A COUNTEREXAMPLE TO UNIQUE CONTINUATION IN DIMENSION TWO

Niculae Mandache

University of Loughborough, Department of Mathematical Sciences,

Loughborough, Leicestershire, LE11 3TU, UK. E-mail:

               N.P.Mandache@lboro.ac.uk

On leave from: Institute of Mathematics, Romanian Academy, P.O. Box 1-764, 70700 Bucharest, Romania. E-mail: mandache@stoilow.imar.ro

Abstract. We construct a non-zero solution u ∈ C_{0}^{∞}(ℂ) of the equation \overline{∂}u = V u for a certain V which belongs to L^{p} for any p < 2. The same is done in arbitrary dimension d ≥ 2 for the Laplace equation with a first order term Δv + W · ∇v = 0 and for the Dirac equation Dw + W w = 0, with W ∈ L^{p} for any p < 2. The construction is based on a Weierstrass product in the unit ball. Although its poles accumulate at the boundary, it is flat at the boundary if we remove small disjoint discs around the poles.

1. Introduction.

Let Ω ⊂ ℝ^{n} be a connected open set. We say that an equation P (x, ∂/∂x)u = 0 has the unique continuation property in Ω if any solution u which vanishes in a non-empty open subset vanishes identically. We are interested in the equation

1991 Mathematics Subject Classification: Primary 35A05, Secondary 35B60.

\overline{∂}u = V u in ℂ = ℝ^{2},  (1.1)

where \overline{∂} is the Cauchy-Riemann operator, \overline{∂} = \frac{1}{2}(∂/∂x_{1} + i∂/∂x_{2}) and V ∈ L_{loc}^{1}(ℂ). We will also consider here the Laplace equation with a first order term :

Δv + W · ∇v = 0,  (1.2)

and the Dirac equation:

Dw + W w = 0,  (1.3)

with the (zero mass) Dirac operator D = ∑_{i=1}^{d} α_{i}∂/∂x_{i}. The α_{i} are matrices of order m satisfying α_{i}^{∗} = −α_{i} and the Clifford commutation relations: α_{i}α_{j} + α_{j}α_{i} = −2δ_{jk}. m can be chosen 2^{[d/2]}.

There is extensive literature on unique continuation and strong unique continuation properties (in the later, one supposes the solution to vanish to infinite order at a point instead of vanishing on an open set), see [4]. Although the problem of strong unique continuation has been almost settled, it is a long standing problem whether equations (1.1), (1.2) and (1.3) have the unique continuation property when the coefficients are L_{loc}^{1}. We give here a negative answer, constructing solution with compact support for (1.1),(1.2) and (1.3) with coefficients in L^{p} for any p < 2. This is optimal in dimension 2. The best positive result belongs to Wolff [3]: in dimension d, if W ∈ L^{d} then (1.2) and (1.3) have the unique continuation property.

We fix the function F (t) = t^{2} ln^{−3}(t + 2) on [0, ∞). It has the property that F (at) ≤ a^{2}F (t) for a > 1 and F (t)/F̃(t) is bounded, F̃ being the greatest convex function with F̃(t) ≤ F (t) for any t ≥ 0. Therefore the set

L^{F} (ℝ^{d}) = {f : f measurable on ℝ^{d}, ∫ F (|f|)dx < ∞}

coincides with the Orlicz space associated with F̃ on (ℝ^{d}, dx) (see e.g. [1]). It is a Banach space such that L^{p} ⊂ L^{F} for p < 2.

Theorem 1.1. There are u, V which satisfy (1.1) such that u ∈ C_{0}^{∞}(ℂ), supp u = B(0, 1) and V ∈ L^{F} (ℂ). There are a real-valued v ∈ C_{0}^{∞}(ℂ) and W ∈ L^{F} (ℂ) satisfying (1.2), such that supp v = B(0, 1).

Corollary 1.2. For any d ≥ 2, there are non-zero smooth solutions v realvalued and w, for the equations (1.3) and (1.2) respectively, with compact support in ℝ^{d} and such that the coefficients W belong to L^{F} (ℝ^{d}).

Notations. C, C′, C′′ will stand for absolute positive constants, not necessarily the same in different formulae. The constants which will be carried from a formula to another will be numbered C_{1}, C_{2}, …. We write f(x) ∼ g(x) in M if there are absolute constants 0 < c < C such that cf(x) ≤ g(x) ≤ Cf(x) for all x ∈ M.

We use the variable z = x_{1} + ix_{2}, identifying ℂ with ℝ^{2}. If β ∈ ℕ^{2} is a multi-index, ∂^{β} = (∂/∂x_{1})^{β_{1}}(∂/∂x_{2})^{β_{2}} and |β| = β_{1} + β_{2}. B(a, r) is the ball of center a and radius r and B(a, r) is its interior. If x is a real number, we denote by [x] the greatest integer in (−∞, x].

2. Proofs.

We start by constructing a meromorphic function in the unit disk B(0, 1) as a Weierstrass product. It vanishes to infinite order at the boundary, in a sense which is made precise by Lemma 2.1 below:

f(z) = ∏_{j=2}^{∞} \frac{1}{1 − (\frac{z}{1−1/j} )^{j^{2}}} .  (2.1)

Let us denote the poles of f by

z_{k,l} = \frac{k − 1}{k} exp(2πil/k^{2}), for k ≥ 2, 0 ≤ l < k^{2}.

Then each of the sets

M_{k,l} = \{z : \begin{matrix}|z| ∈ [\frac{2k−3}{2k−1}, \frac{2k−1}{2k+1}) , \\ arg z ∈ [(2l − 1)π/k^{2}, (2l + 1)π/k^{2}) \end{matrix} \} , k ≥ 2, 0 ≤ l ≤ k^{2},

contains the corresponding pole of f. Moreover, these sets form a partition of the annulus {z : |z| ∈ [1/3, 1)}. We will also need slightly bigger sets:

M̃_{k,l} = \{z : \begin{matrix}|z| ∈ [\frac{4k−7}{4k−3}, \frac{4k−1}{4k+3}] , \\ arg z ∈ [(4l − 3)π/2k^{2}, (4l + 3)π/2k^{2}] \end{matrix} \} .

We have then

M_{k,l} + B(0, k^{−2}/8) ⊂ M̃_{k,l}.  (2.2)
B(z_{k,l}, k^{−2}/8) ⊂ M_{k,l}.  (2.3)

Lemma 2.1. The infinite product (2.1) converges to a meromorphic function in the unit open disk, and it converges to 0 elsewhere. Let

g_{k,l}(z) = f(z)(z − z_{k,l})

be the analytic function in M_{k,l} obtained by multiplicatively removing the pole in this region. Then the following estimate holds:

|g_{k,l}(z)| ∼ k^{−2} ∏_{j=2}^{k−1}(1 − 1/j)^{j^{2}}|z|^{−k^{3}/3+k^{2}/2} ≤ exp(−k^{2}/6 + 2k), z ∈ M̃_{k,l}.

(2.4)

Proof of Lemma 2.1. We divide the product (2.1) in three parts: the factors with index j < k, the factors with j > k and the factor with j = k.

We estimate first the modulus of (\frac{z}{1−1/j} )^{j^{2}} for z ∈ M̃_{k,l}, for 2 ≤ j ≤ k−1. Since \frac{4k−7}{4k−3} ≤ |z| we have \frac{4j−3}{4j+1} ≤ |z|, so

|(\frac{z}{1 − 1/j} )^{j^{2}} | ≥ (\frac{1 − 1/(j + 1/4)}{1 − 1/j})^{j^{2}} → e^{1/4} as j → ∞, hence:

|(\frac{z}{1 − 1/j} )^{j^{2}} | ≥ C_{1} > 1.  (2.5)

In the same way we obtain for |z| ≤ \frac{4k−1}{4k+3}:

|(\frac{z}{1 − 1/j} )^{j^{2}} | ≤ C_{2} < 1 for j ≥ k + 1.  (2.6)

Since | ln(|1/(1−a)|·|a|)| = | ln(|1−1/a|)| ≤ C_{3}|1/a| for |a| ≥ C_{1} > 1, putting a = (z/(1−1/j))^{j^{2}}, we obtain from (2.5), using again |z| ≥ (4k −7)/(4k −3) as z ∈ M̃_{k,l}:

|ln |^{\begin{matrix}∏ & {k−1 \atop j=2} & \frac{1}{1−( \frac{z}{1−1/j} )^{j^{2}}} \\ ∏ & {k−1 \atop j=2} & (\frac{1−1/j}{z} ) & j^{2} \end{matrix}} | | ≤ C_{3} ∑_{j=2}^{k−1} |\frac{1 − 1/j}{(4k − 7)/(4k − 3)}|^{j^{2}}

≤ C_{3} ∑_{j=2}^{k−1}(1 − 1/j)^{j^{2}}(1 + 1/(k − 7/4))^{j^{2}}

≤ C_{3} ∑_{j=2}^{k−1} exp −\frac{j(k − 7/4 − j)}{k − 7/4}

≤ C_{3}e^{3} + C_{3} ∑_{j=2}^{k−2} exp(− min(j, k − 7/4 − j)/2)

≤ C_{3}e^{3} + 2C_{3} ∑_{j=1}^{∞} exp(−j/2 + 3/8) = C_{4}

Since |z|^{k} ∼ 1 for \frac{4k−7}{4k−3} ≤ |z| ≤ 1, the above gives in this region:

|∏_{j=2}^{k−1} \frac{1}{1 − (\frac{z}{1−1/j} )^{j^{2}}} | ∼ ∏_{j=2}^{k−1} |\frac{1 − 1/j}{z} |^{j^{2}} ∼ |z|^{−k^{3}/3+k^{2}/2} ∏_{j=2}^{k−1}(1 − 1/j)^{j^{2}}. (2.7)

We have the following upper bound for the right hand side above:

|z|^{−k^{3}/3+k^{2}/2} ∏_{j=2}^{k−1}(1 − 1/j)^{j^{2}} ≤ (\frac{4k − 7}{4k − 3})^{−k^{3}/3+k^{2}/2} exp ∑_{j=2}^{k−1} j^{2}(−1/j)

≤ exp ((−k^{3}/3 + k^{2}/2)\frac{−1}{k − 7/4} − ∑_{j=2}^{k−1} j)

≤ exp(k^{2}/3 + k − k(k − 1)/2 + 1)

≤ exp(−k^{2}/6 + 2k), for |z| ≥ \frac{4k−7}{4k−3},  (2.8)

This proves the inequality in (2.4). Since for (2.5) to hold we only need |z| ≥ \frac{4k−7}{4k−3}, we infer from (2.7) and (2.8) that the product (2.1) converges to 0 if |z| ≥ 1.

Now, for the factors with index greater than k in the definition of f, we use that | ln(1 − a)| ≤ C_{5}|a| for |a| ≤ C_{2} < 1. Via the relation (2.6) this can be applied for a = (\frac{z}{1−1/j})^{j^{2}} and we obtain for |z| ≤ \frac{4k−1}{4k+3}:

∑_{j=k+1}^{∞} |ln \frac{1}{1 − (\frac{z}{1−1/j} )^{j^{2}}} | ≤ C_{5} ∑_{j=k+1}^{∞} |\frac{1 − \frac{1}{k+3/4}}{1 − 1/j} |^{j^{2}}

≤ C_{5} ∑_{j=k+1}^{∞} (1 − \frac{1}{k+1})^{j^{2}} (1 + \frac{1}{j−1})^{j^{2}}

≤ C_{5} ∑_{j=k+1}^{∞} exp (\frac{j^{2}}{j−1} − \frac{j^{2}}{k+1})

≤ C_{5}e^{3/2} + C_{5} ∑_{j=k+2}^{∞} exp(k + 2 − j) = C_{6}.

Hence the product ∏_{j=k+1}^{∞} _{\frac{1}{1−( \frac{z}{1−1/j} )^{j^{2}}}} is uniformly absolutely convergent for

|z| ≤ \frac{4k−1}{4k+3}, and

e^{−C_{6}} ≤ | ∏_{j=k+1}^{∞} \frac{1}{1 − (\frac{z}{1−1/j} )^{j^{2}}} | ≤ e^{C_{6}}.  (2.9)

This implies that the product (2.1) is convergent everywhere in B(0, 1). Indeed, for |z| ≤ \frac{4k−1}{4k+3} the product up to the factor k is a meromorphic function and the rest of the product is absolutely convergent.

For the middle factor with j = k in (2.1), notice first that the logarithm is a quasi-isometry in the region 0 < c ≤ |z| ≤ C and −3π/4 ≤ arg z ≤ 3π/4. More generally, |z − z′| ∼ | ln \frac{z}{z′} | for |z|, |z′| ∈ [c, C] and arg \frac{z}{z′} ∈ [−3π/2, 3π/2]. Since z_{k,l}^{k^{2}} = (1 − 1/k)^{k^{2}} we have:

|\frac{1}{1 − (\frac{z}{1−1/k})^{k^{2}}} | ∼ \frac{1}{k^{2} |ln \frac{z}{z_{k,l}} |} ∼ \frac{1}{k^{2}} \frac{1}{|z − z_{k,l}|} for z ∈ M̃_{k,l}.  (2.10)

It remains to multiply the relations (2.7), (2.9) and (2.10) to obtain the first part of (2.4). □

Proof of Theorem 1.1. We choose a function ϕ ∈ C^{∞}(ℂ) such that:

ϕ(z) = ϕ(|z|) = \{ \begin{matrix}1 & for |z| ≥ 1, \\ |z|^{2} & for |z| ≤ 1/2. \end{matrix}

We require ϕ to be increasing on [0, ∞) and ϕ(r)/r to have only one critical point in (0, ∞) and that to be non-degenerate. For a choice of the quantities ε_{k} to be performed later, under the constraint 0 < ε_{k} ≤ k^{−2}/8, we define our function solution of equation (1.1):

u(z) = f(z) ∏_{k=2}^{∞} ∏_{l=0}^{k^{2}−1} ϕ (\frac{z − z_{k,l}}{ε_{k}} ) .  (2.11)

Notice first that the product is well defined, since by (2.3) the sets of points where each of the factors is different from 1 are disjoint. Since the poles z_{k,l} of f are simple and ϕ(z) = z\overline{z} in a neighborhood of 0, u is smooth in B(0, 1). For z ∈ M_{k,l} we have

\overline{∂}u(z) = \overline{∂} (f(z)ϕ (\frac{z − z_{k,l}}{ε_{k}})) = f(z)\overline{∂}ϕ (\frac{z − z_{k,l}}{ε_{k}}) = f(z)\frac{1}{ε_{k}} (\overline{∂}ϕ) (\frac{z − z_{k,l}}{ε_{k}})

and we obtain that the equation (1.1) is satisfied with a potential

V = ∑_{k=2}^{∞} ∑_{l=0}^{k^{2}−1} \frac{1}{ε_{k}} (\frac{\overline{∂}ϕ}{ϕ}) (\frac{z − z_{k,l}}{ε_{k}} ) .

The function \overline{∂}ϕ/ϕ has modulus 2|z|^{−1} for |z| ≤ 1/2 and vanishes for |z| > 1. Since ϕ does not vanish outside 0, we have that |z|\overline{∂}ϕ(z)/ϕ(z) is bounded in B(0, 1). Then

∫ F (|V |)dx_{1}dx_{2} ≤ C ∑_{k=2}^{∞} ∑_{l=0}^{k^{2}−1} ∫_{B(0,ε_{k})} |z|^{−2} ln^{−3}(|z|^{−1} + 2)dx_{1}dx_{2}
≤ C ∑_{k=2}^{∞} k^{2}2π ∫_{0}^{ε_{k}} −x^{−1} ln^{−3} x dx
≤ C′ ∑_{k=2}^{∞} k^{2} ln^{−2} ε_{k}.

We choose

ε_{k} = min(k^{−2}/8, e^{−k^{1.7}}),  (2.12)

and obtain that the series above is convergent, so V ∈ L^{F}. It remains to check that the solution (2.11) is smooth.

Take g̃_{k,l}(z) = z^{k(k−1)(2k−1)/6}g_{k,l}(z). The relation (2.4) from Lemma 2.1 implies that |g̃_{k,l}(z)| ∼ |g̃_{k,l}(z_{k,l})| for z ∈ M̃_{k,l}. From (2.2), each point z ∈ M_{k,l} is the center of a ball of radius k^{−2}/8 contained in M̃_{k,l}. Since g̃_{k,l} is holomorphic in M̃_{k,l}, we can apply the Cauchy representation formula using as contour the boundary of this ball, to obtain |∂^{α}g̃_{k,l}(z)| ≤ C_{α}(k^{2|α|})|g̃_{k,l}(z_{k,l})| for z ∈ M_{k,l}. Then differentiating g_{k,l}(z) = g̃_{k,l}(z)z^{−k(k−1)(2k−1)/6} we obtain for a suitable choice of the constants C′_{α}:

|∂^{α}g_{k,l}(z)| ≤ C′_{α}k^{3|α|}|g_{k,l}(z)| for z ∈ M_{k,l} .  (2.13)

Let us define

ϕ̃(z) = \frac{ϕ(z)}{z} .  (2.14)

Since ϕ̃ is a symbol of order −1, all its derivatives are bounded. Then, with the notation of Lemma 2.1, u(z) = \frac{g_{k,l}(z)}{ε_{k}} ϕ̃ (\frac{z−z_{k,l}}{ε_{k}} ) for z ∈ M̃_{k,l}. Using (2.13), the definition of ε_{k} and then (2.4) we obtain for k big enough:

|∂^{α}u(z)| ≤ C′′_{α}(max(ε_{k}^{−1}, k^{3}))^{|α|}ε_{k}^{−1}|g_{k,l}(z)|
≤ C′′_{α} exp((1 + |α|)k^{1.7})C exp(−k^{2}/6 + 2k), z ∈ M_{k,l}, (2.15)

so all the derivatives of u tend to zero as |z| → 1, hence u is smooth in ℂ.

For the second part of Theorem 1.1, we take v = Re u, where u is given by (2.11), but with one more condition on ε_{k}, which will change the choice (2.12) for finitely many values of k. In order to obtain a lower bound for |∇v| we will need the following

Lemma 2.2. Let h_{0}(z) = Re ϕ̃(z), where ϕ̃ is given by (2.14). Then there are positive constants c, C such that any real valued function h ∈ C^{2}(B(0, 1)) with ||h − h_{0}||_{C^{2}} ≤ c has exactly two critical points z_{1}, z_{2} and we have

|∇h(z)| ≥ C min(|z − z_{1}|, |z − z_{2}|).

Proof. We have h_{0}(z) = \frac{ϕ(|z|)}{|z|} cos(arg z), so ∇h_{0} has two zeroes (r_{ϕ}, 0) and (−r_{ϕ}, 0), where r_{ϕ} is, by the choice of ϕ, the unique zero of (r^{−1}ϕ(r))′ in (0, ∞) and \frac{d^{2}}{dr^{2}} (r^{−1}ϕ(r))|_{r=r_{ϕ}} ≠ 0. Since 0, π are non-degenerate critical points of the function θ → cos θ, we can choose open sets V_{1} ∋ (r_{ϕ}, 0) and V_{2} ∋ (−r_{ϕ}, 0) such that ∇h_{0}|V_{i} is a diffeomorphism and the tangent mapping of its inverse is bounded. We still get these properties after replacing h_{0} by h if ||h − h_{0}||_{C^{2}} ≤ c_{2} and c_{2} is small enough. Taking c_{2} even smaller, we can ensure that h has exactly one critical point in each of V_{1}, V_{2}; let these be z_{1}, z_{2}. Then |∇h|^{−1}|z − z_{i}| is uniformly bounded in V_{i} with respect to different choices of h.

On the other hand, outside V = V_{1} ∪ V_{2} the gradient |∇h_{0}| is bounded away from zero and so is |∇h| if ||h − h_{0}||_{C^{1}} ≤ c_{1} with c_{1} = \frac{1}{2} inf\{|∇h(z)| : z ∈ B(0, 1) \ V }. We choose then c = min(c_{2}, c_{1}). □

Notice that Lemma 2.2 remains valid if we replace ϕ̃ by aϕ̃, with |a| = 1, since this is equivalent to composing h_{0} with a rotation. This allows us to apply it to h(z) = |g_{k,l}(z_{k,l})|^{−1}ε_{k}v(z_{k,l} + ε_{k}z) which is a perturbation of h_{0} = Re \frac{g_{k,l}(z_{k,l})}{|g_{k,l}(z_{k,l})|} ϕ̃. From (2.13) we infer

\frac{|∂^{α}g_{k,l}(z_{k,l} + ε_{k}z)|}{|g_{k,l}(z_{k,l} + ε_{k}z)|} ≤ C′_{α}(ε_{k}k^{3})^{|α|},  (2.16)

so making ε_{k}k^{3} arbitrarily small we can make

||\frac{g_{k,l}(z_{k,l} + ε_{k}z)}{|g_{k,l}(z_{k,l})|} − \frac{g_{k,l}(z_{k,l})}{|g_{k,l}(z_{k,l})|}||_{C^{2}(B(0,1))}

arbitrarily small. Indeed, if ε_{k}k^{3} ≤ C then (2.16) implies that |g_{k,l}(z)| ∼ |g_{k,l}(z_{k,l})| for z ∈ B(z_{k,l}, ε_{k}), so we can replace the denominator in (2.16) with |g_{k,l}(z_{k,l})|. Applying Lemma 2.2 we obtain that there is a constant c_{1} > 0 such that for every k and l, if ε_{k} ≤ c_{1}k^{−3} then there are two points z_{k,l}^{(1)}, z_{k,l}^{(2)} ∈ B(z_{k,l}, ε_{k}) and C > 0 such that :

|g_{k,l}(z_{k,l})|^{−1} ε_{k} |∇v(z)| ≥ Cε_{k}^{−1} min (|\frac{z − z_{k,l}^{(1)}}{ε_{k}} | , |\frac{z − z_{k,l}^{(2)}}{ε_{k}} |) . (2.17)

We take ε_{k} = min(k^{−2}/8, c_{1}k^{−3}, e^{−k^{1.7}}) and W (z) = −(Δv)\frac{∇v}{|∇v|^{2}} . We use again that |g_{k,l}(z)| ∼ |g_{k,l}(z_{k,l})| in B(z_{k,l}, ε_{k}), to obtain in this ball from the first line of (2.15) and from (2.17):

|W (z)| = \frac{|Δv(z)|}{|∇v(z)|} ≤ \frac{(C′′_{(2,0)} + C′′_{(0,2)})ε_{k}^{−3}|g_{k,l}(z)|}{Cε_{k}^{−3}|g_{k,l}(z_{k,l})| min_{i=1,2} |z − z_{k,l}^{(i)}|} ≤ C′(_{\frac{1}{|z−z_{k,l}^{(1)}|}} + _{\frac{1}{|z−z_{k,l}^{(2)}|}})

Since v is harmonic in M_{k,l} \ B(z_{k,l}, ε_{k}) we obtain that W ∈ L^{F}:

∫ F (|W |)dx_{1}dx_{2} ≤ C ∑_{k=2}^{∞} ∑_{{0≤l<k^{2} \atop j=1,2}} ∫_{B(z_{k,l}^{(j)},2ε_{k})} F (|z − z_{k,l}^{(j)}|^{−1}) dx_{1}dx_{2}

≤ C ∑_{k=2}^{∞} 2k^{2}2π ∫_{0}^{2ε_{k}} −x^{−1} ln^{−3} x dx
≤ C′ ∑_{k=2}^{∞} k^{2} ln^{−2} ε_{k} < ∞. □

Proof of Corollary 1.2. The method is a standard one (see e.g. Theorem 2 in [2]) : we wrap the solution around a compact submanifold to make it have compact support. Here the support will be a tube around a sphere of dimension d − 2. We decompose ℝ^{d} = ℝ × ℝ^{d−1}, and use the notation x = (x_{1}, x′) with x′ = (x_{2}, … x_{d}) ∈ ℝ^{d−1}. Let (W_{1}(x_{1}, x_{2}), W_{2}(x_{1}, x_{2})) and v be the coefficient of the gradient and the solution constructed in Theorem 1.1. Then in dimension d ≥ 3 we take

ṽ(x) = v(x_{1}, |x′| − 2)
W̃ (x) = (W_{1}(x_{1}, |x′| − 2), (W_{2}(x_{1}, |x′| − 2) − \frac{d − 2}{|x′|})\frac{x′}{|x′|})

and using polar coordinates in ℝ^{d−1} it is easy to check that (1.2) is satisfied.

For the equation (1.3), we take w(x) = Dṽv_{0} = ∑_{i=1}^{d} ∂_{i}ṽ(x)α_{i}v_{0}, where v_{0} ∈ ℂ^{m} is a fixed non-zero vector, and W_{D} = \frac{W̃ ·∇ṽ}{|∇ṽ|^{2}} ∑_{i=1}^{d} ∂_{i}ṽα_{i}, with the above ṽ, W̃ (in dimension 2 we use v, W provided by Theorem 1.1). □

Remark. It is possible to obtain the above results by the standard procedure going back to Pliś (e.g. [2]). It consists of constructing a basic ’brick’ of the solution and then gluing infinitely many of them. However, the closure under multiplication of the set of solutions of the \overline{∂} equation allowed to avoid this.

References.

[1] L.V. Kantorovich and G.P. Akilov, Functional analysis, Pergamon Press, Oxford, 1982.

[2] A. Pliś, A smooth linear elliptic differential equation without any solution in a sphere, Commun. Pure Appl. Math, Vol. 14 (1961), 599–617.

[3] T. Wolff, A property of measures in ℝ^{n} and an application to unique continuation. Geometric and Functional Analysis, Vol. 2 (1992), 225– [4] T. Wolff, Recent work on sharp estimates in second-order elliptic unique

continuation problems, Journal of Geometric Analysis, Vol. 3 (1993), 621–650.