OBSERVABILITY INEQUALITIES AND MEASURABLE SETS J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG

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1 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG Abstract. This paper presents two observability inequalities for the heat equation over 0, T ). In the first one, the observation is from a subset of positive measure in 0, T ), while in the second, the observation is from a subset of positive surface measure on 0, T ). It also proves the Lebeau-Robbiano spectral inequality when is a bounded Lipschitz and locally star-shaped domain. Some applications for the above-mentioned observability inequalities are provided. 1. Introduction Let be a bounded Lipschitz domain in R n and T be a fixed positive time. Consider the heat equation: t u u = 0, in 0, T ), 1.1) u = 0, on 0, T ), u0) = u 0, in, with u 0 in L 2 ). The solution of 1.1) will be treated as either a function from [0, T ] to L 2 ) or a function of two variables x and t. Two important apriori estimates for the above equation are as follows: 1.2) ut ) L2 ) N, T, D) ux, t) dxdt, for all u 0 L 2 ), where D is a subset of 0, T ), and 1.3) ut ) L 2 ) N, T, J) D J ν ux, t) dσdt, for all u 0 L 2 ), where J is a subset of 0, T ). Such apriori estimates are called observability inequalities. In the case that D = ω 0, T ) and J = Γ 0, T ) with ω and Γ accordingly open and nonempty subsets of and, both inequalities 1.2) and 1.3) where is smooth) were essentially first established, via the Lebeau-Robbiano spectral inequalities in [28] See also [29, 34, 16]). These two estimates were set up to the linear parabolic equations where is of class C 2 ), based on the Carleman inequality provided in [18]. In the case when D = ω 0, T ) and J = Γ 0, T ) 1991 Mathematics Subject Classification. Primary: 35B37. Key words and phrases. observability inequality, heat equation, measurable set, spectral inequality. The first two authors are supported by Ministerio de Ciencia e Innovación grants, MTM and MTM The last two authors are supported by the National Natural Science Foundation of China under grants and and partially by National Basis Research Program of China 973 Program) under grant 2011CB

2 2 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG with ω and Γ accordingly subsets of positive measure and positive surface measure in and, both inequalities 1.2) and 1.3) were built up in [3] with the help of a propagation of smallness estimate from measurable sets for real-analytic functions first established in [43] See also Theorem 4). For D = ω E, with ω and E accordingly an open subset of and a subset of positive measure in 0, T ), the inequality 1.2) with is smooth) was proved in [44] with the aid of the Lebeau- Robbiano spectral inequality, and it was then verified for heat equations where is convex) with lower terms depending on the time variable, through a frequency function method in [39]. When D = ω E, with ω and E accordingly subsets of positive measure in and 0, T ), the estimate 1.2) with is real-analytic) was obtained in [45]. The purpose of this study is to establish inequalities 1.2) and 1.3), when D and J are arbitrary subsets of positive measure and of positive surface measure in 0, T ) and 0, T ) respectively. Such inequalities not only are mathematically interesting but also have important applications in the control theory of the heat equation, such as the bang-bang control, the time optimal control, the null controllability over a measurable set and so on See Section 5 for the applications). The starting point we choose here to prove the above-mentioned two inequalities is to assume that the Lebeau-Robbiano spectral inequality stands on. To introduce it, we write 0 < λ 1 λ 2 λ j for the eigenvalues of with the zero Dirichlet boundary condition over, and {e j : j 1} for the set of L 2 )-normalized eigenfunctions, i.e., { e j + λ j e j = 0, in, e j = 0, on. For λ > 0 we define E λ f = f, e j ) e j and E λ f = f, e j ) e j, where λ j λ f, e j ) = λ j>λ f e j dx, when f L 2 ), j 1. Throughout this paper the following notations are effective: f, g) = fg dx and f L2 ) = f, f) 1 2 ; ν is the unit exterior normal vector to ; dσ is surface measure on ; B R x 0 ) stands for the ball centered at x 0 in R n of radius R; R x 0 ) denotes B R x 0 ) ; B R = B R 0), R = R 0); for measurable sets ω R n and D R n 0, T ), ω and D stand for the Lebesgue measures of the sets; for each measurable set J in 0, T ), J denotes its surface measure on the lateral boundary of R; {e t : t 0} is the semigroup generated by with zero Dirichlet boundary condition over. Consequently, e t f is the solution of Equation 1.1) with the initial state f in L 2 ). The Lebeau-Robbiano spectral inequality is as follows: For each 0 < R 1, there is N = N, R), such that the inequality 1.4) E λ f L 2 ) Ne N λ E λ f L 2 B R x 0)) holds, when B 4R x 0 ), f L 2 ) and λ > 0.

3 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 3 To our best knowledge, the inequality 1.4) has been proved under condition that is at least C 2 [28, 29, 41, 32]. In the current work, we obtain this inequality when is a bounded Lipschitz and locally star-shaped domain in R n See Definitions 1 and 4 in Section 3). It can be observed from Section 3 that bounded C 1 domains, Lipschitz polygons in the plane, Lipschitz polyhedrons in R n, with n 3 and bounded convex domains in R n are always bounded Lipschitz and locally starshaped See Remarks 4 and 6 in Section 3). Our main results related to the observability inequalities are stated as follows: Theorem 1. Suppose that a bounded domain verifies the condition 1.4) and T > 0. Let x 0 and R 0, 1] be such that B 4R x 0 ). Then, for each measurable set D B R x 0 ) 0, T ) with D > 0, there is a positive constant B = B, T, R, D), such that 1.5) e T f L2 ) e B e t fx) dxdt, when f L 2 ). D Theorem 2. Suppose that a bounded Lipschitz domain verifies the condition 1.4) and T > 0. Let q 0 and R 0, 1] be such that 4R q 0 ) is real-analytic. Then, for each measurable set J R q 0 ) 0, T ) with J > 0, there is a positive constant B = B, T, R, J), such that 1.6) e T f L2 ) e B ν et fx) dσdt, when f L 2 ). J The definition of the real analyticity for 4R q 0 ) is given in Section 4 See Definition 5). Theorem 3. Let be a bounded Lispchitz and locally star-shaped domain in R n. Then, verifies the condition 1.4). It deserves mentioning that Theorem 2 also holds when is a Lipschitz polyhedron in R n and J is a measurable subset with positive surface measure of 0, T ) See the part ii) in Remark 11). In Section 5 we explain some applications of the Theorems 1 and 2 in the control theory of the heat equation. In particular, the existence of L D)-interior and L J)-boundary admissible controls, the uniqueness and bang-bang properties of the minimal L -norm control and the uniqueness and bang-bang property for the optimal controls associated to the first type and the second type of the time optimal control problems. In this work we use the new strategy developed in [39] to prove parabolic observability inequalities: a mixing of ideas from [34], the global interpolation inequalitiy in Theorems 6 and 10 and the telescoping series method. This new strategy can also be extended to more general parabolic evolutions with variable time-dependent second order coefficients and with unbounded lower order time-dependent coefficients. To do it one must prove the global interpolation inequalities in Theorems 6 and 10 for the corresponding parabolic evolutions. These can be derived in the more general setting from the Carleman inequalities in [8, 9, 12, 15, 25] or from local versions of frequency function arguments [10, 39]. Here we choose to derive the interpolation inequalities only for the heat equation and from the condition 1.4) because it is technically less involved and helps to make the presentation of the basic ideas more clear.

4 4 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG The rest of the paper is organized as follows: Section 2 proves Theorem 1; Section 3 shows Theorem 3; Section 4 verifies Theorem 2; Section 5 presents the applications of Theorem 1 and Theorem 2 and Section 6 is an Appendix completeting some of the technical details in the work. 2. Interior observability Throughout this section denotes a bounded domain and T is a positive time. First of all, we recall the following observability estimate or propagation of smallness inequality from measurable sets: Theorem 4. Assume that f : B 2R R n R is real-analytic in B 2R verifying α fx) M α! ρr) α, when x B 2R, α N n, for some M > 0 and 0 < ρ 1. Let E B R be a measurable set with positive measure. Then, there are positive constants N = Nρ, E / B R ) and θ = θρ, E / B R ) such that θ 2.1) f L B R ) N f dx) M 1 θ. The estimate 2.1) is first established in [43] See also [37] and [38] for other close results). The reader may find a simpler proof of Theorem 4 in [3, 3], the proof there was built with ideas taken from [33], [37] and [43]. Theorem 4 and the condition 1.4) imply the following: Theorem 5. Assume that verifies 1.4), ω is a subset of positive measure such that ω B R x 0 ), with B 4R x 0 ), for some R 0, 1]. Then, there is a positive constant N = N, R, ω / B R ) such that 2.2) E λ f L 2 ) Ne N λ E λ f L 1 ω), when f L 2 ) and λ > 0. Proof. Without loss of generality we may assume x 0 = 0. Because B 4R and 1.4) stands, there is N = N, R) such that 2.3) E λ f L2 ) Ne N λ E λ f L2 B R ), when f L 2 ) and λ > 0. For f L 2 ) arbitrarily given, define ux, y) = f, e j )e λjy e j. λ j λ One can verify that u+ 2 yu = 0 in B 4R 0, 0) R. Hence, there are N = Nn) and ρ = ρn) such that α x β y u L B 2R 0,0)) N α + β)! Rρ) α +β E B 4R 0,0) u 2 dxdy For the later see [36, Chapter 5], [22, Chapter 3]. function in B 2R, with the estimates: ) 1 2, when α N n, β 1. Thus, E λ f is a real-analytic α x E λ f L B 2R ) N α!rρ) α u L 4,4)), for α N n.

5 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 5 By either extending u as zero outside of R, which turns u into a subharmonic function in R n+1 or the local properties of solutions to elliptic equations [20, Theorems 8.17, 8.25] and the orthonormality of {e j : j 1} in, there is N = N) such that u L 4,4)) N u L 2 5,5)) Ne N λ E λ f L 2 ). The last two inequalities show that α x E λ f L B 2R ) Ne N λ α! Rρ) α E λ f L 2 ), for α N n, with N and ρ as above. In particular, E λ f verifies the hypothesis in Theorem 4 with M = Ne N λ E λ f L 2 ), and there are N = N, R, ω / B R ) and θ = θ, R, ω / B R ) with 2.4) E λ f L B R ) Ne N λ E λ f θ L 1 ω) E λf 1 θ L 2 ). Now, the estimate 2.2) follows from 2.3) and 2.4). Theorem 6. Let, x 0, R and ω be as in Theorem 5. Then, there are N = N, R, ω / B R ) and θ = θ, R, ω / B R ) 0, 1), such that θ 2.5) e t f L 2 ) Ne N t s e t f L ω)) 1 e s f 1 θ L 2 ), when 0 s < t and f L 2 ). Proof. Let 0 s < t and f L 2 ). Since e t E λ f L 2 ) e λt s) e s f L 2 ), when f L 2 ), it follows from Theorem 5 that e t f L 2 ) e t E λ f L 2 ) + e t E λ f L 2 ) Consequently, it holds that Ne N λ e t E λ f L1 ω) + e λt s) e s f L2 ) Ne N λ [ e t f L1 ω) + e t E λ f L ω)] 2 + e λt s) e s [ f L 2 ) ] Ne N λ e t f L1 ω) + e λt s) e s f L2 ). 2.6) e t f L2 ) Ne N λ [ e t f L1 ω) + e λt s) e s f L2 ) Because it follows from 2.6) that for each λ > 0, e t f L 2 ) Ne N t s max λ λt s) λ 0 ea e NA) t s, for all A > 0, ]. [ e Nλt s) e t f L 1 ω) + e λt s)/n e s f L 2 ) Setting ɛ = e λt s) in the above estimate shows that the inequality 2.7) e t [ f L2 ) Ne N t s ɛ N e t f L1 ω) + ɛ e s ] f L2 ), holds, for all 0 < ɛ 1. The minimization of the right hand in 2.7) for ɛ in 0, 1), as well as the fact that e t f L 2 ) e s f L 2 ), when t > s, ].

6 6 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG implies Theorem 6. Remark 1. Theorem 6 shows that the observability or spectral elliptic inequality 2.2) implies the inequality 2.5). In particular, the elliptic spectral inequality 1.4) implies the inequality: θ 2.8) e t f L 2 ) Ne N t s e t f L 2 B R x 0))) e s f 1 θ L 2 ), when 0 s < t, B 4R x 0 ) and f L 2 ). In fact, both 2.2) and 2.5) or 1.4) and 2.8) are equivalent, for if 2.5) holds, take f = λ j λ eλj/ λ a j e j, s = 0 and t = 1/ λ in 2.5) to derive that ) 1 2 Ne N λ L a j e j when a j R, j 1, λ > 0. ω), 1 λ j λ a 2 j λ j λ The interested reader may want here to compare the previous claims, Theorem 3 and [39, Proposition 2.2]. Lemma 1. Let B R x 0 ) and D B R x 0 ) 0, T ) be a subset of positive measure. Set D t = {x : x, t) D}, E = {t 0, T ) : D t D /2T )}, t 0, T ). Then, D t is measurable for a.e. t 0, T ), E is measurable in 0, T ), E D /2 B R and 2.9) χ E t)χ Dt x) χ D x, t), in 0, T ). Proof. From Fubini s theorem, T D = D t dt = 0 E D t dt + D t dt B R E + D /2. [0,T ]\E Theorem 7. Let x 0 and R 0, 1] be such that B 4R x 0 ). Let D B R x 0 ) 0, T ) be a measurable set with D > 0. Write E and D t for the sets associated to D in Lemma 1. Then, for each η 0, 1), there are N = N, R, D / T B R ), η) and θ = θ, R, D / T B R ), η) with θ 0, 1), such that t2 ) θ 2.10) e t2 f L2 ) Ne N/t2 t1) χ E s) e s f L1 D s) ds e t1 f 1 θ L 2 ), when 0 t 1 < t 2 T, E t 1, t 2 ) ηt 2 t 1 ) and f L 2 ). Moreover, 2.11) e N+1 θ t 2 t 1 N t2 t 1 e t2 f L 2 ) e N+1 θ qt 2 t 1) e t1 f L 2 ) t 1 χ E s) e s f L1 D s) ds, when q N + 1 θ)/n + 1). Proof. After removing from E a set with zero Lebesgue measure, we may assume that D t is measurable for all t in E. From Lemma 1, D t B R x 0 ), B 4R x 0 ) and D t / B R D /2T B R ), when t is in E. From Theorem 6, there are N = N, R, D / T B R )) and θ = θ, R, D / T B R )) such that θ 2.12) e t f L2 ) Ne N t s e t f L1 D t)) e s f 1 θ L 2 ),

7 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 7 when 0 s < t, t E and f L 2 ). Let η 0, 1) and 0 t 1 < t 2 T satisfy E t 1, t 2 ) ηt 2 t 1 ). Set τ = t 1 + η 2 t 2 t 1 ). Then 2.13) E τ, t 2 ) = E t 1, t 2 ) E t 1, τ) η 2 t 2 t 1 ). From 2.12) with s = t 1 and the decay property of e t f L2 ), we get 2.14) e t2 f L2 ) Ne N t 2 t 1 e t f L 1 D t)) θ e t 1 f 1 θ L 2 ), t E τ, t 2). The inequality 2.10) follows from the integration with respect to t of 2.14) over E τ, t 2 ), Hölder s inequality with p = 1/θ and 2.13). The inequality 2.10) and Young s inequality imply that 2.15) e t2 f L 2 ) ɛ e t1 f L2 ) + ɛ 1 θ θ Ne N t 2 t 1 t2 t 1 χ E s) e s f L1 D s) ds, when ɛ > 0. Multiplying first 2.15) by ɛ 1 θ θ e N t 2 t 1 and then replacing ɛ by ɛ θ, we get that ɛ 1 θ e N t 2 t 1) e t2 f L2 ) ɛ e N t 2 t 1) e t1 f L2 ) N Choosing ɛ = e 1 t 2 t 1 t2 t 1 χ E s) e s f L 1 D s) ds, when ɛ > 0. in the above inequality leads to e N+1 θ t 2 t 1) e t2 f L 2 ) e N+1 t 2 t 1) e t1 f L 2 ) N t2 t 1 χ E s) e s f L 1 D s) ds. This implies 2.11), for q N+1 θ N+1. The reader can find the proof of the following Lemma 2 in either [31, pp ] or [39, Proposition 2.1]. Lemma 2. Let E be a subset of positive measure in 0, T ). Let l be a density point of E. Then, for each z > 1, there is l 1 = l 1 z, E) in l, T ) such that, the sequence {l m } defined as verifies l m+1 = l + z m l 1 l), m = 1, 2,, 2.16) E l m+1, l m ) 1 3 l m l m+1 ), when m 1. Proof of Theorem 1. Let E and D t be the sets associated to D in Lemma 1 and l be a density point in E. For z > 1 to be fixed later, {l m } denotes the sequence associated to l and z in Lemma 2. Because 2.16) holds, we may apply Theorem 7, with η = 1/3, t 1 = l m+1 and t 2 = l m, for each m 1, to get that there are N = N, R, D / T B R )) > 0 and θ = θ, R, D / T B R )), with θ 0, 1), such

8 8 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG that 2.17) e N+1 θ lm l m+1 e l m f L 2 ) e lm N N+1 θ qlm l m+1 ) e lm+1 f L 2 ) χ E s) e s f L 1 D s) ds, when q N + 1 θ l m+1 N + 1 Setting z = 1/q in 2.17) which leads to 1 < z N+1 N+1 θ ) and recalling that we have N+1 θ γ z t) = e z 1)l 1 l)t, t > 0, l m l m+1 = z m z 1) l 1 l), for m 1, and m ) γ z z m ) e lm f L 2 ) γ z z m 1 ) e lm+1 f L 2 ) N lm l m+1 χ E s) e s f L 1 D s) ds, when m 1. Choose now z = N + 1 ). 2 N + 1 θ The choice of z and Lemma 2 determines l 1 in l, T ) and from 2.18), 2.19) with γz m ) e lm f L 2 ) γz m 1 ) e lm+1 f L 2 ) N lm l m+1 χ E s) e s f L 1 D s) ds, when m 1. γt) = e A/t and A = A, R, E, D / T B R )) = Finally, because of e T f L 2 ) e l1 f L 2 ), sup e t f L 2 ) < +, t 0 2 N + 1 θ)2 θ l 1 l) lim γt) = 0, t 0+ and 2.9), the addition of the telescoping series in 2.19) gives e T f L2 ) Ne za e t fx) dxdt, for f L 2 ), D [l,l 1]) which proves 1.5) with B = za + log N.. Remark 2. The constant B in Theorem 1 depends on E because the choice of l 1 = l 1 z, E) in Lemma 2 depends on the possible complex structure of the measurable set E See the proof of Lemma 2 in [39, Proposition 2.1]). When D = ω 0, T ), one may take l = T/2, l 1 = T, z = 2 and then, B = A, R, ω / B R )/T. Remark 3. The proof of Theorem 1 also implies the following observability estimate: sup sup m+1a e t f L 2 ) N e t fx) dxdt, e z m 0 l m+1 t l m D [l,l 1])

9 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 9 for f in L 2 ), and with z, N and A as defined along the proof of Theorem 1. Here, l 0 = T. 3. Spectral inequalities Throughout this section, is a bounded domain in R n and ν q is the unit exterior normal vector associated to q. Definition 1. is a Lipschitz domain sometimes called strongly Lipschitz or Lipschitz graph domains) with constants m and ϱ when for each point p on the boundary of there is a rectangular coordinate system x = x, x n ) and a Lipschitz function φ : R n 1 R verifying 3.1) φ0 ) = 0, φx 1) φx 2) m x 1 x 2, for all x 1, x 2 R n 1, p = 0, 0) on this coordinate system and 3.2) Z m,ϱ = {x, x n ) : x < ϱ, φx ) < x n < 2mϱ}, Z m,ϱ = {x, φx )) : x < ϱ}, where Z m,ϱ = B ϱ 2mϱ, 2mϱ). Definition 2. is a C 1 domain when it is a Lipschitz domain and the functions ϕ associated to points p in satisfying 3.1) and 3.2) are in C 1 R n 1 ). Under such condition, there is with θ : [0, + ) [0, + ), with θ nondecreasing, lim θt) = 0, t ) q p) ν q p q θ p q ), when p, q. Definition 3. is a lower C 1 domain when it is a Lipschitz domain and q p) ν q 3.4) lim inf q, q p q p 0, for each p. Remark 4. Lipschitz polygons in the plane, convex domains in R n See [36, p. 72, Lemma 3.4.1] for a proof that convex domains in R n are Lipschitz domains or strongly Lipschitz domains, to keep pace with Morrey s definition)) or Lipschitz polyhedron in R n, n 3, are lower C 1 domains. In fact, in all these cases and for p, there is r p > 0 such that q p) ν q 0, for a.e. q B rp p). When is convex, r p = +. In general, a Lipschitz domain is a lower C 1 domain when the Lipschitz functions φ describing its boundary can be discomposed as the sum of two Lipschitz functions, φ = φ 1 + φ 2, with φ 1 convex over R n 1 and φ 2 satisfying lim ɛ 0 + sup φ 2 x 1) φ 2 x 2) = 0. x 1 x 2 <ɛ In particular, C 1 domains are lower C 1 domains See 3.3)). Definition 4. A Lipschitz domain in R n is locally star-shaped when for each p there are x p in and r p > 0 such that 3.5) p x p < r p and B rp x p ) is star-shaped with center x p.

10 10 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG In particular, q x p ) ν q 0, for a.e. q B rp x p ). Remark 5. The compactness of shows that when is locally star-shaped, there are a finite set A, 0 < ɛ, ρ 1 and a family of positive numbers 0 < r x 1, x A, such that 3.6) Here, x A B rx x), B 1+ɛ)rx x) is star-shaped, A 4ρ and \ 4ρ x A B rx x). η = {x : dx, ) > η}, when η > 0. Remark 6. Theorem 8 below shows that Lispchitz polygons in the plane, C 1 domains, convex domains and Lipschitz polyhedron in R n, n 3 are locally star-shaped. Lipschitz domains in R n with Lipschitz constant m < 1 2 are also locally star-shaped. Recall that not all the polygons in the plane or polyhedra in R 3 are Lipschitz domains See [42, p. 496, pp ] or the two-brick domain of [26, p. 303]). Theorem 8. Let be a lower C 1 domain. Then, is locally star-shaped. Proof. Let x, x n ) and φ be accordingly the rectangular coordinate system and Lipschitz function associated to p, satisfying 3.1) and 3.2). Let then, x p = p + δ e, e = 0, 1), and r p = 2δ, where δ > 0 will be chosen later. Clearly x p is in, p x p < r p and B rp x p ) Z m,ϱ, when 0 < δ < min { ϱ 2, 2mϱ 3 }. Moreover, for almost every q in B rp x p ), it holds that q = x, φx )) for some x in B ρ, and ν q = φx ), 1) 1 + φx ) 2 and q p r p + δ = 3δ From 3.4) there is s p > 0 such that q x p ) ν q = q p) ν q δ e ν q δ q p) ν q m 2 q p 3.7) q p) ν q m, when q and q p < s p. 2 Thus, 3.5) holds for the choices we made of x p, r p and s p, provided that δ is chosen with 0 < δ < min { ϱ 2, 2mϱ 3, sp 3 }. The proof of Theorem 3 will follow from the following Lemmas 3 and 4. Lemma 3. Let be a Lipschitz domain in R n, R > 0 and assume that B R x 0 ) is star-shaped with center at some x 0. Then, 3.8) u L2 B r2 x 0) ) u θ L 2 B r1 x 0) ) u 1 θ L 2 B r3 x 0) ) r3 log r, with θ = 2 when u = 0 in B R x 0 ), u = 0 on R x 0 ) and 0 < r 1 < r 2 < r 3 R. log r3 r 1,

11 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 11 Remark 7. See [27, Lemma 3.1] for a proof of Lemma 3 and [1] and [2] for related results but with spheres replacing balls. The relevance of the assumption that B R x 0 ) is star-shaped in the proof of [27, Lemma 3.1] is that q x 0 ) ν q 0, for a.e. q in B R x 0 ) and certain terms arising in the arguments can be drop because of their nonnegative sign. In fact, 3.8) is the logarithmic convexity of the L 2 -norm of u over B r x 0 ) for 0 < r R with respect to the variable log r. Lemma 3 extends up to the boundary the classical interior three-spheres inequality for harmonic functions, first stablished for complex analytic functions by Hadamard [21] and extended for harmonic functions by several authors [19]: r3 log 3.9) u L 2 B r2 x 0)) u θ L 2 B r1 x 0)) u 1 θ r L 2 B r3 x 0)), with θ = 2, log r3 r 1 when u = 0 in, B R x 0 ) and 0 < r 1 < r 2 < r 3 R. When B R x 0 ), B R x 0 ) is star-shaped, and a proof for 3.9) is within the one for [27, Lemma 3.1]. Lemma 4. Let be a Lipischitz domain with constants m and ϱ, 0 < r < ϱ/10 and q be in. Then, there are N = Nn, m) and θ = θn, m), 0 < θ < 1, such that the inequality u L2 B rq) ) Nr 3θ 2 u ν θ L 2 6rq)) u 1 θ L 2 B 8rq) ), holds for all harmonic functions u in B 8r q) satisfying u = 0 on B 8r q). To prove Lemma 4 we use the Carleman inequality in Lemma 5. As far as the authors know, the first L 2 -type Carleman inequality with a radial weight and whose proof was worked out in Cartesian coordinates appeared first in [24, Lemma 1]. Lemma 5 borrows ideas from [24, Lemma 1] but the proof here is somehow simpler. In [1, p. 518] also appears an interpolation inequality similar to the one in Lemma 4 but with the L 2 -norms replaced by L 1 -norms. The inequality in [1, p. 518] holds though its proof in [1] is not correct. It follows from Lemma 4 and properties of harmonic functions. Lemma 5. Let be a Lipschitz domain in R n with B R and τ > 0. Assume that 0 is not in. Then, x 2τ u 2 dx R2 4τ 2 x 2τ+2 u) 2 dx R2 q ν q 2τ ) u 2 2τ ν dσ, for all u in C 2 ) satisfying u = 0 on. Proof of Lemma 5. Let u be in C 2 ), u = 0 on and define f = x τ u. Then, ) 3.10) x 1 τ u = x f + τ 2 x f + 2τ 2 x x f + n 2 2 f). Square both sides of 3.10) to get 3.11) x 2 2τ u) 2 = 4τ 2 x 2 ) x f + n 2 2 f) x 2 f + τ 2 x f 2 + 4τ x f + n 2 2 f) f + τ 2 x 2 f ).

12 12 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG Observe that x 2 x f + n 2 2 f) f dx = 1 q ν 2 q f 2 dσ = 0, 2 and integrate 3.11) over. Then we get the identity 3.12) x 2 2τ u) 2 dx = 4τ 2 x 2 x f + n 2 + x 2 f + τ 2 x f 2 The Rellich-Nečas or Pohozaev identity and the identity give the formula 2 f) 2 dx ) 2 dx + 4τ x f + n 2 2 f) f dx. [x f 2 2 x f) f ] + 2 x f) f = n 2) f 2 f 2) = 2f f + 2 f 2, [x f 2 2 x f) f ] + 2 x f + n 2 2 f) f = n 2 2 f 2 ). The integration of this identity over implies the formula 3.13) 4 x f + n 2 2 f) f dx = 2 q ν f ν ) 2 dσ, and plugging 3.13) into 3.12) gives the identity 3.14) x 2 2τ u) 2 dx = 4τ 2 x 2 x f + n 2 2 f) 2 dx ) 2 + x 2 f + τ 2 x f dx + 2τ q ν 2 Next, the identity x f + n 2 2 f) f dx = f 2 dx, the Cauchy-Schwarz s inequality and B R show that 3.15) R 2 f 2 dx x 2 x f + n 2 2 f) 2 dx. Then, Lemma 5 follows from 3.14) and 3.15). f ν ) 2 dσ. Proof of Lemma 4. After rescaling and translation we may assume that q = 0, r = 1 and that there is a Lipschitz function φ : R n 1 R verifying 3.16) φ0 ) = 0, φx 1) φx 2) m x 1 x 2, for all x 1, x 2 R n 1, Z m,10 = {x, x n ) : x < 10, φx ) < x n < 20m}, Z m,10 = {x, φx )) : x < 10}, where Z m,10 = B 10 20m, 20m). Recalling that e = 0, 1). Thus, e / and 3.17) B 1 B 2 e) B 5 e) B 6. Choose ψ in C 0 B 5 e)) with ψ 1 in B 3 e) and ψ 0 outside B 4 e). From 3.16), there is β 0, 1), β = βm), such that 3.18) B 2β e) =.

13 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 13 By translating the inequality in Lemma 5 from 0 to e we find that x + e τ f L2 B x + e τ+1 f 5 e) ) L2 B 5 e) ) 3.19) + q + e 1 2 τ f, ν L2 B 5 e) ) when f is in C0B 2 5 e) ), f = 0 on B 5 e) and τ 20. Let then u be harmonic in B 8 with u = 0 on B 8 and take f = uψ in 3.19). We get x + e τ uψ L x + e 1 τ u ψ + 2 ψ u) 2 B 5 e) ) L 2 B 5 e) ) 3.20) + q + e 1 2 τ ψ u, when τ 20. ν L2 B 5 e) ) Next, because ψ 1 on B 3 e) and 3.17), we have 3.21) 2 τ u L2 B 1 ) x + e τ f L, when τ 1. 2 B 5 e) ) Also, it follows from from 3.18) that 3.22) q + e 1 2 τ ψ u ν L 2 B 5 e) ) β τ u ν L 2 6), when τ 1. Now, ψ + ψ is supported in B 4 e) \ B 3 e), where x + e 3 and x + e 1 τ u ψ + 2 ψ u) L 2 B 5 e) ) 3.23) N3 τ u + u L 2 B 4 e)\b 3 e) ) N3 τ u L 2 B 6 ), when τ 1, with N = Nn). Putting together 3.20), 3.21), 3.22) and 3.23), we get [ u L 2 B 1 ) N 2 τ 2 3) u L 2 B 6 ) + β ) τ u ν L 2 6) The later inequality shows that there is N = Nn, m) 1 such that ], when τ ) u L2 B 1 ) N [ ɛ u L2 B 8 ) + ɛ N u ν L 2 6)], when 0 < ɛ ). Finally, the minimization of 3.24) shows that Lemma 4 holds for q = 0 and r = 1 and completes the proof. Proof of Theorem 3. Without loss of generality we may assume that x 0 = 0 and B 4R for some fixed 0 < R 1. To prove that 1.4) holds with x 0 = 0, we first show that under the hypothesis of Theorem 3, there are ρ and θ in 0, 1) and N 1 such that the inequality 3.25) u L2 [ 1,1]) N u θ L 2 2ρ [ 2,2]) u 1 θ L 2 [ 4,4]) holds when 3.26) { u + yu 2 = 0, in R, u = 0, on R. To prove 3.25), we recall that 3.6) holds and A = {x 1, x 2,..., x l } for some x i in 4ρ, i = 1,..., l, for some l 1. Because B 1+ɛ)rxi x i ) is star-shaped with center x i, the same holds for B 1+ɛ)rxi x i, τ) R and with center x i, τ), i = 1,..., l, for all τ R. Then, from Lemma 3, it follows that 3.27) u L 2 B rxi x i,τ) R) u θ L 2 B ρx i,τ) R) u 1 θ L 2 B 1+ɛ)rxi x i,τ) R),

14 14 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG for i = 1,..., l with 3.28) θ = min Also, from 3.9) { 1 + ɛ) rxi log 1 + ɛ)/ log ρ ) } : i = 1,..., l. 3.29) u L 2 B ρx,τ)) u θ log 2 L 2 B ρ x,τ)) u 1 θ L 2 B 4 2ρx,τ)), with θ = log 8 when x is in the closure of 4ρ and τ R. Because 3.6), 3.27), 3.29) hold and there are z 1,..., z m in the closure of 4ρ with 4ρ m i=1b ρ z i ), it is now clear that there are N = Nl, m, ρ, A) 1 and a new θ = θl, m, ρ, A) in 0, 1) such that 3.25) holds. From 3.9) with r 1 = r 4, r 2 = r, r 3 = 2r, we see that 3.30) u L2 B rx 1,y 1)) u θ L 2 B r 4 x 1,y 1)) u 1 θ L 2 B 2rx 1,y 1)), θ = log 2 log 8, when B 2r x 1, y 1 ) [ 4, 4]. From 3.30) and the fact that we find that B r 4 x 1, y 1 ) B r x 2, y 2 ), when x 2, y 2 ) B r 4 x 1, y 1 ), 3.31) u L2 B rx 1,y 1)) u θ L 2 B rx 2,y 2)) u 1 θ L 2 [ 4,4]), θ = log 2 log 8, when B 2r x 1, y 1 ) [ 4, 4], x 2, y 2 ) is in B r 4 x 1, y 1 ) and r > 0. Because ρ is now fixed and is compact, there are 0 < r 1 and k 2, which depend on ρ, the geometry of and R, such that for all x 0, y 0 ) in 2ρ [ 2, 2] there are k points x 1, y 1 ), x 2, y 2 ),..., x k, y k ) in 2ρ [ 2, 2] with B 2r x i, y i ) [ 4, 4], x i+1, y i+1 ) B r x 4 i, y i ), for i = 0,..., k 1, x k, y k ) = 0, R 16 ) and B r0, R 16 ) B+ R 0, 0). 8 The later and 3.31) show that 3.32) u L 2 B rx i,y i)) u θ L 2 B rx i+1,y i+1)) u 1 θ L 2 [ 4,4]), for i = 0,..., k 1, while the compactness of and the iteration of 3.32) imply that for some new N = Nl, m, k, ρ, A, R) 1 and 0 < θ = θl, m, k, ρ, A, R) < ) u L 2 2ρ [ 2,2]) N u θ L 2 B + R 0,0)) u 1 θ L 2 [ 4,4]). 8 Putting together 3.25) and 3.33), it follows that when u satisfies 3.26), 3.34) u L 2 [ 1,1]) N u θ L 2 B + R 0,0)) u 1 θ L 2 [ 4,4]), 8 with N and θ as before. Finally, proceeding as in [28], for λ > 0 we take 3.35) ux, y) = sinh λ j y) a j e j x), λj λ j λ with {a j } a sequence of real numbers. It satisfies 3.26) and also ux, 0) 0 in. Then, we may assume that R 8 ϱ 10 and Lemma 4 implies 3.36) u L 2 B R8 0,0) + ) N u y, 0) θ L 2 B R ) u 1 θ L 2 [ 4,4]).

15 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 15 Combining 3.34) with 3.36) leads to u L2 [ 1,1]) N u y, 0) θ L 2 B R ) u 1 θ L 2 [ 4,4]). This, along with 3.35) and standard arguments in [29], shows that satisfies the condition 1.4). Remark 8. The reader can now easily derive with arguments based on the Lemmas 3, 4 and Theorem 3, the following spectral inequality: Theorem 9. Suppose that is Lipschitz and verifies 1.4). Let q 0, τ R and R 0, 1]. Then, there is N = N, R, τ) such that the inequalities 1 2 a 2 j + b 2 j e N λ a j e λj y + b j e λ j y ) ej ν, λ j λ λ j λ L2 B R q 0,τ) R) hold for all sequences {a j } and {b j } in R. In particular, the above inequality holds when is a bounded Lipschitz and locally star-shaped domain. 4. Boundary observability Throughout this section is a bounded Lipschitz domain in R n and T is a positive time. We first study quantitative estimates of real analyticity with respect to the space-time variables for caloric functions in 0, T ) with zero lateral Dirichlet boundary conditions. Let 4.1) Gx, y, t) = j 1 e λjt e j x)e j y) be the Green s function for t on with zero lateral Dirichlet boundary condition. By the maximum principle and 0 Gx, y, t) 4πt) n 2 e x y 2 4t, 4.2) Gx, x, t) = j 1 e λjt e j x) 2 4πt) n 2. Definition 5. Let q 0 and 0 < R 1. We say that 4R q 0 ) is real-analytic with constants ϱ and δ if for each q 4R q 0 ), there are a new rectangular coordinate system where q = 0, and a real-analytic function φ : B ϱ R n 1 R verifying φ0 ) = 0, α φx ) α!δ α 1, when x B ϱ, α N n 1, B ϱ = B ϱ {x, x n ) : x B ϱ, x n > φx )}, B ϱ = B ϱ {x, x n ) : x B ϱ, x n = φx )}. Here, B ϱ denotes the open ball of radius ϱ and with center at 0 in R n 1. Lemma 6. Let q 0 and R 0, 1]. Assume that 4R q 0 ) is real-analytic with constants ϱ and δ. Then, there are N = Nϱ, δ), ρ = ρϱ, δ), with 0 < ρ 1, such that 4.3) x α β t e t fx) N t n s) 4 e 8R2 /t s) α! β! Rρ) α t s) /4) β e s f L2 ),

16 16 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG when x B 2R q 0 ), 0 s < t, α N n and β 0. Proof. It suffices to prove 4.3) for the case when s = 0. Let f = j 1 a je j. Set then ux, t, y) = + j=1 ux, t) = ux, t, 0) = a j e λjt+ λ jy e j x), x, t > 0, y R, + j=1 a j e λjt e j x) = e t fx), x, t > 0. We have { y 2 β t u + β t u = 0, in R, β t u = 0, on R, 4.4) β t ux, t, y) = 1) β j 1 a j λ β j e λjt+ λ jy e j x). Because 4R q 0 ) is real-analytic, there are N = Nϱ, δ) and ρ = ρϱ, δ) such that 4.5) x α β t u, t, ) L B 2R q 0,0) R) N α! β Rρ) α t ux, t, y) 2 dxdy B 4R q 0,0) R ) 1 2, when α N n, β 0. For the later see [36, Chapter 5] and [22, Chapter 3]. Now, it follows from 4.4) that ) β 1 t ux, t, y) a 2 2 j λ 2β j e 2λjt+2 λ jy e j x) 2) 1 2. Also, it stands that j 1 j 1 j 1 λ 2β j e 2λjt+2 λ jy e j x) 2 = t 2β e y2 /t λ j t) 2β e λjt e j x) 2 e λjt y 2 t j 1 t 2β e y2 /t j 1 λ j t) 2β e λjt e j x) 2 = t 2β e y2 /t j 1 λ j t) 2β e λjt/2 e λjt/2 e j x) 2. Next Stirling s formula shows that max x 0 x2β e x/2 = 4 4β β 2β e 2β 4 4β β!) 2, when β 0. Finally, the above three inequalities, as well as 4.2), imply that ) 4.6) β n ) 1 4 t ux, t, y) 2πt e y2 /2t 2 t/4) β β!, for t > 0, x, y R and β 0. The later inequality and 4.5) imply that x α β t ux, t) Nt n 4 e 8R2 /t α! β! ) 1 Rρ) α t/4) β a 2 2 j, when x B 2R q 0 ), t > 0, α N n and β 0. The next caloric interpolation inequality plays the same role for the boundary case as the inequality 2.8) for the interior case. j 1 a 2 j j 1

17 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 17 Theorem 10. Let be a bounded Lipschitz domain in R n with constants m, ϱ) and satisfy the condition 1.4). Then, given 0 < R 1, 0 t 1 < t 2 T 1 and q, there are N = N, R, m, ϱ) 1 and θ = θm, ϱ), with θ 0, 1), such that ) e t2 f L 2 ) Ne N θ t 2 t 1 ν et f L 2 R q) t 1,t 2)) e t 1 f 1 θ L 2 ), f L2 ). To prove Theorem 10, we need first some lemmas. We begin with the following Carleman inequality See [9] and [11]). Lemma 7. Let be a bounded Lipschitz domain in R n with constants m, ϱ), 0 / and σt) = te Mt, M > 0. Then τm σt) τ e x 2 /8t h L 2 0,+ )) t 1/2 σt) τ e x 2 /8t + t )h L 2 0,+ )) + q 1/2 σt) τ e q 2 /8t h ν L 2 0,+ )), when τ 1 and h C 0 [0, + )) with h = 0 on [0, + ). Proof. First, let f = σt) τ e x2 /8t h. Then σt) τ e x 2 /8t + t )h = σt) τ e x 2 /8t + t )σt) τ e x 2 /8t f) Thus = f x 2 16t 2 f + τ t log σ)f + t f + x 2t f + n 4t f. 4.7) t 1/2 σt) τ e x 2 /8t + t )h 2 L 2 0,+ )) = t 1/2 f x 2 16t 2 f + τ t log σ)f) 2 L 2 0,+ )) + t 1/2 t f + x 2t f + n 4t f) 2 L 2 0,+ )) + 2t f x 2 16t f + τ 2 t log σ)f) t f + x 2t f + n 4t f) dxdt. 0,+ ) Next, integrating by parts we have the following two identities: 2t t f f dxdt 4.8) 4.9) 0,+ ) = 0,+ ) t t f 2 + 2t t f f) dxdt = 0,+ ) = = 0,+ ) 0,+ ) 2t t f The Rellich-Nečas or Pohosaev identity ) x 2 16t f + τ 2 t log σ)f 0,+ ) dxdt x 2 16t tf 2 + τt t log σ) t f 2 dxdt x 2 16t 2 f 2 τ t t t log σ)) f 2 dxdt. [ x f 2) 2 x f) f ] = n 2) f 2 2 x f) f = n 2) f 2 2 x f + n 2 f) f + nf f = n 2 f 2 ) 2 f 2 2 x f + n 2 f) f f 2 dxdt.

18 18 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG gives the formula x f + n 2 f) f = [x f) f x 2 f 2] + n 4 f 2 ) f 2. Integrating the above identity in, we get that for each t > 0, ) ) x f + n 4 f) f dx = 1 2 q ν f ν dσ On the other hand, 4.11) 0,+ ) = + n 2 = 0,+ ) 0,+ ) 0,+ ) ) x f + n 2 f) x 2 16t f + τ 2 t log σ)f dxdt x 2 f 2) ) x 2 16t + τ 2 t log σ) dxdt τ t log σ) x 2 16t 2 ) f 2 dxdt x 2 16t 2 f 2 dxdt. f 2 dx. Combining 4.7) and the identities 4.8), 4.9), 4.10) and 4.11), we get t 1/2 σt) τ e x 2 /8t + t )h 2 L 2 0,+ )) τ t t t log σ)) f 2 dxdt + 0,+ ) 0,+ ) ) q ν f ν dσdt. Choose then, σt) = te Mt, M > 0, t t t log σ)) = M, and it leads to the desired estimate. In Lemmas 8, 9, 10 and 11, we assume that is a Lipschitz domain with constants m, ϱ) and 0. In Lemmas 8 and 11 we also assume that is near the origin, the region above the graph, x n = φx ), with φ as in 3.1) and 3.2), so that ρ e is not in, when 0 < ρ mϱ, e = 0, 1). Lemma 8. Let σt) = te t. Then, for 0 < ρ mϱ, τ σt) τ e x+ρe 2 /8t h L2 0,+ )) t 1/2 σt) τ e x+ρe 2 /8t + t ) h L2 0,+ )) + q + ρe 1/2 σt) τ e q+ρe 2 /8t h ν L 2 0,+ )), when τ 1 and h C 0 [0, + )) with h = 0 on [0, + ). Proof. It follows from Lemma 7 by translation. Lemma 9. There is N = Nn) such that u L 2 B R 0,T/2)) N R 2 + T 1 u L 2 B 4R/3 0,T )), when u satisfies t u + u = 0 in B 4R/3 [0, T ], u = 0 on 4R/3 [0, T ], for some R, T > 0.

19 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 19 Proof. Let ψ C0 R n ) and α C R) be such that ψ = 1 in B R, ψ = 0 outside B 7R/6,0 ψ 1, ψ NR 2 ; α = 1 in, T/2], α = 0 in [ 3T 4, + ), 0 α 1 and t α N/T. Then 2 u 2 dxdt ψx)αt) + t ) u 2 ) dxdt B R 0,T/2)) = 0,+ ) NR 2 + T 1 ) 0,+ ) t )ψx)αt))u 2 x, t) dxdt B 4R/3 0,T ) u 2 dxdt. ψx)u 2 x, 0) dx Lemma 10. Let 0 < ρ 1 and T > 0. Then, there is N = Nn) such that 4.12) ut) L2 B 3ρ ) 1 2 u0) L 2 B ρ ), when T 4.13) 0 < t min 2, ρ 2 N log N u + L 2 B 4 [0,T ]) ρ u0) L 2 Bρ ) and u verifies t u + u = 0 in B 4 [0, T ], u = 0 on 4 [0, T ]. Here, log + x = max{log x, 0} and 1 0 = +. Proof. Let ψ C 0 B 3ρ ) verify ψ = 1 in B 2ρ, 0 ψ 1 and ρ ψ +ρ 2 D 2 ψ N. Set fx, t) = ux, t)ψx). Then Define f + t f N ρ 2 u + ρ 1 u ) χ B3ρ \B 2ρ, in [0, T ]. Ht) =! f 2 x, t)gx y, t) dx, when t > 0, y B ρ, with Gx, t) = 4πt) n 2 e x 2 4t. We have d dt Ht) = 2 f f + t f) + f 2) Gx y, t) dx N ρ 2 u 2 + u 2) Gx y, t) dx B 3ρ\B 2ρ) Nt n/2 e ρ2 /4t Nρ n e ρ2 /8t B 3ρ B 3ρ ρ 2 u 2 + u 2 dx ρ 2 u 2 + u 2 dx, for 0 < t < T. Integrating the above inequality in 0, t), f 2 x, t)gx y, t) dx u 2 y, 0) Nρ n e ρ2 /8t B 3ρ t 0 ρ 2 u 2 + u 2 dτdx

20 20 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG This, along with Lemma 9 with R = 3ρ, T 2 = t, shows that u 2 x, t)gx y, t) dx u 2 y, 0) B 3ρ Nρ n ρ 2 /t)e ρ2 /8t Nρ n 2 e ρ2 /16t B 4ρ B 4ρ 2t 0 2t 0 u 2 x, τ) dxdτ, u 2 x, τ) dxdτ when y B ρ, 0 < 2t T. Integrating the above inequality for y B ρ and recalling that Gx y, t) dy = 1, for all x R n, t > 0, R n we get u 2 x, t) dx u 2 y, 0) dy Nρ 2 e ρ2 /16t u 2 L 2 B, 4ρ [0,2t]) B 3ρ B ρ when 0 < 2t T. The last inequality shows that 4.12) holds when t verifies 4.13) with N = Nn). Lemma 11. There are ρ 0, 1), N = Nm, ϱ) 1 and θ = θm, ϱ) with θ 0, 1), such that ) θ 4.14) u0) L 2 B ρ ) Ne N/T u ν L 2 8 [0,T ]) u 1 θ L 2 B 8 [0,T ]), when u verifies t u + u = 0 in B 8 [0, T ], u = 0 on 8 [0, T ] The readers can find a similar interpolation inequality to 4.14) in [6, Theorem 4.6] though not with optimal T dependency, so that its application to observability boundary inequalities does not imply optimal cost constants. Proof. Let ψ C0 R n ) and α C R) be such that ψ = 1 in B 4, ψ = 0 outside B 5, 0 ψ 1, ψ + D 2 ψ N; α = 1 in, 1], α = 0 in [2, + ), 0 αt) 1 and t αt) 1. Let τ is a positive number verifying 4 τ min{t, 1}. Take hx, t) = ux, t)ψx + ρe)ατt) in Lemma 8. Here, ρ 0, 1) will be fixed later. Then, h + t h N [1 + τ) u + u ] χ E x, t), with E = B 5 ρe) [0, 2 τ ] \ B 4 ρe) [0, 1 τ ]. Since t e σt) t in 0, 1), where σ is as in Lemma 8, we have t 1/2 σt) τ e x+ρe 2 /8t e τ t 1/2 τ e x+ρe 2 /8t e τ τ τ 1/2, when x, t) E. The later inequality shows that 4.15) t 1/2 σt) τ h + t h)e x+ρe 2 /8t L2 0,+ )) Ne τ τ τ 1/2 1 + τ) u + u L2 B 6 [0, 2 τ ]). From Lemma 9 with R = 6 and T = 2 τ, we get 4.16) 1 + τ) u + u L 2 B 6 [0, 2 τ ]) Nτ u L 2 B 8 [0,T ]).

21 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 21 From 4.15) and 4.16), it follows that 4.17) t 1/2 σt) τ h + t h)e x+ρe 2 /8t L2 0,+ )) e τ τ τ+ 1 2 N u L 2 B 8 [0,T ]), N = Nn). Next, because is Lipschitz, there is a positive number β = βm, ϱ) such that B 2βρ ρe) =. Then Thus q + ρe 1/2 σt) τ e q+ρe 2 /8t 8 e τ t τ e β2 ρ 2 2t ) τ 8 β 2 ρ 2 τ τ, on B 5 ρe) 0, 1). 4.18) q + ρe 1/2 σt) τ e q+ρe 2 /8t h 2 ν L 2 0,+ )) From 4.17) and 4.18) and Lemma 8, it follows that 4.19) for 4 τ τ 1/2 t τ e x+ρe 2 /8t u L 2 B 4 ρe) [0, 1 τ [ ]) N e τ τ τ+1/2 u L2 B 8 0,T )) + min{1, T } and with N = Nm, ϱ, n). Because ) τ 2 8 β 2 ρ τ τ u 2 ν L 2 8 0,T )). ) τ 2 β 2 ρ τ τ u 2 ν L 2 8 0,T )) B 3ρ [ ρ2 2τ, ρ2 τ ] B 4ρ ρe) [ ρ2 2τ, ρ2 τ ] B 4ρ ρe) [0, 1 τ ], we have τ 1/2 t τ e x+ρe 2 /8t τ τ+1/2 ρ 2τ e 4τ, for x, t) B 4ρ ρe) [ ρ2 2τ, ρ2 τ ]. Hence, the left hand side of 4.19) verifies 4.20) τ 1/2 t τ e x+ρe 2 /8t u L 2 B 4 ρe) [0, 1 τ ]) τ τ+1/2 ρ 2τ e 4τ u L2 B 3ρ [ ρ2 2τ, ρ2 ]). τ Also, from Lemma 10, when u L 2 B 3ρ [ ρ2 2τ, ρ2 τ ]) ρ 8 τ 1/2 u0) L2 B ρ ), 1 τ min T, 1 N log + N u L 2 B8 [0,T ]) This, along with 4.19) and 4.20), shows that ρ u0) L 2 Bρ ) «τ τ ρ 2τ e 4τ ρ u0) L2 B ρ ) [ N e τ τ τ+1/2 u L 2 B 8 [0,T ]) + when ) τ min Thus, there is N = Nm, ϱ, n) such that. 1 4, T 4, 1 N u L «N log + 2 B8 [0,T ]) ρ u0) L 2 Bρ ) ], ) τ 2 β 2 ρ τ τ u 2 ν L 2 8 0,T )). 4.22) ρ u0) L 2 B ρ ) e 7 ρ 2) τ N 2 u L 2 B 8 [0,T ]) + N τ u ν L 2 8 [0,T ]), ],

22 22 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG when τ satisfies 4.21). Fix now ρ so that e 7 ρ 2 = e 1 and choose ) 4.23) τ = T + N log+ N u L 2 B8 [0,T ]) ρ u0). L 2 Bρ ) Clearly, τ verifies 4.21). Moreover, e 7 ρ 2) τ N 2 u L 2 B 8 [0,T ]) = e τ N 2 u L 2 B 8 [0,T ]) ρ u0) L 2 B ρ ) ρ 2 u0) L 2 B ρ ). This, together with 4.22) shows that From 4.23), it follows that ρ u0) L 2 B ρ ) 2N τ u ν L 2 8 [0,T ]). N τ = e τ log N = e 4+ 4 T ) log N and then we get that [ ρ u0) L2 B ρ )] 1+N log N In particular, N u L 2 B8 [0,T ]) ρ u0) L 2 Bρ ) ) ) ρ u0) L2Bρ ) ) N log N e 4+ 4 T ) log N N log N N u L2 B 8 [0,T ])) u N log N Ne N T u L2 B 8 [0,T ])) u ν L 2 8 [0,T ]). ν L 2 8 [0,T ]) u0) L 2 B ρ ) ) θ Ne N/T u ν L 2 8 [0,T ]) u 1 θ L 2 B 8 [0,T ]), θ = N log N., By translation and rescaling, Lemma 11 is equivalent to the following: Lemma 12. Let be a Lipschitz domain with constants m, ϱ), 0 < R 1, q, and T > 0. Then, there are ρ 0, 1], N = Nm, ϱ, n) 1 and θ = θm, ϱ, n), with θ 0, 1), such that u0) L 2 B ρr q) ) Ne NR2 /T R 1 2 u ν L 2 8R q) [0,T ])) θ R 1 u L2 B 8R q) [0,T ])) 1 θ, when u verifies t u + u = 0 in B 8R q) [0, T ] and u = 0 on 8R q) [0, T ]. Proof of Theorem 10. It suffices to prove Theorem 10 when t 2 = T, t 1 = 0 and q = 0. From Theorem 6 or 2.8), there are N = N, R) and 0 < θ 1 < 1, θ 1 = θ 1, R) such that ) θ1 4.24) e T f L2 ) Ne N/T e T f L 2 B ρr 0, ρr 2 )) f 1 θ1 L 2 ), f L2 ). 20

23 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 23 with ρ 0, 1) as in Lemma 12. On the other hand, it follows from Lemma 12 that e T f L2 B ρr ) ) Ne N/T ν et t) θ2 f L2 B 8R [0,T ]) e T t) f 1 θ2 L 2 B 8R [0,T ]). In particular, e T f L 2 B ρr ) ) Ne N/T θ2 ν et f L 2 8R [0,T ]) f 1 θ 2 L 2 ). This, together with 4.24) and B ρr 0, ρr 20 2 ) B ρr, leads to the desired estimate. Remark 9. It follows from Theorem 10 that there are constants N = N, R, n) and θ = θ, n), θ 0, 1), such that ) θ 4.25) e T f L 2 ) Ne N/[ɛ2 ɛ1)t ] ν et f L 2 R q) [ɛ 1T,ɛ 2T ]) f 1 θ L 2 ), when f L 2 ) and 0 < ɛ 1 < ɛ 2 < 1. Lemma 13 below is a rescaled and translated version of [3, Lemma 2]. Lemma 13. Let f be analytic in [a, a+l] with a R and L > 0, F be a measurable set in [a, a + L]. Assume that there are positive constants M and ρ such that 4.26) f k) x) Mk!2ρL) k, for k 0, x [a, a + L]. Then, there are N = Nρ, F /L) and γ = γρ, F /L) with γ 0, 1), such that γ f L a,a+l) N f dτ) M 1 γ. Lemma 14. Let q 0 and J R q 0 ) 0, T ) be a subset with J > 0. Set J t = {x : x, t) J}, E = {t 0, T ) : J t J /2T )}, t 0, T ). Then, J t R q 0 ) is measurable for a.e. t 0, T ), E is measurable in 0, T ), E J /2 R q 0 ) ) and χ E t)χ Jt x) χ J x, t) over 0, T ). Proof. From Fubini s theorem, T J = J t dt = J t dt + 0 E F [0,T ]\E J t dt R x 0 ) E + J /2. Theorem 11. Suppose that verifies the condition 1.4). Assume that q 0 and R 0, 1] such that 4R q 0 ) is real-analytic. Let J be a subset in R q 0 ) 0, T ) of positive surface measure on 0, T ), E and J t be the measurable sets associated to J in Lemma 14. Then, for each η 0, 1), there are N = N, R, J /T R q 0 ) ), η) and θ = θ, R, J /T R q 0 ) ), η) with θ 0, 1), such that the inequality t2 ) θ 4.27) e t2 f L 2 ) Ne N/t2 t1) χ E t) ν et f L1 J t) dt e t1 f 1 θ L 2 ), t 1

24 24 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG holds, when 0 t 1 < t 2 T with t 2 t 1 < 1, E t 1, t 2 ) ηt 2 t 1 ) and f L 2 ). Moreover, 4.28) e N+1 θ t 2 t 1 N t2 t 1 e t2 f L 2 ) e N+1 θ qt 2 t 1) e t1 f L 2 ) χ E t) ν et f L 1 J t) dt, when q N+1 θ N+1. Proof. Because 4R x 0 ) is real-analytic, according to Lemma 6, there are N = Nϱ, δ) and ρ = ρϱ, δ), 0 < ρ 1, such that 4.29) x α β t e t fx) N t n s) 4 e 8R2 /t s) α! β! Rρ) α t s) /4) β e s f L 2 ), when x 2R x 0 ), 0 s < t, α N n and β 0. Let 0 t 1 < t 2 T with t 2 t 1 < 1, J be a subset in R x 0 ) 0, T ) with positive surface measure in 0, T ). Let J t and E be the measurable sets associated to J in Lemma 14. Assume that η 0, 1) satisfies Define Then, we have E t 1, t 2 ) ηt 2 t 1 ). τ = t 1 + η 20 t 2 t 1 ), t 1 = t 1 + η 8 t 2 t 1 ), t 2 = t 2 η 8 t 2 t 1 ), τ = t 2 η 20 t 2 t 1 ). t 1 < τ < t 1 < t 2 < τ < t 2 and E t 1, t 2 ) 3η 4 t 2 t 1 ). Taking T = t 2 t 1 This T is different from the time T in Theorem 11 and only used here), ɛ 1 = η 20, ɛ 2 = 1 η 20 and replacing f by et1 f in 4.25), we get 4.30) e t2 f L 2 ) Ne N/t2 t1) ν et f θ L 2 R x 0) τ, τ)) et1 f 1 θ L 2 ), with N = N, R, η) and θ = θ) 0, 1). From 4.29) with β = 0, α = 1 and s = t 1, there is N = N, R, η) such that 4.31) ν et f L 2R x 0) τ, τ)) Ne N/t2 t1) e t1 f L2 ). Next, 4.30), the interpolation inequality and 4.31) yield ν et f L2 R x 0) τ, τ)) ν et f 1/2 L 1 R x 0) τ, τ)) ν et f 1/2 L R x 0) τ, τ)) 4.32) e t2 f L2 ) Ne N/t2 t1) e t1 f 1 θ/2 L 2 ) ν et f θ/2 L 1 R x 0) τ, τ)), with N = N, R, η) and θ = θ) 0, 1). Next, setting vx, t) = ν et fx), for x 2R x 0 ), t > 0, we have 4.33) v L1 R x 0) τ, τ)) τ τ) R x 0) vx, ) L τ, τ) dσ.

25 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 25 Write [τ, τ] = [a, a + L], with a = τ and L = τ τ = 1 η 10 )t 2 t 1 ). Also, 4.29) with α = 1 and s = t 1 shows that there is N = N, R, η) such that for each fixed x 2R x 0 ), t [τ, τ] and β 0, 4.34) with β t vx, t) NeN/t t1) β! t t 1 ) /4) β et1 f L 2 ) NeN/t2 t1) β! ηt 2 t 1 )/80) β et1 f L 2 ) = Mβ! 2ρL) β, M = Ne N/t2 t1) e t1 f L2 ) and ρ = η η). From 4.34), Lemma 13 with F = E t 1, t 2 ) and observing that 15η 2 10 η) F 5 4 η) L 2 10 η), we find that for each x in 2R x 0 ) 4.35) ) γ 1 γ vx, ) L τ, τ) vx, t) dt Ne N/t2 t1) e t1 f L2 )), E t 1, t 2) with N = N, R, η) and γ = γη) in 0, 1). Thus, it follows from 4.33) and 4.35) that 4.36) v L 1 R x 0) τ, τ)) R x 0) E t 1, t 2) γ ) 1 γ vx, t) dt) dσ Ne N/t2 t1) e t1 f L2 ) E t 1, t 2) ) γ vx, t) dσdt Ne N/t2 t1) e t1 f 1 γ R x 0) L 2 ), with N and γ as above. In the second inequality in 4.36) we used Hölder s inequality. Next, from 4.29) with β = 0 and s = t 1, we have that for t t 1, t 2 ), and t t 1 t 1 t 1 = η 8 t 2 t 1 ) α x v, t) L 2R x 0)) NeN/t2 t1) α! Rρ) α e t1 f L2 ), when α N n and with N = N, R, η). Also, J t J / 2T ), when t E. By the obvious generalization of Theorem 4 to real-analytic hypersurfaces, there are N = N, R, η, J / T R x 0 ) )) and ϑ = ϑ, J / T R x 0 ) )) 0, 1) such that ) ϑ ) 1 ϑ 4.37) vx, t) dσ vx, t) dσ Ne N/t2 t1) e t1 f L2), R x 0) J t

26 26 J. APRAIZ, L. ESCAURIAZA, G. WANG, AND C. ZHANG when t E t 1, t 2 ). Both 4.36) and 4.37), together with Hölder s inequality, imply that ) ϑγ v L 1 R x 0) τ, τ)) Ne N/t2 t1) vx, t) dσdt e t1 f 1 ϑγ L 2 ). E t 1, t 2) J t This, along with 4.32) and the definition of v leads to the first estimate in this theorem. The second estimate in the theorem can be proved with the method we used in the proof of the second part of Theorem 7. Proof of Theorem 2. Let E and J t be the sets associated to J in Lemma 14 and l be a density point in E. For z > 1 to be fixed later, {l m } denotes the sequence associated to l and z in Lemma 2. Because of 2.16) and from Theorem 11 with η = 1/3, t 1 = l m+1 and t 2 = l m, with m 1, there are N = N, R, J / T R q 0 ) )) > 0 and θ = θ, R, J / T R q 0 ) )), with θ 0, 1), such that e N+1 θ lm l m+1 e l m f L 2 ) e lm N N+1 θ qlm l m+1 ) e lm+1 f L 2 ) χ E s) ν es f L1 J s) ds, when q N + 1 θ l m+1 N + 1 and m 1. Let z = N + 1 ). 2 N + 1 θ Then, we can use the same arguments as those in the proof of Theorem 1 to verify Theorem 2. Remark 10. The proof of Theorem 2 also implies the following observability estimate: sup sup m+1a e t f L2 ) N ν et fx) dσdt, e z m 0 l m+1 t l m J [l,l 1]) for f in L 2 ), with A = 2N + 1 θ) 2 /[θl 1 l)] and with z, N and θ as given along the proof of Theorem 2. Here, l 0 = T. Remark 11. i) In Theorem 2, one may relax more the hypothesis on and to allow 4R q 0 ) to be piece piecewise analytic or simply to require that {q 4R q 0 ) : 4r q) is not real-analytic for some r R} = 0. ii) In particular, Theorem 2 holds when is a Lipschitz polyhedron in R n and J is a measurable subset with positive surface measure on 0, T ). For if is a Lipschitz polyhedron, Theorem 8 shows that verifies the condition 1.4). Also, J must have a boundary density point q, τ), q, τ 0, T ), with q in the interior of one the open flat faces of. Thus, we can find R > 0 such that R q) τ R, τ + R) J R q) τ R, τ + R) 1 2, with 4R q) contained in a flat faces of. Then, replace the original set J by R q) τ R, τ + R) J R q) 0, T ), set q 0 = q and apply Theorem 2 as stated. iii) Theorem 2 improves the work in [35].

27 OBSERVABILITY INEQUALITIES AND MEASURABLE SETS 27 Remark 12. When J = Γ 0, T ), Γ R q 0 ) is a measurable set, one may take l = T/2, l 1 = T, z = 2 and the constant B in Theorem 2 becomes B = A, R, Γ / R q 0 ) )/T. Remark 13. Theorem 10 also implies the following: if is only a bounded Lipschitz domain and verifies the condition 1.4), the heat equation is null controllable with L Γ 0, T )) controls, when Γ is an open subset of. This seems to be a new result and we give its proof in the Appendix in section Applications Throughout this section, we assume that T > 0, is a bounded Lipschitz domain verifying the condition 1.4) and we show several applications of Theorem 1 and Theorem 2 to some control problems for the heat equation. First of all, we will show that Theorems 1 and 2 imply the null controllability with controls restricted over measurable subsets in 0, T ) and 0, T ) respectively. Let D be a measurable subset with positive measure in B R x 0 ) 0, T ) with B 4R x 0 ). Let J be a measurable subset with positive surface measure in R q 0 ) 0, T ), where q 0, R 0, 1] and 4R q 0 ) is real-analytic. Consider the following controlled heat equations: t u u = χ D v, in 0, T ], 5.1) u = 0, on [0, T ], u0) = u 0, in, and 5.2) t u u = 0, in 0, T ], u = g χ J, on [0, T ], u0) = u 0, in, where u 0 L 2 ), v L 0, T )) and g L 0, T )) are controls. We say that u is the solution to Equation 5.2) if v u e t u 0 is the unique solution defined in [14, Theorem 3.2] See also [4, Theorems 8.1 and 8.3]) to t v v = 0, in 0, T ), v = gχ J, on 0, T ), v0) = 0, in, with g in L p 0, T )) for some 2 p. From now on, we always denote by u ; u 0, v) and u ; u 0, g) the solutions of Equations 5.1) and 5.2) corresponding to v and g respectively. Corollary 1. For each u 0 L 2 ), there are bounded control functions v and g with v L 0,T )) C 1 u 0 L2 ), g L 0,T )) C 2 u 0 L 2 ), such that ut ; u 0, v) = 0 and ut ; u 0, g) = 0. Here C 1 = C, T, R, D) and C 2 = C, T, R, J). Proof. We only prove the boundary controllability. Let E be the measurable set associated to J in Lemma 14. Write J = {x, t) : x, T t) J} and Ẽ = {t : T t E}.