.

Weak interaction.

Weak interaction.

◮ The support of f 1 n , f 2 n get further and further from each other.

Weak interaction.

◮ The support of f 1 n , f 2 n get further and further from each other. ◮ Counts how many planes interset both the support of f 1 n and

the support of f 2

n .

Weak interaction.

◮ The support of f 1 n , f 2 n get further and further from each other. ◮ Counts how many planes interset both the support of f 1 n and

the support of f 2

n . ◮ There are not so many!

Weak interaction.

◮ The support of f 1 n , f 2 n get further and further from each other. ◮ Counts how many planes interset both the support of f 1 n and

the support of f 2

n . ◮ There are not so many! ◮ Thus if T f 1 n is relatively big, then T f 2 n is really small and

• conversely. The interaction between T f 1

n and T f 2 n is weak.

Weak interaction.

◮ The support of f 1 n , f 2 n get further and further from each other. ◮ Counts how many planes interset both the support of f 1 n and

the support of f 2

n . ◮ There are not so many! ◮ Thus if T f 1 n is relatively big, then T f 2 n is really small and

• conversely. The interaction between T f 1

n and T f 2 n is weak. ◮ Thus |T fn|4 4 ≡ |T f 1 n |4 4 + |T f 2 n |4 4.

Ruling out dichotomy.

Ruling out dichotomy.

◮ Define

Sα = sup{|T f |4

4, |f |p p = α}.

Ruling out dichotomy.

◮ Define

Sα = sup{|T f |4

4, |f |p p = α}. ◮ By convexity relations S1 > Sα + S1−α for every α ∈ (0, 1).

Ruling out dichotomy.

◮ Define

Sα = sup{|T f |4

4, |f |p p = α}. ◮ By convexity relations S1 > Sα + S1−α for every α ∈ (0, 1). ◮ Using the previous slide,

Sα + S1−α ≥ |T f 1

n |4 4 + |T f 2 n |4 4 ≡ |T fn|4 4 ≡ S1.

Ruling out dichotomy.

◮ Define

Sα = sup{|T f |4

4, |f |p p = α}. ◮ By convexity relations S1 > Sα + S1−α for every α ∈ (0, 1). ◮ Using the previous slide,

Sα + S1−α ≥ |T f 1

n |4 4 + |T f 2 n |4 4 ≡ |T fn|4 4 ≡ S1. ◮ Thus S1 ≤ Sα + S1−α. Contradiction.

Ruling out dichotomy.

◮ Define

Sα = sup{|T f |4

4, |f |p p = α}. ◮ By convexity relations S1 > Sα + S1−α for every α ∈ (0, 1). ◮ Using the previous slide,

Sα + S1−α ≥ |T f 1

n |4 4 + |T f 2 n |4 4 ≡ |T fn|4 4 ≡ S1. ◮ Thus S1 ≤ Sα + S1−α. Contradiction. ◮ Hence dichotomy does not occur.

Tightness is winning.

Tightness is winning.

◮ By Lions’ lemma, tightness occurs with yn = 0.

Tightness is winning.

◮ By Lions’ lemma, tightness occurs with yn = 0. That is:

lim

R→∞

• B(0,R)

|fn|p = 1 uniformly in n.

Tightness is winning.

◮ By Lions’ lemma, tightness occurs with yn = 0. That is:

lim

R→∞

• B(0,R)

|fn|p = 1 uniformly in n.

◮ That is fn ≡ 1B(0,R)fn for R large enough.

Tightness is winning.

◮ By Lions’ lemma, tightness occurs with yn = 0. That is:

lim

R→∞

• B(0,R)

|fn|p = 1 uniformly in n.

◮ That is fn ≡ 1B(0,R)fn for R large enough. ◮ The truncated operator T 1B(0,R) is compact.

Tightness is winning.

◮ By Lions’ lemma, tightness occurs with yn = 0. That is:

lim

R→∞

• B(0,R)

|fn|p = 1 uniformly in n.

◮ That is fn ≡ 1B(0,R)fn for R large enough. ◮ The truncated operator T 1B(0,R) is compact. ◮ Thus T fn ≡ T 1B(0,R)fn → g.

Tightness is winning.

◮ By Lions’ lemma, tightness occurs with yn = 0. That is:

lim

R→∞

• B(0,R)

|fn|p = 1 uniformly in n.

◮ That is fn ≡ 1B(0,R)fn for R large enough. ◮ The truncated operator T 1B(0,R) is compact. ◮ Thus T fn ≡ T 1B(0,R)fn → g. ◮ If fn ⇀ f then |f |p ≤ 1.

Tightness is winning.

◮ By Lions’ lemma, tightness occurs with yn = 0. That is:

lim

R→∞

• B(0,R)

|fn|p = 1 uniformly in n.

◮ That is fn ≡ 1B(0,R)fn for R large enough. ◮ The truncated operator T 1B(0,R) is compact. ◮ Thus T fn ≡ T 1B(0,R)fn → g. ◮ If fn ⇀ f then |f |p ≤ 1. g must be T f and have maximal

norm among functions of the form T h with |h|p ≤ 1.

Tightness is winning.

◮ By Lions’ lemma, tightness occurs with yn = 0. That is:

lim

R→∞

• B(0,R)

|fn|p = 1 uniformly in n.

◮ That is fn ≡ 1B(0,R)fn for R large enough. ◮ The truncated operator T 1B(0,R) is compact. ◮ Thus T fn ≡ T 1B(0,R)fn → g. ◮ If fn ⇀ f then |f |p ≤ 1. g must be T f and have maximal

norm among functions of the form T h with |h|p ≤ 1. Thus f is an extremizer.

Tightness is winning.

◮ By Lions’ lemma, tightness occurs with yn = 0. That is:

lim

R→∞

• B(0,R)

|fn|p = 1 uniformly in n.

◮ That is fn ≡ 1B(0,R)fn for R large enough. ◮ The truncated operator T 1B(0,R) is compact. ◮ Thus T fn ≡ T 1B(0,R)fn → g. ◮ If fn ⇀ f then |f |p ≤ 1. g must be T f and have maximal

norm among functions of the form T h with |h|p ≤ 1. Thus f is an extremizer.

◮ Thus |f |p = 1 and then fn → f .

Conclusions.

Conclusions.

◮ In dimension 3, radial extremizing sequences for the

Radon transform converge modulo dilations.

Conclusions.

◮ In dimension 3, radial extremizing sequences for the

Radon transform converge modulo dilations.

◮ The proof uses a concentration compactness argument

rather than an iteration argument.

Conclusions.

◮ In dimension 3, radial extremizing sequences for the

Radon transform converge modulo dilations.

◮ The proof uses a concentration compactness argument

rather than an iteration argument.

◮ It works for all the endpoint inequalities for the k-plane

transform restricted to radial functions.

Possible extension.

Possible extension.

◮ Generalize this proof to the k-plane transform inequalities

with no radial restriction, i.e. prove that the extremizing sequences converge modulo the affine group.

Possible extension.

◮ Generalize this proof to the k-plane transform inequalities

with no radial restriction, i.e. prove that the extremizing sequences converge modulo the affine group.

◮ This was proved by Christ for k = d − 1.

Possible extension.

◮ Generalize this proof to the k-plane transform inequalities

with no radial restriction, i.e. prove that the extremizing sequences converge modulo the affine group.

◮ This was proved by Christ for k = d − 1. ◮ This is much harder than this talk because of the size of the

group of symmetries.

Thanks for your attention!