The principle of concentration-compactness and an application.

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 f n X = 1 and Tf n Y A and show that f n 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 f n X = 1 and Tf n Y A and show that f n 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 f n X = 1 and Tf n Y A and show that f n 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 φ n 0 on R d with φ n 1 = 1. Then there exists a subsequence of φ n, still noted φ n with one of the following:

The concentration compactness principle. Lemma Let φ n 0 on R d with φ n 1 = 1. Then there exists a subsequence of φ n, still noted φ n with one of the following: (tightness) There exists y n R d such that uniformly in n, lim φ n = 1. R B(y n,r) (vanishing) For all R, lim sup φ n = 0. n y R d B(y,R) (dichotomy) There exist 0 < α < 1 and φ n φ 1 n, φ 2 n 0 with d(supp(φ 1 n), supp(φ 2 n)) and φ n φ 1 n φ 2 n 1 0, φ 1 n 1 α, φ 2 n 1 1 α.

Qualitative description. Let φ n satisfying the assumptions of the Lemma. Then one of the following happens:

Qualitative description. Let φ n satisfying the assumptions of the Lemma. Then one of the following happens: (tightness) There exists y n R d such that uniformly in n, lim φ n = 1. R B(y n,r)

Qualitative description. Let φ n satisfying the assumptions of the Lemma. Then one of the following happens: (tightness) There exists y n R d such that uniformly in n, lim φ n = 1. R B(y n,r) φ n is mostly supported on a ball of radius R whose center y n moves around.

Qualitative description. Let φ n satisfying the assumptions of the Lemma. Then one of the following happens:

Qualitative description. Let φ n satisfying the assumptions of the Lemma. Then one of the following happens: (vanishing) For all R, lim sup φ n = 0. n y R d B(y,R)

Qualitative description. Let φ n satisfying the assumptions of the Lemma. Then one of the following happens: (vanishing) For all R, lim sup φ n = 0. n y R d B(y,R) φ n is not really concentrated anywhere and somehow dissipates.

Qualitative description. Let φ n satisfying the assumptions of the Lemma. Then one of the following happens:

Qualitative description. Let φ n satisfying the assumptions of the Lemma. Then one of the following happens: (dichotomy) There exist 0 < α < 1 and φ n φ 1 n, φ 2 n 0 with d(supp(φ 1 n), supp(φ 2 n)) and φ n φ 1 n φ 2 n 1 0, φ 1 n 1 α, φ 2 n 1 1 α.

Qualitative description. Let φ n satisfying the assumptions of the Lemma. Then one of the following happens: (dichotomy) There exist 0 < α < 1 and φ n φ 1 n, φ 2 n 0 with d(supp(φ 1 n), supp(φ 2 n)) and φ n φ 1 n φ 2 n 1 0, φ 1 n 1 α, φ 2 n 1 1 α. φ n 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 L p (R +, u 2 du) L 4 (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 L p (R +, u 2 du) L 4 (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 L p (R +, u 2 du) L 4 (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 L p (R +, u 2 du) L 4 (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 f n with f n p = 1 and T f n 4 T.

The case of f n. Fix now 0 f n with f n p = 1 and T f n 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 f n with f n p = 1 and T f n 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 f n with f n p = 1 and T f n 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 fn (still called fn ) so that fn converges weakly to a non-zero function.

The case of f n. Fix now 0 f n with f n p = 1 and T f n 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 fn (still called fn ) so that fn converges weakly to a non-zero function. Using Lieb s lemma f n converges in L p.

What about f n?

What about f n? Rescale fn f n. so that it converges and use the same rescaling for

What about f n? Rescale fn f n. f n so that it converges and use the same rescaling for and f n have the same distribution function.

What about f n? Rescale fn f n. f n so that it converges and use the same rescaling for and f n have the same distribution function. Since f n converges there exists a set E n with f n 1 En and E n 1.

What about f n? Rescale fn f n. f n so that it converges and use the same rescaling for and f n have the same distribution function. Since f n converges there exists a set E n with f n 1 En and E n 1. So vanishing does not occur!

The set E n does not move away from 0.

The set E n does not move away from 0. Assume that E n has a large part F n with d(f n, 0).

The set E n does not move away from 0. Assume that E n has a large part F n with d(f n, 0). There are not so many planes that have a big intersection with F n.

The set E n does not move away from 0. Assume that E n has a large part F n with d(f n, 0). There are not so many planes that have a big intersection with F n. Hence T f n cannot be so big.

The set E n does not move away from 0. Assume that E n has a large part F n with d(f n, 0). There are not so many planes that have a big intersection with F n. Hence T f n cannot be so big. Then f n cannot be an extremizing sequence.

The set E n does not move away from 0. Assume that E n has a large part F n with d(f n, 0). There are not so many planes that have a big intersection with F n. Hence T f n cannot be so big. Then f n cannot be an extremizing sequence. Thus E n mainly remains close from 0.

f n converges weakly to a non-zero function.

f n converges weakly to a non-zero function. E n remains in a ball B(0, R).

f n converges weakly to a non-zero function. E n remains in a ball B(0, R). f n converges weakly to f. Can f be zero?

f n converges weakly to a non-zero function. E n remains in a ball B(0, R). f n converges weakly to f. Can f be zero? E n f n 1 B(0,R) f.

f n converges weakly to a non-zero function. E n remains in a ball B(0, R). f n converges weakly to f. Can f be zero? E n f n 1 B(0,R) f. Thus the weak limit of f n is non-zero.

f n does not blow up like a Dirac mass.

f n does not blow up like a Dirac mass. f n converge in L p.

f n does not blow up like a Dirac mass. f f n n converge in L p. and f n have the same distribution function.

f n does not blow up like a Dirac mass. f f n n converge in L p. and f n have the same distribution function. Hence we have. uniformly in R. 0 = lim R f fn R n p = lim R f n R f n p

f n does not blow up like a Dirac mass. f f n n converge in L p. and f n have the same distribution function. Hence we have. uniformly in R. 0 = lim R f fn R n p = lim R f n R Thus f n cannot blow up like a Dirac mass. f n p

Break.

Break. We ruled out vanishing.

Break. We ruled out vanishing. We proved that E n remains close from 0. Thus if tightness occurs then y n (the center of the ball) is 0.

Break. We ruled out vanishing. We proved that E n remains close from 0. Thus if tightness occurs then y n (the center of the ball) is 0. Blow-up like a Dirac mass cannot occur. Thus think about f n as somehow uniformly locally in L p.

Break. We ruled out vanishing. We proved that E n remains close from 0. Thus if tightness occurs then y n (the center of the ball) is 0. Blow-up like a Dirac mass cannot occur. Thus think about f n as somehow uniformly locally in L p. Usual process now: construct E 1 n, E 2 n,... where f n is concentrated. Instead we follow a concentration compactness argument.

Break. We ruled out vanishing. We proved that E n remains close from 0. Thus if tightness occurs then y n (the center of the ball) is 0. Blow-up like a Dirac mass cannot occur. Thus think about f n as somehow uniformly locally in L p. Usual process now: construct E 1 n, E 2 n,... where f n is concentrated. Instead we follow a concentration compactness argument. What s the ennemy now?

Dichotomy

Dichotomy (dichotomy) There exist 0 < α < 1 and φ n φ 1 n, φ 2 n 0 with d(supp(φ 1 n), supp(φ 2 n)) and φ n φ 1 n φ 2 n 1 0, φ 1 n 1 α, φ 2 n 1 1 α.

Dichotomy (dichotomy) There exist 0 < α < 1 and φ n φ 1 n, φ 2 n 0 with d(supp(φ 1 n), supp(φ 2 n)) and φ n φ 1 n φ 2 n 1 0, φ 1 n 1 α, φ 2 n 1 1 α. Here φ n = f n p.

Dichotomy (dichotomy) There exist 0 < α < 1 and φ n φ 1 n, φ 2 n 0 with d(supp(φ 1 n), supp(φ 2 n)) and φ n φ 1 n φ 2 n 1 0, φ 1 n 1 α, φ 2 n 1 1 α. Here φ n = f n p. Assume dichotomy happens and write f n fn 1 + fn 2 with d(supp(fn 1 ), supp(fn 2 )) and f n 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 φ n φ 1 n, φ 2 n 0 with d(supp(φ 1 n), supp(φ 2 n)) and φ n φ 1 n φ 2 n 1 0, φ 1 n 1 α, φ 2 n 1 1 α. Here φ n = f n p. Assume dichotomy happens and write f n fn 1 + fn 2 with d(supp(fn 1 ), supp(fn 2 )) and f n 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 fn 1, fn 2 get further and further from each other.

Weak interaction. The support of fn 1, fn 2 get further and further from each other. Counts how many planes interset both the support of fn 1 and the support of fn 2.

Weak interaction. The support of fn 1, fn 2 get further and further from each other. Counts how many planes interset both the support of fn 1 and the support of fn 2. There are not so many!

Weak interaction. The support of fn 1, fn 2 get further and further from each other. Counts how many planes interset both the support of fn 1 and the support of fn 2. There are not so many! Thus if T fn 1 is relatively big, then T fn 2 is really small and conversely. The interaction between T fn 1 and T fn 2 is weak.

Weak interaction. The support of fn 1, fn 2 get further and further from each other. Counts how many planes interset both the support of fn 1 and the support of fn 2. There are not so many! Thus if T fn 1 is relatively big, then T fn 2 is really small and conversely. The interaction between T fn 1 and T fn 2 is weak. Thus T f n 4 4 T f n 1 4 4 + T f n 2 4 4.

Ruling out dichotomy.

Ruling out dichotomy. Define S α = sup{ T f 4 4, f p p = α}.

Ruling out dichotomy. Define S α = sup{ T f 4 4, f p p = α}. By convexity relations S 1 > S α + S 1 α for every α (0, 1).

Ruling out dichotomy. Define S α = sup{ T f 4 4, f p p = α}. By convexity relations S 1 > S α + S 1 α for every α (0, 1). Using the previous slide, S α + S 1 α T f 1 n 4 4 + T f 2 n 4 4 T f n 4 4 S 1.

Ruling out dichotomy. Define S α = sup{ T f 4 4, f p p = α}. By convexity relations S 1 > S α + S 1 α for every α (0, 1). Using the previous slide, S α + S 1 α T f 1 n 4 4 + T f 2 n 4 4 T f n 4 4 S 1. Thus S 1 S α + S 1 α. Contradiction.

Ruling out dichotomy. Define S α = sup{ T f 4 4, f p p = α}. By convexity relations S 1 > S α + S 1 α for every α (0, 1). Using the previous slide, S α + S 1 α T f 1 n 4 4 + T f 2 n 4 4 T f n 4 4 S 1. Thus S 1 S α + S 1 α. Contradiction. Hence dichotomy does not occur.

Tightness is winning.

Tightness is winning. By Lions lemma, tightness occurs with y n = 0.

Tightness is winning. By Lions lemma, tightness occurs with y n = 0. That is: lim f n p = 1 R B(0,R) uniformly in n.

Tightness is winning. By Lions lemma, tightness occurs with y n = 0. That is: lim f n p = 1 R B(0,R) uniformly in n. That is f n 1 B(0,R) f n for R large enough.

Tightness is winning. By Lions lemma, tightness occurs with y n = 0. That is: lim f n p = 1 R B(0,R) uniformly in n. That is f n 1 B(0,R) f n for R large enough. The truncated operator T 1 B(0,R) is compact.

Tightness is winning. By Lions lemma, tightness occurs with y n = 0. That is: lim f n p = 1 R B(0,R) uniformly in n. That is f n 1 B(0,R) f n for R large enough. The truncated operator T 1 B(0,R) is compact. Thus T f n T 1 B(0,R) f n g.

Tightness is winning. By Lions lemma, tightness occurs with y n = 0. That is: lim f n p = 1 R B(0,R) uniformly in n. That is f n 1 B(0,R) f n for R large enough. The truncated operator T 1 B(0,R) is compact. Thus T f n T 1 B(0,R) f n g. If f n f then f p 1.

Tightness is winning. By Lions lemma, tightness occurs with y n = 0. That is: lim f n p = 1 R B(0,R) uniformly in n. That is f n 1 B(0,R) f n for R large enough. The truncated operator T 1 B(0,R) is compact. Thus T f n T 1 B(0,R) f n g. If f n 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 y n = 0. That is: lim f n p = 1 R B(0,R) uniformly in n. That is f n 1 B(0,R) f n for R large enough. The truncated operator T 1 B(0,R) is compact. Thus T f n T 1 B(0,R) f n g. If f n 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 y n = 0. That is: lim f n p = 1 R B(0,R) uniformly in n. That is f n 1 B(0,R) f n for R large enough. The truncated operator T 1 B(0,R) is compact. Thus T f n T 1 B(0,R) f n g. If f n 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 f n 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!