Potential Analysis https://doi.org/10.1007/s11118-023-10071-y Removable Singularities for Solutions of the Fractional Heat Equation in Time Varying Domains Joan Mateu1 Laura Prat1 Received: 25 May 2022 / Accepted: 13 February 2023 © The Author(s) 2023 Abstract In this paper we study removable singularities for solutions of the fractional heat equation in the spacial-time space. We introduce associated capacities and we study some of its metric and geometric properties. Keywords Fractional heat equation Capacity Removability Mathematics Subject Classification (2010) 35K55 42B20 31C45 28A75 1 Introduction In the present paper we study removable singularities for solutions of the fractional heat equation in time varying domains. Our main motivation comes from the paper [13] where we studied removable singularities for regular 1 1 2 Lipschitz solutions of the heat equation in time varying domains. Our ambient space is 1 with a generic point denoted as 1 , where and . For a smooth function depending on 1 , the heat equation is just 0 (1.1) 2 2 with fundamental solution 4 exp 4 for 0 and 0 if 0. We set 0 1 and let denote the fractional heat operator, . Then, for a smooth function depending on 1, 0 Laura Prat laurapb@mat.uab.cat Joan Mateu mateu@mat.uab.cat 1 Departament de Matemàtiques, Universitat Autònoma de Barcelona, Bellaterra, 08193, Catalonia, Barcelona, Spain J. Mateu, L. Prat is just the fractional heat equation. The pseudodifferential operator ,0 1, is the fractional Laplacian with respect to the variable. It may be defined through 2 its Fourier transform, , or by its representation p.v. 2 where is a normalization constant (see [10] and [17] for its basic properties). and p.v. stands for the principal value. The standard Laplace operator, , is recovered when taking the limit 1 (see [4, Section 4]), but there is a big difference between the local operator , that appears in the classical heat equation and represents Brownian motion, and the non-local family ,0 1, which are generators of Lévy processes in Stochastic PDEs (see [1] i [2]). Given and , with , , we consider the -parabolic distance in 1 defined by 1 2 dist max . Sometimes we also write instead of dist . Notice that for 1 2, dist is comparable to the usual distance. We denote by an -parabolic ball (i.e., in the distance dist ) centered at with radius . By an -parabolic cube of side length , we mean a set of the form 1 1 where 1 are intervals in with length , and 1 is another interval with length 2 . We write . We say that a Borel measure in 1 has upper parabolic growth of degree 2 1, 0 1, if there exists some constant such that 2 1 1 for all 0. (1.2) Clearly, this is equivalent to saying that any -parabolic cube 1 satisfies 2 1 . Now, let 1 2. We say that a compact set 1 is removable for the 1 2 fractional heat equation (or removable for bounded 1 2 caloric functions) if any bounded function 1 , i.e. 1 , satisfying the 12 fractional heat equation in 1 , also satisfies the 1 2 fractional heat equation in the whole space 1. Given a compact set 1 , we define its 1 2 fractional caloric capacity by 1 2 sup 1 supp and 1 1 (1.3) where is the space of distributions in 1 and denotes the fundamental solution of the 1 2 fractional heat equation in 1, that is 1 if 0 2 2 2 . (1.4) 0 if 0 The fractional heat equation can be solved (by applying Fourier transform) for all 0 1 by means of the fundamental solution , which is the inverse transform of the function 2 . It is worth mentioning that only in the particular cases 1 and 1 2 the kernel is known to be explicit. In the first case we get the Gaussian kernel for the standard heat 2 2 4 equation 4 and in the fractional case 1 2 we have Eq. 1.4. Removable Singularities for Solutions of the Fractional Heat... For 0 1, there is no explicit fundamental solution for the fractional heat equation. In the 1960s, the probabilists Blumenthal and Getoor [3, Theorem 2.1] generalized in a precise way the power-like tail behaviour observed in the case 1 2 to the other values of 0 1, that is, they established that the fundamental solution, , of the fractional heat equation, for 0 1, satisfies 2 1 2 2 when 0 (and 0 for 0). Here means that is uniformly bounded above and below by a constant. Notice the marked difference with the Gaussian kernel of the heat Eq. 1.1 (case 1). The behaviour as tends to infinity of is power-like while has exponential spatial decay. In the paper, we will also introduce and study some -fractional caloric capacities , 0 1, associated with the kernels of the -fractional heat equation. We shall now give a brief description of the main results in the paper. Section 2 includes some estimates of the kernels (0 1) and its derivatives, that will be needed in the rest of the paper. The four next sections deal with the case 1 2. More concretely, in Section 3 we prove a localization result for 1 2, that is, for a distribution , we localize the potential in the norm. We think that the localization property for the potential ,0 1, should also hold, but the arguments would be much more involved and we don’t have a proof of it. The localization method for the Cauchy potential 1 in the plane is a basic tool developed by A.G. Vitushkin in the theory of rational approximation in the plane. This was later adapted in [14] for Riesz potential in and used in problems of 1 harmonic approximation. These localization results have also been essential to prove the semiadditivity of analytic capacity and of Lipschitz harmonic capacity, see [19] and [22] respectively (see also [15] for other related capacities). In Section 4 we study the connection between 12 fractional caloric removability and 1 2 1 is 1 fractional the capacity . In particular, we show that a compact set 2 1 2 1 2 caloric removable if and only if 0. We also compare the capacity to the Hausdorff content and we prove that if has zero dimensional Hausdorff measure, 1 2 i.e., 0, then 0 too. In the converse direction, we show that if has 1 2 Hausdorff dimension larger than , then is positive. Hence, the critical dimension 1 1 for 2 fractional caloric capacity (and thus for 2 fractional caloric removability) occurs in dimension , in accordance with the classical case. Also by analogy with the classical case, 1 2 1 2 one should expect that 0 if and only if 0 or even that both capacities 1 2 are comparable (the definition of is the same as (1.3) but instead of distributions one considers positive measures). However, there is a big obstacle when trying to follow this approach. Namely, the kernel is not antisymmetric and thus, if is such that is in 1 , apparently one cannot get any useful information regarding ( being the dual operator to ). This prevents any direct application of the usual 1 or theorems from Calderón-Zygmund theory, which are essential tools in the case of analytic capacity and Lipschitz harmonic capacity. Hence, In Section 4, due to the lack of antisymmetry of 1 2 1 2 our kernel , we also introduce a new capacity . We set sup the supremum taken over all positive measures supported on with 1 1 and 1 2 1 2 1 2 1 2 1 1. Clearly . We show that the capacity can be characterized in terms of the 2 -norm of . J. Mateu, L. Prat In Sections 5 and 6, we give some concrete examples of 12 caloric removable and non- removable sets with Hausdorff dimension . In particular, in the first section we construct a self similar Cantor set in 1 with positive and finite Hausdorff -dimensional measure which is 12 caloric removable and in the second one we show that on the plane, the capacity 1 2 vanishes on vertical segments and is positive on horitzontal ones. These examples allow us to deduce that, in dimension two, neither analytic capacity nor newtonian capacity 1 2 are comparable to the 1 2 fractional caloric capacity (although these three capacities have the same critical dimension). Sections 7 and 8 are devoted to the study of the fractional capacities , when 1 2 1 and 0 1 2 respectively. In the first case, namely 1 2 1 we are able to show that the critical dimension for the -capacity is 2 1. We have had several technical problems when trying to prove this statement for the case 0 1 2 but we can show that sets with zero 2 1 dimensional Hausdorff measure, i.e., 2 1 0, have 0. Some comments about the notation used in the paper: as usual, the notation means that there is a positive absolute constant (that may depend on and the dimension ) such that . Also, is equivalent to . The gradient symbol refers to , with and . 2 Some Preliminary Estimates on the Kernels P and Ps In the next two lemmas we will obtain upper bounds for the kernels , and its derivatives. We will need them later. Lemma 2.1 For any 1, and , the following holds: 1 (a) for all 0. 1 1 (b) 1 for all 0 and 1 for all 0. (c) For all 1 such that 2, 0, 1 . Notice that the kernel P is not differentiable with respect to at the points 0 . Proof The estimate in (a) comes from the explicit expression of . For the estimates in (b) we compute 1 3 2 2 2 1 and so we get 1 . On the other hand, for 0 1 1 2 1 3 . 2 2 2 2 2 2 Removable Singularities for Solutions of the Fractional Heat... 1 Then 1 for all 0 . Finally, (c) will be obtained using (b). Indeed, given 1 such that 2, write . Then (2.1) The first term in the above inequality is bounded by sup 1 1 . For the second term in Eq. 2.1 let . If 0, then sup 1 1 . If 0, 0 and if 0 and 0, then 0 sup 1 . 0 The following lemma shows some growth properties of our kernel . Before stating it, we need to introduce a definition. For a function 1 and 0 1 , the fractional derivative with respect to (recall that and ) is defined as 1 when the integral is finite. Now the lemma reads as follows Lemma 2.2 Let 0 1 . For any 1, and , the following holds: 1 (1) for all 0. 1 (2) For 0 1, 2 . 1 (3) 2 for all 0. 1 (4) 1 for all 0. 1 1 1 2 1 (5) For 2 1, 1 2 . J. Mateu, L. Prat Proof The first property follows from the definition of the kernel. For 0, 0 and for 0, 1 2 1 1 2 1 2 . 2 2 2 2 We prove now the second one. Applying Fourier transform to the fractional heat equa- tion with respect to the space variable , and calling the new variable , we get the equation 2 , that allows to solve the initial-value problem in Fourier space by means 2 of the formula 0 . Applying the inverse transform, the fractional heat equation can be solved for all 0 1 by means of the fundamental solution, , 2 which is the inverse transform of the function . Hence the kernel has Fourier trans- 2 form . It is well known that for 0 (recall that for 0 0) it has the self-similar form 1 2 2 (2.2) 2 for some positive function , radially decreasing and satisfying 1 2 2 (see [3] and [21]). Using Eq. 2.2 we deduce that 1 2 2 2 2 2 where . Since , then and using the expression of the inverse Fourier transform of a radial function (see [7, Section B.5] or [18, Section IV.I] for a proof) we obtain 1 2 2 2 2 2 1 0 where is the classical Bessel function of order . Thus [16, Lemma 1] gives us the decay 2 , for large. Since is bounded we have 2 2 1 2 (2.3) which implies 1 1 2 2 1 2 2 2 1 2 the second estimate in the statement of the lemma. Observe that for this is 2 . (2.4) Therefore using that is the fundamental solution of the -fractional heat equation 1 1 2 2 0 (2.5) and the fact that 2 2 1 2 (2.6) (which is estimate Eq. 2.3 with ) we get 1 1 1 2 1 2 2 2 1 2 which is the third estimate in the statement of the lemma. Removable Singularities for Solutions of the Fractional Heat... Next we will estimate the spatial derivative . Clearly 1 1 1 2 2 2 2 . If we can show that 1 2 1 2 (2.7) then we are done because 1 2 1 1 1 . 1 2 1 2 In order to estimate , we consider the equation for that comes from Eq. 2.5, that is 2 0. Notice that it implies that . 2 Since is bounded, from 1 2 2 and Eq. 2.6 we deduce Eq. 2.7. To prove the last estimate in the statement of the lemma, that is 1 1 2 1 1 2 1 2 1 (2.8) we claim that for 0, 1 2 (2.9) for some function 2 . Eq. 2.9 can be proved using Eq. 2.2, that is 1 1 2 1 2 2 2 for some with 1 2 2 . In fact, 1 1 1 2 2 1 1 2 2 2 2 1 2 the last equality being a definition for . Hence 1 2 2 2 2 2 2 2 1 1 2 2 1 1 2 1 1 2 and claim Eq. 2.9 is proved. Notice that we have 1 1 2 1 1 1 2 1 1 1 2 2 2 1 2 . Hence, to show Eq. 2.8 it is enough to see that for all , 1 1 2 1 min 1 . (2.10) Once Eq. 2.10 is available we obtain Eq. 2.8 easily: 1 1 2 1 1 1 2 2 1 min 1 2 1 1 2 . max J. Mateu, L. Prat To show Eq. 2.10, we write 1 1 2 1 1 2 2 2 2 2 1 1 2 2 2 2 2 2 2 1 2 3. We begin with 1. Notice that here . Then, 1 1 1 11 12 . 2 2 2 2 If 1, then 1 11 1 1 2 1 1 2 2 2 2 0 1 2 2 0 1 2 1 2 1. If 1, then we can write 1 1 1 11 1 1 2 1 1 2 . 2 2 2 0 1 2 2 1 1 2 Since 1 1 1 1 1 1 2 2 1 2 1 1 2 1 1 1 2 2 2 2 2 1 1 2 2 1 2 2 then if 1, 1 1 1 11 1 . 2 2 1 Hence 11 min 1 . Since 0 for 0, to estimate 12 we only need to consider positive , therefore 1 2 2 1 12 1 1 1 2 1 2 min 1 . 1 2 2 2 1 2 1 2 This finishes the estimate of 1 . To deal with 2 , we distinguish two cases, according to whether has the same sign as or not. In the first case we write , and in the second one, . In the case , with 2 2 , it turns out that , and thus 1 2 2 1 1 1 2 1 . 2 2 1 2 2 2 2 0 1 2 2 Observe that this last expression is very similar to the ones in 11 and 12 . Then, by almost the same arguments we deduce that 1 2 min 1 . Removable Singularities for Solutions of the Fractional Heat... To deal with the case when the sign of is the same as the one of (i.e., ), we take into account that sup . Notice that Eq. 2.9 tells us that if 1, then . Hence due to property (3) of this lemma and since in this case 2 2 , it is immediate to check that for this we have 1 1 2 . 1 2 Thus, 2 2 2 2 21 1 1 2 2 1 21 1 2 1 2 1 1 2 min 1 . 1 2 Finally we will deal with 3. 3 1 1 31 32 . 2 2 2 2 2 2 Notice that 2 . If 1, then 1 31 1 1 2 1 1 2 . 1 1 2 1 2 1 2 1 2 Since 1 1 1 1 1 2 1 1 2 1 1 1 2 1 2 2 and 1 1 2 1 1 1 2 1 2 we have 31 1 when 1. If 1, then 1 1 31 1 1 2 1 1 . 1 2 1 2 1 2 2 2 1 Hence 31 min 1 . Finally 1 2 1 32 1 2 1 1 2 min 1 . 2 1 2 2 1 2 3 Localization Estimates for s 1 2 The main objective of this section is to show the following localization result, that will be used for the proof of Theorem 4.2. J. Mateu, L. Prat Theorem 3.1 Let be a distribution in 1 with 1. Let be a 1 function supported on a cube 1 such that 1. Then 1. In the statement the operator refers to . Before proving Theorem 3.1 we need several lemmas and definitions. We say that a 1 function is admissible for 1 if it is supported on 1 and satisfies 1 2 and . (3.1) 1 Recall that 1 2 1 with , 1 , being the Riesz transforms with Fourier multiplier (here is a constant depending on ). Then setting 1 with 1 and , one can write 1 2 1 1 . 1 Therefore if satisfies and 1 for 1 then is admissible for . Lemma 3.2 Let be a 1 function supported on a cube 1 1, with 1 and , and such that (1) 1 . (2) 1 2 1 , for 1 2 . Then is admissible for . Proof Due to the above explanation, we only have to show that 1 2 1 . In fact, 1 2 1 1 2 1 1 1 2 1 1 1 1 where the forelast inequality is obtained by integrating on annulus, for example. Removable Singularities for Solutions of the Fractional Heat... Corollary 3.3 A 1 function supported on a cube 1 with 1 is admissible for . Proof The first condition in Lemma 3.2 is clearly fulfilled. To show the second condition in Lemma 3.2 we will use the Cauchy-Schwarz inequality and the 2 boundedness of the Riesz transforms as follows: 2 1 2 2 2 1 1 2 2 2 1 Therefore, standard test functions supported on are admissible for . It is worth mentioning here that if is a standard 2 test function supported on 1 satisfying 1 and 2 , then is admissible for . The first condition in Eq. 3.1 is clear and for the second one, notice that if , 1 with 1 and denoting the convolution on the variable, then taking the Fourier transform with respect to , we get: 1 2 , for a suitable constant 0. Then, integrating on annuli and using that 2, 1 2 1 1 1 2 1 1 1 (3.2) 2 1 1 1 2 1 . 2 1 The following lemma shows an growth condition that every distribution fulfilling the hypothesis of Theorem 3.1 satisfies. Lemma 3.4 Let be a distribution in 1 with 1. If is a 1 function admissible for 1 , then . Proof Since is the fundamental solution of 1 2, we can write 1 2 1 2 because 1 and is admissible for , so it satisfies Eq. 3.1. We will say that a distribution has -growth if for any 1 and any 1 admissible for , . Lemma 3.5 Let be a distribution in 1 with -growth. If is a 1 function supported on a cube 1 with 1 , then the distribution is a locally integrable function and there exists a point 0 14 such that 0 1. (3.3) J. Mateu, L. Prat Proof We will show that the mean of on 14 is bounded by a constant. Hence at many Lebesgue points of the inequality Eq. 3.3 holds. We only need to show that is integrable on 2 . In fact, we will prove a stronger 1 statement, namely that is in 2 for each 1 . Indeed, fix any 1 satisfying 1 and let be its dual exponent, so that 1 . We need to estimate the action of on test functions supported on 2 in terms of . We clearly have . We claim that the test function 1 (3.4) satisfies the inequalities in Lemma 3.2 and hence it is and admissible function. Once this is proved we get 1 1 and so 1 2 . Hence 1 4 1 1 1 4 4 1 1 1 4 which completes the proof of the lemma. To prove the claim, we need to show that satisfies the inequalities in Lemma 3.2, that is 1 2 and 1 2 1 1 . Or equivalently 1 1 1 and 1 2 1 (3.5) for 1 . For the first inequality in Eq. 3.5 apply Hölder’s inequality to obtain 1 1 2 2 1 2 2 1 1 2 . Removable Singularities for Solutions of the Fractional Heat... Since 1 and by Hölder again, 1 1 1 2 2 1 1 2 2 2 1 1 1 1 2 2 . Applying [5, Theorem 4.12], for example, we deduce that the singular integral with ker- nel is bounded in 1 , 1 . Therefore 2 2 . The estimate of the norm of in 2 is analogous. Finally 2 2 so the first inequality in Eq. 3.5 is proven. For the second one write 3 4. 2 2 2 Using Hölder, the boundedness of the Riesz transform and arguing similar to what we have just done for the term 1 , 1 1 3 2 1 1 1 2 1 1 1 1 2 . 2 2 Analogously, 1 1 1 4 2 2 by an argument similar to the one of 2. 3 Proof of Theorem 3.1 Take 2 and write . We have to show that 1. Write . Let be a 1 function such that 1 in 2 , 0 in 4 and 1. We need to resort a standard regularization process. Take 0 1 such that 1 and set and . We want to estimate (3.6) uniformly on and . Since, as tends to zero, Eq. 3.6 tends to for almost all 1 , this allows the transfer of uniform estimates. 1 . J. Mateu, L. Prat To estimate term , we will show that is admissible for 4 and then apply Lemma 3.4. Since the function satisfies 1 , for we have and by the mean value theorem and Lemma 2.1, 1 we have 1 1 4 4 which is condition 1 in Lemma 3.2. To show 2 in Lemma 3.2 for the function we have to prove 1 8 1 1 2 . (3.7) 4 Applying Hölder’s inequality for some 1 to be chosen later and using the - boundedness of the Riesz transform , we have 1 8 1 4 4 8 1 4 . 4 To estimate the last integral, write 1 2. Using that 1 and 1 1 1 2 (3.8) we get 1 1 4 4 4 1 1 1 1 2 4 4 1 1 2 4 Removable Singularities for Solutions of the Fractional Heat... 1 chosing 1 1 2 . Notice that 1 for 0. Therefore, using the mean value theorem and arguing as above we get 2 4 4 1 4 4 1 1 1 1 2 4 4 1 1 2 4 . Since Eq. 3.7 holds, is admissible for 4 and by Lemma 3.4 we get 1. To estimate term 1 we will use Lemma 3.5, i.e. the fact that there exists 0 such that 0 1. Since 1, we clearly have 1 0 . The analogous inequality holds as well for the regularized potentials appearing in , uniformly in , and therefore 1 1 0 . To estimate 1 , we decompose 1 into a union of rings 0 1 1 2 2 and consider functions in 1 1 supported in 1 1 2 2 2 1 and such that 1 3 such that 2 1 in . Since 2 , the smallest ring intersecting 2 is 3 . Hence 1 0 1 0 3 1 0 1 0 being the set of indices 3 such that supp 4 and denoting the remaining indices (i.e. 3 with 0 on 4 ). Notice that the cardinality of is bounded by a dimensional constant. Set 1 0 and for , 2 2 0 . J. Mateu, L. Prat We will show now that we can apply Lemma 3.2 to and . Once this is available, Lemma 3.4 will give us 1 0 2 2 1 2 1. Notice first that the support of is contained in a cube for some depending only on . On the other hand, the support of is contained in 2 2 . To apply Lemma 3.2 we have to show that 1 1 2 1 1 (3.9) and for , 1 2 3 2 1 2 3 1 2 1 2 2 (3.10) We check first Eq. 3.9. Using Lemma 2.1, 1 1 0 0 supp supp 1 1 . supp supp The estimate for 1 is analogous. For the second inequality in Eq. 3.9, let 1 and be its dual exponent. Apply Hölder’s inequality, the fact that Riesz transforms preserve and argue as in the estimates of the integrals of 1 and 2 (using Eq. 3.8) to obtain: 1 2 2 1 1 2 1 . To show the first inequality in Eq. 3.10 we have to prove 1 2 3 2 and 2 ,1 , . For the 1 norm of we will 1 2 3 use 0 2 1 . (3.11) 2 Notice that Eq. 3.11 comes from a gradient estimate and Lemma 2.1. Hence 1 2 2 2 2 2 which is 1 2 3 2 ,1 , . Removable Singularities for Solutions of the Fractional Heat... To show 1 2 3 2 , we consider different cases. Write and 0 0 0 . If 0 and 0 0 a gradient estimate together with Lemma 2.1 gives us 0 2 . 2 If 0 and 0 0 then 0 0 . If and 0 have different signs, say 0 and 0 0 for example, then for , 1 2 0 1 3 1 1 1 2 1 . 2 2 2 Notice that this last case only happens in a set of measure smaller or equal than 2 . Putting this estimates together we get 1 2 1 2 3 2 2 2 2 2 2 1 2 . So the first inequality in Eq. 3.10 holds. Moreover 3 1 2 3 1 2 2 3 1 2 2 2 2 3 2 2 3 1 2 2 1 2 2 2 2 2 which is the second inequality in Eq. 3.10. 3 Take 2 ( ). Then consider 1 because the function is supported on and satisfies 1 1 which, by Corollary 3.3, implies that is admissible for and we may apply Lemma 3.4 to obtain also in this case. 4 Capacities and Removable Singularities Given a compact set 1, we define 1 2 sup 1 (4.1) where the supremum is taken over all distributions supported on such that 1 1. (4.2) J. Mateu, L. Prat 1 2 We call the 12 -fractional caloric capacity of . We also define the capacity 1 2 , in the same way as in (4.1), but with the supremum restricted to all positive measures supported on satisfying (4.2). Clearly, 1 2 1 2 . We say that a compact set 1 is removable for bounded 1 -fractional caloric 2 1 1 functions (or 2 fractional caloric removable) if any bounded function 1 1 satisfying the 2 fractional heat equation in , also satisfies the heat equation in the whole space 1. Denote by the the -Hausdorff content of the compact set 1. Lemma 4.1 For every Borel set 1, 1 2 1 2 and 1 2 dim 0. 1 2 1 2 1 2 Proof Just by the previous definitions, . To prove first notice that we can assume to be compact. Let be a distribution supported on satisfying Eq. 4.2 and let be a collection of sets in 1 which cover and such that diam 2 . For each , let be an open ball centered in with diam , so that . By the compactness of we can assume to be finite. By means of the Harvey-Polking lemma [8, Lemma 3.1], we can construct functions , , satisfying: supp 2 for each , 1 , 1 in , Hence, by Corollary 3.3 and Lemma 3.4 1 diam . 1 2 Since this holds for any distribution supported on satisfying (4.2), . To prove the second assertion in the lemma, let 1 be a Borel set satisfy- ing dim . We may assume to be bounded and we can apply the well known Frostman lemma. Then, it follows that there exists some non-zero positive measure supported on satisfying for all 1 and all 0. Thus, by Lemma 2.1, we deduce that for all 1 1 diam . Therefore, 1 2 0. 1 Removable Singularities for Solutions of the Fractional Heat... We say that a compact set 1 is removable for bounded 1 -fractional caloric 2 1 1 functions (or 2 fractional caloric removable) if any bounded function 1 1 satisfying the 2 fractional heat equation in , also satisfies the heat equation in the whole space 1 . 1 1 Theorem 4.2 A compact set is 2 fractional caloric removable if and only if 1 2 0. 1 1 2 Proof It is clear that if is 2 fractional caloric removable, then 0. Con- versely, suppose that 1is not 12 fractional caloric removable. So there exists some function 1 satisfying 1 and 1 2 0 in 1 but 1 2 0 in 1 (in the distributional sense). Since is compact there exists some (open) cube such that and 1 2 0 in . Consider the distribution 1 2 . Since does not vanish identically in , there exists some function supported on such that 0. Now take . By Theorem 3.1, 1 and thus, since supp , 1 2 1 0. 1 1 From the preceding lemmas, it is clear that, for any compact set 1, if dim , then is not removable for bounded 12 fractional caloric functions, if 0 (in particular if dim ), then is removable for bounded 1 2 fractional caloric functions. 1 1 2 Thus the critical Hausdorff dimension for 2 fractional caloric removability (and for ) is . We consider the operator defined over distributions in 1. When is a finite measure, one can easily check that is defined for -a.e. 1by the integral . For 0, we also consider the truncated operator whenever the integral makes sense, and for a function 1 , we write . J. Mateu, L. Prat We also denote sup sup . 0 0 We say that is bounded in 2 if the operators are bounded in 2 uniformly on 0. Given 1, we define the capacity 1 2 sup (4.3) where the supremum is taken over all positive measures supported on such that 1 1 1 1. (4.4) Here is dual of . That is, . Notice that by definition, 1 2 1 2 1 2 In the next theorem we characterize in terms of the positive measures supported on such that is bounded in 2 . Theorem 4.3 For any set 1, 1 2 sup supp 2 2 1 with the implicit constant in the above estimate independent of . Proof Denote sup supp 2 2 1. 1 2 The arguments to show that are standard. Indeed, let be a posi- tive measure supported on such that 2 , 1 1 and 1 1. By a Cotlar type inequality analogous to the one in [12, Lemma 5.4], say, one deduces that 1 1 uniformly on 0. To obtain the boundedness of the operator in 2 we will use a suitable 1 theorem with respect to a measure which may be non-doubling (see for example [20, Th 3.21]). Since and are bounded in as a consequence of the result in [9], to apply the 1 theorem in our case it is enough to check that the weak boundedness property is satisfied for balls with thin boundaries, that is, 2 , if is a ball with thin boundary. A ball of radius is said to have thin boundary if dist 2 . (4.5) Let’s consider a function with compact support in 2 such that 1 on . Then . Removable Singularities for Solutions of the Fractional Heat... Using Theorem 3.1 one can see that the first term in the right hand side is bounded by . To get a bound of the second term we will use that B has a thin boundary. Using the boundedness of , property (a) in Lemma 2.1, 2 . 0 dist 2 Given and such that dist 2 since is a measure with -growth, one has 2 2. 2 2 1 1 Therefore, by 4.5 2 2 dist 2 2 2 . 2 1 0 Consequently, the weak boundedness property is satified and by the 1 theorem it follows that is bounded in 2 , with 2 2 1. So we deduce that . To prove the converse estimate, let be a positive measure supported on be such that 2 2 1 and 2 . From the 2 boundedness of , one deduces that and are bounded from the space of finite signed measures 1 to 1 . That is, there exists some constant 0 such that for any measure 1 , any 0, and any 0, 1 and the same holds replacing by . The proof of this fact is analogous to the one of Theorem 2.16 in [20]. Then, by a well known dualization of these estimates (essentially due to Davie and Øksendal) and an application of Cotlar’s inequality, one deduces that there exists some function 0 1 such that 1 1 1 1. See Theorem 4.6 and Lemma 4.7 from [20] for the analogous arguments in the case of analytic capacity and also Lemma 4.2 from [12] for the precise vectorial version of the dualization of the weak 1 1 estimates required in our situation, for example. So we have 1 2 . 5 The Existence of Removable Sets with Positive Hausdorff Measure HN In this section we will construct a self-similar Cantor set 1 with positive and finite measure which is 12 caloric removable. To obtain this result we need the following theorem, showing that certain distributions are actually measures. J. Mateu, L. Prat Theorem 5.1 Let 1 be a compact set with and let be a distribution supported on such that 1. Then is a signed measure, absolutely continuous with respect to 1 and there exists a Borel function such that and 1. In what follows, we say that a distribution in 1 has dimensional growth, if for any cube 1 and any 1 function admissible for , . Theorem 5.1 is a consequence of Lemma 3.4 and the following result. Lemma 5.2 Let 1 be a compact set with . Let be a distribution supported on wich has -dimensional growth. Then is a signed measure, absolutely continuous with respect to and there exists a Borel function such that and 1. Proof First we will show that is a signed measure. By the Riesz representation theorem, it is enough to show that, for any function with compact support, (5.1) where is some constant depending on . To prove (5.1), we fix 0 and we consider a family of open cubes , , such that , for all , and . Since is compact, we can assume that is finite. By standard arguments, we can find a family of non-negative functions , , such that each is supported on 2 and is admissible for 2 , for some absolute constant 0, 1 on , and in particular on . Indeed, to construct the family of functions we can cover each cube by a bounded number (depending on ) dyadic cubes 1 with side length 8 and then apply the usual Harvey-Polking lemma ([8, Lemma 3.1] to the family of cubes . We write . For each , consider the function where . We claim that is admissible for 2 , for some absolute constant 0. To check this, just note that is supported on 2 and satisfies 1 . Hence, 1 . Removable Singularities for Solutions of the Fractional Heat... So, by Corollary 3.3, the claim above holds and, consequently, by the assumptions in the lemma, . From the preceding estimate, we deduce that . Since for each , we obtain that . Letting 0, we get which gives (5.1) and proves that is a finite signed measure, as wished. It remains to show that there exists some Borel function such that , with 1. To this end, let be the density of with respect to its variation , so that with 1 for -a.e. 1 . We will show that lim sup 1 for -a.e. 1. (5.2) 0 This implies that for some non-negative function 1. This fact is well known (see Theorem 6.9 from [11]). So to complete the proof of the lemma it suffices to show (5.2) (since then we will have with 1). Notice that, by the Lebesgue differentiation theorem, 1 1. lim 0 for -a.e. 0 Let be a Lebesgue point for with 1, let 0 to be chosen below, and let 0 0 be small enough such that, for 0 0, 1 . Suppose first that 3 2 2 (5.3) and let be some non-negative function supported on 2 which equals 1 on such that is admissible for the smallest cube containing 2 , so that . Now observe that 2 3 2 2 . J. Mateu, L. Prat Thus, if is chosen small enough, we deduce that 2 . Therefore, using again that 1 on , we get . (5.4) To get rid of the doubling assumption (5.3), notice that for -a.e. there exists 1 a sequence of balls , with 0, satisfying (5.3) (we say that the balls are -doubling). Further, we may assume that 2 , for some . The proof of this fact is analogous to the one of Lemma 2.8 in [20]. So for such a point , by the arguments above, we know that there exists some 0 0 such that for 0, assuming also that is a -Lebesgue point for the density . Given an arbitrary 0 0 , let be the smallest integer 2 , and let 2 be the smallest such that the ball 2 is -doubling (i.e., (5.3) holds for this ball). Observe that 2 0 . Then, taking into account that the balls 2 are non-doubling for and applying (5.4) for 2 , we obtain 3 3 2 2 2 2 2 2 . Hence, (5.2) holds and we are done. Next we will construct a self-similar Cantor set 1 with positive and finite mea- 1 sure which is removable for the 2 fractional heat equation. Our example is inspired by the typical planar 1 4 Cantor set in the setting of analytic capacity (see [6] or [20, p. 35], for example). We construct the Cantor set as follows: we let 0 be the unit cube, i.e. 0 0 1 0 1 1 . Next we replace 0 by 2 1 disjoint closed cubes 1 with side length 2 such that they are all contained in 0 and each one contains a vertex of 0 . 1 We proceed inductively: In each generation , we replace each cube of the previous 1 1 generation by 2 1 cubes with side length 2 which are contained in and 1 1 1 located in the same relative position to as the cubes 1 with respect 1 2 to 0 . 1 Notice that in each generation there are 2 1 closed cubes with side length 2 . We denote by the union of all these cubes from the -th generation. By construction, 1 . We let . (5.5) 0 1 It is easy to check that dist 2 for , and if , are contained 1 1 in the same cube , then dist 2 . Taking into account that, for each 0, 2 1 1 1 2 2 1 1 by standard arguments it follows that 0 . Removable Singularities for Solutions of the Fractional Heat... Further, coincides, modulo a constant factor, with the probability measure sup- ported on which gives the same measure to all the cubes of the same generation , that is 2 1 . 1 Theorem 5.3 The Cantor set defined in (5.5) is removable for 2 fractional caloric functions. Proof We will suppose that is not removable and we will reach a contradiction. By Theorem 4.2, there exists a distribution supported on such that 1 0 and 1 1. By Theorem 5.1, is a signed measure of the form 1 where is the probability measure supported on such that 2 1 for all . It is easy to check that (and thus ) has upper growth of degree . Recall that denotes the operator , defined over distributions in 1 . Then, arguing as in [12, Lemma 5.4], it follows that there exists some constant such that for all 1. (5.6) For , let the cube that contains . Then we consider the auxiliary operator sup 1 . 0 By the separation condition between the cubes , the upper growth of , and the condition (5.6), it follows easily that for all , (5.7) for some fixed constant . We will contradict the last estimate. To this end, consider a Lebesgue point 0 (with respect to ) of the density such that 0 0. The existence of this point is guarantied by the fact that 0. Given 0 small enough to be chosen below, consider a cube containing 0 such that 1 0 . Given 1, to be fixed below too, it is easy to check that if is chosen small enough (depending on and on 0 ), then the above condition ensures that every cube contained in such that satisfies 1 0 2 0 . (5.8) 2 Notice also that, writing , since 0 0, 0 . J. Mateu, L. Prat Let be one of the upper corners of (i.e., maximal in ). Since 2 , we have 1 1 1 . 2 2 2 Using the fact that dist , we get 1 2 . 2 1 To estimate from below, recall that our kernel is 1 0 . 2 2 2 Then, by the choice of , it follows that 0 for all . (5.9) We write 1 . 1 1 Taking into account (5.9) and the fact that, for 1, contains a 1 cube such that for all , 1 0 using also (5.8), we deduce 1 1 1 0 1 0 1 Thus, 1 0 . Together with the previous estimate for , this tells us that 1 1 0 2 for some fixed 0. It is clear that if we choose big enough and then small enough, depending on , this lower bound contradicts (5.7), as wished. 6 Neither Analytic Capacity nor Newtonian Capacity are Comparable 1 2 to 1 2 In this section we will show that, in the plane, the capacity is not comparable to analytic capacity nor to Newtonian capacity; two classical capacities related to complex analysis 1 2 and potential theory with critical dimension 1, as . More precisely, we will construct 1 2 two sets, the first one will be a compact set 1 with 1 0 but vanishing Removable Singularities for Solutions of the Fractional Heat... Newtonian capacity. The second example we will be a compact set 2 with positive analytic capacity but removable for the 12 fractional heat equation. First we recall some definitions. For a compact set , the analytic capacity of is defined as sup where the supremum is taken over all analytic functions with 1 in and lim . The Newtonian capacity of a compact set can be defined as 1 sup spt 1 for . It is well-known that if is a segment of length , then 4 and 0 (see [20, Proposition 1.4] and [6, Corollary 3.6] respectively). 1 2 We will prove that if 1 is a horizontal segment then 1 0. On the 1 2 other hand, if 2 is a vertical segment, 2 0. Consequently 1 is not 1 2 1 2 comparable to 1 and 2 is not comparable to 2 . The result reads as follows. Proposition 6.1 Let be a positive number. Set 0 2 0 and 1 0 2 0 . Then 2 1 2 a) 1 0. 1 2 b) 2 0. Proof a) Let 1 be the characteristic function of 1 and consider the positive measure 1 and the function 1 where is the fundamental solution of the 12 -fractional heat equation in 2, that is 2 2 0 . Clearly 0 if 0. For 0 we have 2 2 arctan arctan 0 so that 1 2 1 1 0. 1 2 b) Assume that 2 0. Then there exists a distribution supported on 2 such that 1. Approximating the distribution by signed measures (take for instance the convolution of with approximations of the identity in the variable ), we can assume that there exists a signed measure supported on the segment 0 for some small 0, satisfying 2. Since is a non negative kernel, there exists a positive measure supported on with 0 such that 2. (6.1) J. Mateu, L. Prat To get a contradiction we will show that Eq. 6.1 implies 0. Since, by Lemma 3.4, the measure has linear growth, given 0 we can take such that 0 . Set 0 . By Eq. 6.1, for any 0 there exists 0 0 such that 2 . Hence, since is an interval, there exists a finite number of almost disjoint intervals 0 1 such that and 1 . Consequently, 1 1 1 1 which means 0 and we get the desired contradiction. 7 s-Capacities and s-Growth for s (1 2, 1) Let denote the norm of the parabolic BMO space: sup where the supremum is taken over all parabolic cubes 1, stands for the Lebesgue measure in 1 and is the mean of with respect to . Now we introduce the capacities for 1 2 1 . Given a compact set 1, we define sup 1 (7.1) where the supremum is taken over all distributions supported on satisfying 1 1 1 2 2 1 and 1. (7.2) 1 2 Notice that for 1 2 we obtain . The capacity is called the -fractional caloric capacity of . We set as in Eq. 7.1 but with the supremum restricted to all positive measures supported on satisfying Eq. 7.2. Clearly . We are now ready to describe the basic relationship between the capacity and parabolic Hausdorff content (the -dimensional parabolic Hausdorff content will be denoted by ). Theorem 7.1 Let 1 2 1 . For every Borel set 1, 2 1 Removable Singularities for Solutions of the Fractional Heat... and dim 2 1 0. To prove Theorem 7.1 we need to study the growth conditions satisfied by positive measures and distributions with properties Eq. 7.2. 7.1 The Case of Positive Measures We state first two lemmas concerning the case when the distribution is a positive measure. But before going to the next lemma, recall that a function defined in 1 is Lip , 0 1, in the variable if sup . Lemma 7.2 Let 1 2 1 and let be a positive measure in 1 with upper parabolic growth of degree 2 1 with constant 1. Then, 1 1 1. 2 1 Proof Let , , and 0 2 . Then, writing , we split 0 2 0 2 1 2. 1 To shorten notation, we write 2 . Then we have 1 sup . 1 1 2 0 2 Since by property 3 in Lemma 2.2 1 1 2 2 if , 0 2 , 2 we deduce that 2 1 1 0 1 1 2 2 . 2 1 Next we deal with 2. Writing 0 0 2 , we have 2 0 0 . Observe now that by Lemma 2.2 1 0 0 0 1 2 1 1 1 2 . 4 J. Mateu, L. Prat The last estimate follows by splitting the integral into parabolic annuli and using the parabolic growth of order 2 1 of , for example. The same estimate holds replacing by . Then gathering all the estimates above, the lemma follows. Lemma 7.3 Let 1 2 1 and let be a positive measure in 1 which has upper parabolic growth 2 1, with constant 1. Then 1 1 2 1. Proof Let 1 be a fixed -parabolic cube. We have to show that there exists some constant (to be chosen below) such that 1 1 2 2 . To this end consider a test function 5 which equals 1 on 5 and vanishes in 6 . We also denote 5 1 5 . Then we split 1 1 1 2 1 2 5 1 1 2 5 1 2. To deal with the integral 1, we use Lemma 2.2 and write 1 1 2 1 1 5 1 2 5 1 2 . 6 Taking into account that, for , 1 1 1 1 2 1 2 3 4 and writing 1 , where 1 is a cube with side length in and is an interval of length 2 , we deduce that for 6 , 1 1 1 1 2 1 2 3 4 1 1 2 2 1 4 . Thus, 1 1 2 2 5 6 . 1 1 2 Next we will estimate the integral 2, taking 5 , where is the center of . To show that 2 2 , it suffices to prove 1 1 1 2 1 2 5 5 1. In turn, to prove this it is enough to show that 1 1 1 2 1 2 5 5 1 (7.3) for 1 in the following two cases: Case 1: of the form , . Removable Singularities for Solutions of the Fractional Heat... Case 2: of the form , . Proof of 7.3 in Case 1. Let 5 . We split 1 1 1 2 1 2 1 1 2 2 2 2 1 22 2 2 2 1 22 2 2 2 1 22 2 2 2 1 2 . First we will estimate the term 1 . For such that 22 2 , we write 1 22 . Observe now that for each 1 22 (see property 3 in Lemma 2.2) 1 1 2 5 by splitting the last domain of integration into parabolic annuli and using the growth condition of order 2 1 of . Thus, . Plugging this into the integral that defines 1, we obtain 1 2 2 1 1 1. 22 2 2 2 By exactly the same arguments, just writing in place of above, we deduce also that 2 1. Concerning the term , we write 1 22 2 2 2 1 1 2. 22 2 2 2 By Lemma 2.2, it follows that for , 2 2 1 . 5 Therefore, 2 1 . J. Mateu, L. Prat Hence 1 1. For 2 , we split the integral into annulus 2 2 and write 2 1 22 2 2 2 1 . 2 2 For using Lemma 2.2 we get 5 3 5 3 5 3 1 5 3 2 1 2 2 2 1 . Therefore 2 1 2 1 2 1 1 1. 22 2 2 2 2 1 2 So (7.3) holds in this case. Proof of (7.3) in Case 2. As in Case 1, we write 1 1 1 2 1 2 1 1 2 2 2 2 1 22 2 2 2 1 22 2 2 2 1 1 22 2 2 2 2 2 1 2 . The terms 1 and 2 can be estimated exactly in the same way as the terms 1 and 2 in Case 1, so that 1 2 1. Removable Singularities for Solutions of the Fractional Heat... Concerning , we have 1 1 1 1 22 2 2 2 2 2 1 1 22 2 2 2 Taking into account that, for 22 2 , 1 1 2 1 1 1 1 2 2 2 2 3 2 3 2 1 and that, by Lemma 7.2, is Lip 1 2 in the variable , we deduce that 1 1 2 2 1 1 1 1 1 2 1. 22 2 3 2 22 2 2 2 1 1 2 This concludes the proof of 1 for 1 2 1. 7.2 Growth Condition and Admissible Functions When 1 2 1 we say that a 1 function is admissible for an -parabolic cube , 1 and , if it is supported on 1 and satisfies 1 1 2 2 1 (7.4) 1 and 1 1 2 . (7.5) 1 2 As in the case 1 2 (see (3.1)) recall that with being the 1 Riesz transforms with Fourier multiplier . Then, for an -parabolic cube 1 1 with 1 and , one can write 1 2 1 1 1 Therefore if satisfies (7.5) and 2 1 for 1 then is admissible for the -parabolic cube , 1 2 1. Let’s remark that the analogues of Lemmas 3.2 and 3.3 also hold in this context and if is a standard test function supported on an parabolic cube 1 with 1 J. Mateu, L. Prat 1 and 2 2, then is admissible for . In fact, arguing as in Eq. 3.2 we have that 1 2 1 1 1 2 1 1 1 2 1 1 2 2 2 1 2 1 1 . 2 1 For these values of , the analogous growth condition to Lemma 3.4 reads as follows. 1 Lemma 7.4 Let 2 1 and a distribution such that 1 1 1 2 2 1 and 1. If is an admissible function for an parabolic cube 1, then 2 1 . Proof 1 2. 1 Notice that since 2 1, if we consider the Fourier transform of , we get 2 2 1 . 1 1 Hence we can write 2 2 . Therefore, since is admissible for the -parabolic cube , (7.4) is satisfied and so 1 12 1 2 12 1 2 2 1 . We deal now with 2. Notice that 1 1 2 1 1 2 for some constant . This identity can be easily seen by taking the Fourier transform (on the 1 variable ). Set 1 2 . Then 1 1 2 2 . Removable Singularities for Solutions of the Fractional Heat... Write 1 , with 1 and , and let be the center of the interval . Because of the zero mean of (integrating with respect to ) it is easy to check that 1 1 decays at most like 2 at infinity. Indeed, for 2 we have 1 1 1 1 1 (7.6) 2 2 2 2 1 1 1 1 2 1 2 1 2 where is the center of . And for 4 , 1 1 1 1 2 2 . (7.7) 1 1 2 Since 0, writing , we have 2 2 1 2 21 22 . Hence, due to Eq. 7.7, 2 1 2 1 21 . For 22 , we split the domain of integration in annuli. Write 2 2 1 for 1. Remark that for an -parabolic cube 1 , we denote 2 2 1 22 so that 2 is an -parabolic cube too (notice that if is centered at the origin and we consider the parabolic dilation 2 , 0, we have 2 ). Then, 2 using the decay of given by (7.6), we get 2 22 1 2 2 2 . 2 supp supp 1 (7.8) To estimate the first integral on the right hand side, recall that supp 1 . Using Hölder’s inequality with some exponent 0 to be chosen in a moment and the fact that (together with John-Nirenberg), then we get: 1 1 2 2 supp 2 supp 2 2 1 2 2 2 2 2 . For the last integral on the right hand side of (7.8), we write 2 2 2 supp 2 . supp J. Mateu, L. Prat Therefore, 2 2 2 22 1 2 2 2 2 1 2 2 1 1 2 2 1 . 2 1 Choosing , we get 2 1 22 . 7.3 Proof of Theorem 7.1 The inequality comes from the definition of the capacities. For the second inequality we assume to be compact and let be a covering of by - parabolic cubes 1 with disjoint interiors. By a parabolic version of a well known lemma of Harvey and Polking (see [8]), there exist functions 0 2 satisfying 1 in a neighborhood of , Eqs. 7.4 and 7.5. Let be a distribution with compact support contained in such that Eq. 7.2 hold. Then, using Lemma 7.4 2 1 1 . Thus, 2 1 . The second assertion in the theorem follows a standard argument that we reproduce for the reader’s convenience. Suppose 1 is a Borel set with dim 2 1. By a parabolic version of Frostman’s Lemma (that can be proved by arguments analogous to classical ones replacing the usual dyadic lattice in 1 by a parabolic dyadic lattice), there exists a non-zero positive measure supported on such that for all 1 and 0. We have to show estimates Eq. 7.2 with replaced by . The parabolic BMO-estimate in Eq. 7.2 is just a consequence of Lemma 7.3. Therefore 1 1 2 1. 1 To prove 2 1, apply the second property in Lemma 2.2, 1 2 2 1 2 1 0 1 2 1 0 1 2 1 (7.9) 0 1 2 2 1 2 0 0 2 1 . Removable Singularities for Solutions of the Fractional Heat... 8 s Capacities and s-Growth in Case s (0, 1 2) Now we introduce the capacities for 0 1 2 . Given a compact set 1, we define sup 1 (8.1) where the supremum is taken over all distributions supported on satisfying properties Eq. 8.2 listed below. Before we write them, to get some intuition, let’s consider the case 1 4 1 2 . Then 2 1 2 1 . In this case, the second condition in Eq. 7.2, that is 1 1 2 1 1, has 1 2 1 0 and can be rewritten as (via Fourier transform 1 with respect to the variable) 1 with 1 . 2 2 Hence, the distributions admissible for the capacity ,0 1 2, will be the ones supported on and satisfying 1 1 and 1. (8.2) Here is the Calderón-Zygmund operator with kernel 2 , the symbol denotes convolution with respect to the variable and is the kernel 1 1 1 1 1 1 (8.3) 1 2 1 2 1 where 2 (for a real number stands for its integer part) and we convolve -times the kernel. Notice that for 0 1 2, . (8.4) 1 We say that a 1 function is admissible for an parabolic cube 1, 0 1 2, (recall that we write 1 with and ) if 1 (8.5) and for 0 1 1 1 2 . (8.6) Now the growth condition in this case reads as follows. 1 1 1 Lemma 8.1 Let 0 1 2, 2 and set 1 1 1 1 . 1 2 1 2 Suppose that is a distribution satisfying 1 1 and 1. Then, if is an admissible function for an parabolic cube 1, we have 2 1 . J. Mateu, L. Prat Notice that if 1 4 1 2 , for example, then 2 1 2 1 and 1. In this case 1 1 and the condition 1 can be rewritten (via Fourier transform 2 2 1 2 1 1 with respect to the variable) as 1 (here 1 2 1 0 ), which is the same condition as in Theorem 7.4. Proof 1 2. Using (8.4) and the admissibility of , 2 1 1 . 1 To estimate 2 , write 1 , with 1 and , and let be the center of the interval . We claim that with 0, 1 4 . (8.7) and 2 1 2 . (8.8) 1 2 Once Eqs. 8.7 and 8.8 are available, since 1, arguing as in the proof of theorem 7.4, we get 2 1 2 . Indeed, since 0, writing , we have 2 2 1 2 21 22 . Due to Eq. 8.7, 2 1 2 1 21 . For 22 , we split the domain of integration in annuli. Write 2 2 1 for 1. Remark that for an -parabolic cube 1 , we denote 2 2 1 22 so that 2 is a parabolic cube too (notice that if is centered at the origin and we consider the parabolic dilation 2 , 0, we have 2 ). Then, using 2 the decay of given by (8.8), we get 2 22 1 2 2 2 . 2 supp supp 1 (8.9) Removable Singularities for Solutions of the Fractional Heat... To estimate the first integral on the right hand side, observe that by definition supp 1 . Using Hölder’s inequality with some exponent 0 to be chosen in a moment and the fact that (together with John-Nirenberg), then we get: 1 1 2 2 supp 2 supp 2 2 1 2 2 2 2 2 . For the last integral on the right hand side of (8.9), we write 2 2 2 supp 2 . supp Therefore, 2 2 2 22 1 2 2 2 2 1 2 2 1 1 2 2 . 2 1 Choosing , we get 2 1 22 . We still have to show claims Eqs. 8.7 and 8.8. Notice that (by taking Fourier transform with respect to ), 1 2 2 1 1 2 This identity can be understood better if we distinguish two cases. (1) Case even. We can write 1 1 1 . 2 1 1 Set 1 . Then . The zero mean of (integrating 2 1 1 with respect to ) implies that decays at most like 2 at infinity. To see this notice that for 2 we have 1 1 1 1 2 2 1 1 1 1 (8.10) 2 2 2 1 1 1 . 1 2 1 2 1 2 And for 4 , 1 1 1 1 1 2 2 . J. Mateu, L. Prat (2) Case odd. We can write 1 2 the last equality being a definition for . Notice that for 2 , 1 1 1 1 1 . 2 2 2 So arguing as in Eq. 8.10, 2 1 . 1 2 And for 4 , 5 1 1 5 1 1 1 2. 2 2 To estimate 1, apply the mean value theorem and Eq. 8.6 to get 1 1 1 1 1 1 2 . 5 2 Integrating and using Eq. 8.6 we get 1 1 1 2 1 2 . 5 2 1 Therefore for 4 , 1 . Therefore the claims Eqs. 8.7 and 8.8 are proved and also the lemma. Now we can state a result analogous to the first statement of Theorem 7.1 but for 0 1 2. Theorem 8.2 Let 0 1 2 . For every Borel set 1, 2 1 . The proof is analogous to the one of Theorem 7.1 but using Lemma 8.1 instead of Lemma 7.4. We leave it for the reader. Although we think that the second statement of Theorem 7.1 should hold for the case 0 1 2, we are still having some technical problems to show it. Funding Open Access Funding provided by Universitat Autonoma de Barcelona. L.P. was supported by PID2020-114167GB-I00 (MINECO, Spain) and J.M. was supported by PID2020-112881GB-I00 (MINECO, Spain). 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Removable Singularities for Solutions of the Fractional Heat... References 1. Applebaum, D.: Lévy processes and stochastic calculus. Second edn. Cambridge studies in advanced mathematics, vol. 116, Cambridge University Press, Cambridge (2009) 2. Bertoin, J.: Lévy processes, Cambridge tracts in mathematics, vol. 121, Cambridge University Press, Cambridge (1996) 3. Blumenthal, R.M., Getoor, R.K.: Some theorems on stable processes. Trans. Amer. Math. Soc. 95(2), 263–273 (1960) 4. Di Nezza, E., Palatucci, G., Valdinoci, E.: Hitchhiker’s guide to the fractional Sobolev spaces. Bull. Sci Math. 136(5), 521–573 (2012) 5. Duandikoetxea Zuazo, J.: Fourier Analysis Graduate Studies in Mathematics, 29. American Mathemati- cal Society, Providence (2001) 6. Garnett, J.: Positive length but zero analytic capacity. Proc. Amer. Math. Soc. 24, 696–699 (1970) 7. Grafakos, L.: Classical Fourier analysis, Second edn., Graduate Texts in Math. 249. Springer, New York (2008) 8. Harvey, R., Polking, J.: Removable singularities of solutions of linear partial differential equations. Acta Math. 125, 39–56 (1970) 9. Hytönen, T., Martikainen, H.: Non-homogeneous theorem and random dyadic cubes on metric measure spaces. J. Geom. Anal. 22(4), 1071–1107 (2012) 10. Landkof, N.S.: Foundations of Modern Potential Theory, Die Grundlehren Der Mathematischen Wissenschaften Band, vol. 180. Springer, New York (1972) 11. Mattila, P.: Geometry of Sets and Measures in Euclidean Spaces. Cambridge Studies in Advanced Mathematics, vol. 44. Cambridge University Press, Cambridge (1995) 12. Mattila, P., Paramonov, P.V.: On geometric properties of harmonic Lip1-capacity. Pacific J. Math. 171(2), 469–491 (1995) 13. Mateu, J., Prat, L., Tolsa, X.: Removable singularities for Lipschitz caloric functions in time varying domains. Rev. Mat Iberoam. 38(2), 547–588 (2022) 14. Paramonov, P.: On harmonic approximation in the 1 norm. Math. USSR Sbornik 71(1), 183–207 (1992) 15. Prat, L.: On the semiadditivity of the capacities associated with signed vector valued Riesz kernels. Trans. Amer. Math. Soc. 364(11), 5673–5691 (2012) 16. Pruitt, W.E., Taylor, S.J.: The potential kernel and hitting probabilities for the general stable process in . Trans. Amer. Math. Soc. 146, 299–321 (1969) 17. Stein, E.M.: Singular Integrals and Differentiability Properties of Functions, Princeton Mathematical Series, vol. 30. Princeton University Press, Princeton (1970) 18. Stein, E.M., Weiss, G.: Fourier Analysis on Euclidean Spaces, Princeton Mathematical Series, vol. 31. Princeton University Press, Princeton (1971) 19. Tolsa, X.: Painleve’s problem and the semiadditivity of analytic capacity. Acta Math. 190(1), 105–149 (2003) 20. Tolsa, X.: Analytic capacity, the Cauchy Transform, and Non-Homogeneous Calderón-Zygmund Theory, vol. 307 of Progress in Mathematics. Basel, Birkhäuser (2014) 21. Vázquez, J.L.: Asymptotic behaviour for the fractional heat equation in the Euclidean space. Complex Var. Elliptic Equ. 63(7-8), 1216–1231 (2018) 22. Volberg, A.: Calderón-Zygmund capacities and Operators on Nonhomogeneous Spaces. Vol. 100 of CBMS Regional Conference Series in Mathematics. Published for the Conference Board of the Mathematical Sciences, Washington, DC (2003) Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.