Heisenberg Uniqueness Pairs for the Parabola

Abstract

Let Γ denote the parabola y=x ² in the plane. For some simple sets Λ in the plane we study the question whether (Γ,Λ) is a Heisenberg uniqueness pair. For example we shall consider the cases where Λ is a straight line or a union of two straight lines.

1 Introduction

Let μ denote a finite complex-valued Borel measure in ℝ². The Fourier transform of μ is defined by

$$\hat{\mu}(x,y)=\int_{\mathbb {R}^2} e^{-i(x\xi+y\eta)} d\mu(\xi,\eta) \quad\text {for $(x,y)\in \mathbb {R}^{2}$.} $$

Let Γ denote the parabola y=x ² in ℝ². We assume that $\operatorname {supp}\mu\subset\varGamma$ and that μ is absolutely continuous with respect to the arc length measure on Γ. Also let Λ be a subset of ℝ². Following Hedenmalm and Montes-Rodríguez [3] we say that (Γ,Λ) is a Heisenberg uniqueness pair (or only uniqueness pair) if $\hat{\mu}(x,y)=0$ for (x,y)∈Λ implies that μ is the zero measure.

The case where Γ is a hyperbola was discussed in [3], and Sjölin [5] and Lev [4] studied the case where Γ is a circle. For further results see also Canto-Martin, Hedenmalm, and Montes-Rodríguez [1].

We shall here let Γ denote the parabola y=x ².

If μ has the above properties it is clear that there exists a measurable function f on ℝ such that

$$\int_\mathbb {R}\bigl|f(t) \bigr|\sqrt{1+4t^2} dt < \infty $$

and $\int_{\mathbb {R}^{2}} h d\mu= \int_{\mathbb {R}}h(t,t^{2}) f(t) \sqrt{1+4t^{2}} dt$ if h is continuous and bounded in ℝ². Thus we have

$$\hat{\mu}(x,y)=\int_\mathbb {R}e^{-i(xt+yt^2)} f(t) \sqrt{1+4t^2} dt, \quad (x,y)\in \mathbb {R}^2. $$

We shall prove the following theorem, where Γ denotes the parabola y=x ².

Theorem 1

• (i) Let Λ=L where L is a straight line. Then (Γ,Λ) is a uniqueness pair if and only if L is parallel to the x-axis.

• (ii) Let Λ=L ₁∪L ₂, where L ₁ and L ₂ are different straight lines. Then (Γ,Λ) is a uniqueness pair.

• (iii) Assume that L ₁ and L ₂ are different straight lines, which are not parallel to the x-axis. Also assume that E ₁⊂L ₁, and E ₂⊂L ₂ and that E ₁ and E ₂ have positive one-dimensional Lebesgue measure. Set Λ=E ₁∪E ₂. Then (Γ,Λ) is a uniqueness pair.

Remark

When we talk about the one-dimensional Lebesgue measure of a subset E of a straight line L in the plane, we identify L with ℝ.

We also remark that Heisenberg uniqueness pairs are somewhat related to the notion of annihilating pairs (see Havin and Jöricke [2]). To give the definition of this concept we let S and Σ be subsets of ℝ. Following [2] we say that the pair (S,Σ) is mutually annihilating if ψ∈L ²(ℝ), $\operatorname {supp}\psi\subset S$, $\operatorname {supp}\hat{\psi }\subset\varSigma$ implies that ψ=0. Here $\hat{\psi}$ denotes the Fourier transform of ψ. We refer to [2] for results on annihilating pairs.

2 Lemmas and Proofs

We let the function f be defined as in the Introduction and set

$$g(t)=f(t)\sqrt{1+4t^2}, \quad t\in \mathbb {R}. $$

Then g∈L ¹(ℝ) and

$$ \hat{\mu}(x,y)=\int_\mathbb {R}e^{-i(xt+yt^2)} g(t) dt, \quad(x,y)\in \mathbb {R}^2. $$

(1)

We also set $Tg(x,y)=\hat{\mu}(x,y)$ so that T is a mapping from L ¹(ℝ) to C(ℝ²). It is then easy to see that if we set U(x,y)=Tg(x,−y), then U satisfies the Schrödinger equation

$$i\frac{\partial U}{\partial y}=\frac{\partial ^2 U}{\partial x^2}. $$

We shall use the following theorem, which can be found in Havin and Jöricke [2], p. 36.

Theorem A

Assume that φ∈L ¹(ℝ) and that $\operatorname {supp}\varphi$ is a subset of [0,∞). Also assume that

$$\int_\mathbb {R}\log \bigl|\hat{\varphi }(x) \bigr| \frac{dx}{1+x^2} = -\infty. $$

Then φ=0 almost everywhere.

Here $\hat{\varphi }$ denotes the Fourier transform of φ.

We shall then state and prove three lemmas. We let |E| denote Lebesgue measure of a set E.

Lemma 1

Assume that g∈L ¹(ℝ) and that E⊂ℝ with |E|>0. Then

$$ \int_\mathbb {R}e^{-iyt^2} g(t) dt = 0 \quad \text{\textit{for every} $y\in E$} $$

(2)

if and only if g is an odd function.

Proof

It is obvious that (2) holds if g is odd. We have to prove the converse and assume that (2) holds. Denoting the integral in (2) by I and performing a change of variable we obtain

where F(t)=g(t)+g(−t) for t≥0. It is clear that F∈L ¹(0,∞) and setting u=t ² we obtain

$$I=\int_0^\infty e^{-iyu} F(\sqrt{u}) \frac{1}{2\sqrt{u}} du = 0 \quad \text{for $y\in E$}. $$

We have $F(\sqrt{u})/\sqrt{u}\in L^{1}(0,\infty)$ and we set $\varphi (u)=F(\sqrt{u})\frac{1}{2\sqrt{u}}$ for u>0 and φ(u)=0 for u≤0. It follows that φ∈L ¹(ℝ) and $I=\hat{\varphi }(y)=0$ for y∈E. Since |E|>0 we conclude that

$$\int_\mathbb {R}\log \bigl|\hat{\varphi }(x) \bigr|\frac{dx}{1+x^2} = -\infty $$

and Theorem A implies that φ=0 almost everywhere. Hence F(t)=0 almost everywhere on [0,∞). It follows that g(−t)=−g(t) for almost every t, i.e. g is odd. □

Lemma 2

Let g∈L ¹(ℝ) and let γ and δ denote different real numbers. Assume that g(u−γ) and g(u−δ) are odd as functions of u. Then g=0 almost everywhere.

Proof

Using first the fact that g(u−γ) is odd and then the fact that g(u−δ) is odd we obtain

$$g(u-\gamma)=-g(-u-\gamma)=-g\bigl((-u+\delta-\gamma)-\delta\bigr)=g(u-\delta + \gamma-\delta)=g(u+\gamma-2\delta) $$

for almost every u. Hence g(u)=g(u+2γ−2δ), that is g has period 2γ−2δ≠0. Since g∈L ¹(ℝ) it follows that g=0 almost everywhere. □

We shall need one more lemma.

Lemma 3

Assume that g∈L ¹(ℝ) and that $\hat{\mu}$ is given by (1). Also assume that E⊂ℝ and |E|>0. Then the following holds.

• (i) Assume x ₀∈ℝ. Then $\hat{\mu}(x_{0},y)=0$ for y∈E if and only if $e^{-i x_{0} t}g(t)$ is odd as a function of t.

• (ii) Assume α∈ℝ and α≠0. Then $\hat{\mu}(x,\alpha x)=0$ for x∈E if and only if g(u−1/2α) is odd as a function of u.

• (iii) Assume that α and b are real numbers and α≠0 and b≠0. Then $\hat{\mu}(x,\alpha x+b)=0$ for x∈E if and only if the function $h(t)=e^{-ibt^{2}}g(t)$ has the property that h(u−1/2α) is odd as a function of u.

Proof

To prove (i) we invoke Lemma 1 and observe that

$$\hat{\mu}(x_0,y)=\int_\mathbb {R}e^{-iyt^2}e^{-ix_0 t} g(t) dt = 0 \quad\text {for $y\in E$} $$

if and only if $e^{-ix_{0}t}g(t)$ is an odd function.

To obtain (ii) we write

$$\hat{\mu}(x,\alpha x)=\int_\mathbb {R}e^{-i(xt+\alpha x t^2)} g(t) dt = \int_\mathbb {R}e^{-ix(t+\alpha t^2)} g(t) dt. $$

However,

$$t+\alpha t^2=\alpha \biggl(t^2+\frac{1}{\alpha}t \biggr)=\alpha \biggl[ \biggl(t+\frac{1}{2\alpha} \biggr)^2- \frac{1}{4\alpha^2} \biggr]=\alpha \biggl(t+\frac{1}{2\alpha} \biggr)^2- \frac{1}{4\alpha} $$

and hence

$$\hat{\mu}(x,\alpha x)=e^{ix/4\alpha}\int_\mathbb {R}e^{-ix\alpha(t+1/2\alpha )^2}g(t) dt. $$

Setting u=t+1/2α we then obtain

$$\hat{\mu}(x,\alpha x)=e^{ix/4\alpha}\int_\mathbb {R}e^{-ix\alpha u^2} g(u-1/2\alpha) du $$

and (ii) follows from an application of Lemma 1.

It remains to prove (iii). We have

where $h(t)=e^{-ibt^{2}}g(t)$, and (iii) follows from (ii). Thus Lemma 3 has been proved. □

Before proving Theorem 1 we mention the well-known fact that if Λ is a subset of ℝ² and Λ ₁ is a translate of Λ, then (Γ,Λ) is a uniqueness pair if and only if (Γ,Λ ₁) is a uniqueness pair. This follows from elementary properties of the Fourier transform.

Finally we shall prove Theorem 1.

Proof of Theorem 1

We first prove (i) and assume that L is parallel to the x-axis. Using the above remark we may assume that L is the x-axis. We assume that g∈L ¹(ℝ) and that

$$\hat{\mu}(x,0)=\int_\mathbb {R}e^{-ixt} g(t) dt = 0 \quad \text{for every $x$.} $$

Hence $\hat{g}(x)=0$ everywhere and we conclude that g=0 almost everywhere. It follows that (Γ,L) is a uniqueness pair.

We then assume that L is not parallel to the x-axis. It then follows directly from Lemma 3 that (Γ,L) is not a uniqueness pair.

For example, if L is also not parallel to the y-axis, we may assume that L is the line y=αx where α≠0. Then take φ as a non-zero odd function in L ¹(ℝ) and set g(t)=φ(t+1/2α). It then follows from (ii) in Lemma 3 that $\hat{\mu}=Tg$ vanish on L and thus (Γ,L) is not a uniqueness pair.

Thus we have proved (i). We then observe that (ii) follows from (i) and (iii), and therefore it only remains to prove (iii). We suppose that L ₁,L ₂,E ₁,E ₂ and Λ have the properties in the statement of (iii) and shall prove that (Γ,Λ) is a uniqueness pair.

We first study the case where L ₁ and L ₂ intersect. Performing a translation we may assume that the point of intersection is the origin. First assume that L ₁ and L ₂ are the lines y=α ₁ x and y=α ₂ x, where α ₁ and α ₂≠0. We assume that $\hat{\mu}$ is given by (1) and that $\hat{\mu }$ vanishes on Λ=E ₁∪E ₂. It then follows from (ii) in Lemma 3 that g(u−1/2α ₁) and g(u−1/2α ₂) are odd. Lemma 2 then implies that g=0 almost everywhere.

We then treat the case where L ₁ is the y-axis and L ₂ is the line y=αx with α≠0. Assuming that $\hat{\mu}$ vanishes on Λ we conclude from (i) and (ii) in Lemma 3 that g and g(u−1/2α) are odd. We then invoke Lemma 2 to conclude that g=0 almost everywhere.

It remains to study the case where L ₁ and L ₂ are parallel lines. First suppose that these lines are parallel to the y-axis. We may assume that L ₁ is the line x=0 and L ₂ the line x=x ₀ where x ₀≠0. We also assume that $\hat{\mu}$ is given by (1) and that $\hat {\mu}$ vanishes on Λ=E ₁∪E ₂. It follows from (i) in Lemma 3 that the functions g and $e^{-ix_{0}t}g(t)$ are odd. Hence

$$g(-t)=-g(t) \quad\text{and}\quad e^{ix_0t}g(-t)=-e^{-ix_0t}g(t). $$

We conclude that

$$-e^{ix_0t}g(t)=-e^{-ix_0t}g(t) $$

and

$$e^{i2x_0t}g(t)=g(t). $$

We obtain

$$\bigl(e^{i2x_0t}-1\bigr)g(t)=0 $$

for almost every t. It is clear that $e^{i2x_{0}t}=1$ only for t=πn/x ₀ where n is an integer. We conclude that g=0 almost everywhere.

We shall finally study the case where L ₁ and L ₂ are two parallel lines which are not parallel to the coordinate axes. We may assume that L ₁ is the line y=αx and L ₂ the line y=αx+b where α≠0 and b≠0. We assume again that $\hat{\mu}$ is given by (1) and that $\hat {\mu}$ vanishes on Λ. According to (ii) in Lemma 3 it follows that g(u−1/2α) is odd. Setting $h(t)=e^{-ibt^{2}}g(t)$ we also conclude from (iii) in Lemma 3 that h(u−1/2α) is odd. Setting γ=1/2α we then have g(−u−γ)=−g(u−γ) and h(−u−γ)=−h(u−γ) for almost every u.

The above equality for h can be written

$$e^{-ib(u+\gamma)^2} g(-u-\gamma)=-e^{-ib(u-\gamma)^2}g(u-\gamma). $$

Using the fact that g(u−γ) is odd we obtain

$$-e^{-ib(u+\gamma)^2}g(u-\gamma)=-e^{-ib(u-\gamma)^2}g(u-\gamma) $$

and

$$e^{ib(u-\gamma)^2-ib(u+\gamma)^2} g(u-\gamma) = g(u-\gamma). $$

Hence

$$\bigl(e^{ib(u-\gamma)^2-ib(u+\gamma)^2} - 1\bigr)g(u-\gamma) = 0 $$

and

$$\bigl(e^{-ib4u\gamma}-1\bigr)g(u-\gamma)=0. $$

It is clear that e ^−ib4uγ=1 only for u=2πn/4bγ, where n is an integer, and we conclude that g=0 almost everywhere. The proof of the theorem is complete. □