Heisenberg Uniqueness pairs

Heisenberg Uniqueness pairs Philippe Jaming Bordeaux Fourier Workshop 2013, Renyi Institute Joint work with K. Kellay

Heisenberg Uniqueness Pairs µ : finite measure on R 2 µ(x, y) = R 2 e i(sx+ty) dµ(s, t). Γ : finite union of disjoint curves M(Γ) : measures supported on Γ AC(Γ) : µ M(Γ) absolutely continuous w.r.t arc length. Λ R 2 : set of lines Definition (Γ, Λ) Heisenberg Uniqueness Pair (HUP) if µ AC(Γ), µ = 0 on Λ µ = 0.

References H. HEDENMALM & A. MONTES-RODRÍGUEZ Heisenberg uniqueness pairs and the Klein-Gordon equation. Ann. of Math. (2), 173 (2011), 1507-1527. N. LEV Uniqueness theorems for Fourier transforms. Bull. Sci. Math., 135 (2011), 134-140. P. SJÖLIN Heisenberg uniqueness pairs and a theorem of Beurling and Malliavin. Bull. Sci. Math., 135 (2011), 125-133. P. SJÖLIN Heisenberg uniqueness pairs for the parabolla. Jour. Four. Anal. Appl., to appear.

Aim Aim Get rid of explicit computation get a hint on the geometric nature of the problem Unify proofs.

HUP 2 Inv 1 Fix (s 0, t 0 ), (x 0, y 0 ) R 2. Then ( Γ, Λ ) is a HUP if and only if ( Γ (s0, t 0 ), Λ (x 0, y 0 ) ) is a HUP. Inv 2 Fix T a linear invertible transformation R 2 R 2 and denote by T its adjoint. Then ( Γ, Λ ) is a HUP if and only if ( T 1 (Γ), T (Λ) ) is a HUP. If Γ = {s, γ(s), s I} (Γ, λ) HUP f L 1 (I) s.t. (x, y) Λ f (s)e i(sx+γ(s)y) ds = 0 f = 0 I

Basic lemma γ(s)

Basic lemma γ(s) t 3 t 1 t 4 t 2 s 1 s 2 s 3 s 4 s

Basic lemma γ(s) t 3 t 1 t t 4 t 2 γ 1 0 s 1 s 2 s 3 s 4 (t) γ 1 1 (t) γ 1 2 (t) γ 1 3 (t) γ 1 s 4 (t)

Basic lemma R f (s)e iγ(s)x ds =

Basic lemma R f (s)e iγ(s)x ds = m sk+1 k=0 s k f (s)e iγ(s)x ds

Basic lemma R f (s)e iγ(s)x ds = = m sk+1 k=0 s k tk+1 m k=0 t k f (s)e iγ(s)x ds ( f γ 1 k (t) ) γ ( γ 1 k (t) )eitx dt

Basic lemma R f (s)e iγ(s)x ds = = = m sk+1 k=0 s k tk+1 m k=0 t k max tk min t k m k=0 f (s)e iγ(s)x ds ( f γ 1 k (t) ) γ ( γ 1 k (t) )eitx dt 1 [tk,t k+1 ](t) f ( γ 1 k (t) ) γ ( γ 1 k (t) )eitx dt.

Basic lemma R f (s)e iγ(s)x ds = = = m sk+1 k=0 s k tk+1 m k=0 t k max tk min t k m k=0 f (s)e iγ(s)x ds ( f γ 1 k (t) ) γ ( γ 1 k (t) )eitx dt 1 [tk,t k+1 ](t) f ( γ 1 k (t) ) γ ( γ 1 k (t) )eitx dt. Thus (Γ, {x = 0}) HUP

Basic lemma R f (s)e iγ(s)x ds = = = Thus (Γ, {x = 0}) HUP m sk+1 k=0 s k tk+1 m k=0 t k max tk min t k m k=0 m k=0 f (s)e iγ(s)x ds ( f γ 1 k (t) ) γ ( γ 1 k (t) )eitx dt 1 [tk,t k+1 ](t) f ( γ 1 k (t) ) γ ( γ 1 k (t) )eitx dt. 1 [tk,t k+1 ](t) f ( γ 1 k (t) ) γ ( γ 1 k (t) ) = 0.

Example 1 γ is one-to-one, then (Γ, {x = 0}) HUP

Example 2 γ has one local extrema.

Example 2 γ has one local extrema. t γ 1 (t) γ 1 + (t)

Example 2 γ has one local extrema. t γ 1 (t) γ 1 + (t) f ( γ+ 1 (t)) γ ( γ+ 1 (t)) = f ( γ 1 (t)) γ ( γ 1 (t)).

Lemma f ( γ+ 1 (t)) γ ( γ+ 1 (t)) = f ( γ 1 (t)) γ ( γ 1 (t)). ( γ(α+ ) ), β = γ 1 ( γ(β+ ) ), and for α +, β + > s 1 and α = γ 1 we have β β+ f (s) ds = f (s) ds α + α

Theorem : two lines γ smooth, a > b, ψ(t) = γ(t) + at and χ(t) = γ(t) + bt ψ (resp. χ) have a unique local minimum in s 1 (resp s 2 ) ψ(t), χ(t) + when t ±. f L 1 (R) s.t. f (s)e iψ(s)x ds = f (s)e iχ(s)x ds = 0 then f = 0. R R

Proof 1

Proof 1 σ 0 σ 1

Proof 1 ψ 1 ( ψ(σ1 ) ) ψ(σ 1 ) σ 0 σ 1

Proof 1 ( χ ψ 1 ( ψ(σ1 ) )) ψ(σ 1 ) ψ 1 ( ψ(σ1 ) ) σ 0 σ 1 σ 2 σ k +.

Proof 2 If f 0, σ k+1 σ k f (s) ds 0 σk+2 + 0 σ k+1 f (s) ds = f (s) ds = k 0 σk+1 σ k σk+1 σ k f (s) ds. f (s) ds = +.

A better picture t = γ(s) σ 2 σ 0 σ 1 t = as

Hyperbola, case 1 t = γ(s) σ 2 σ 1 σ 0 σ 1 σ 0 t = as σ 2 t = γ(s)

Hyperbola, case 2 t = γ(s) σ 1 σ 0 σ 1

Hyperbola, case 3 t = γ(s) J 2 I 2 I 2 I 1 J 1 J 1 J 2 I 1

Closed curve

One local max, one local min

That s all! Thank you for your attention!