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1 t ∂αu(t) := (t s)−αu′(s) ds. t Γ(1 α) − − Z0 Recently, researchers have started to analyze finite element methods with respect to their ability to approximate solutions of fractional differential equations. While this started with classical Galerkin finite element methods for steady-state fractional diffusion equations as in [9, 12], nu- merical methods for time-dependent fractional partial differential equations include time-stepping methods [8, 13, 14], Discontinuous Galerkin methods [19, 20], as well as methods based on the Laplace transform [17]. It goes without saying that this list is far from being exhaustive. We mention here also the numerical approach from [21] which is based on the extension theory by Caffarelli and Silvestre [4]. The aforementioned numerical methods are usually based on a variational formulation of the equation under consideration. Existing works on variational formu- lations of time-fractional parabolic partial differential equations are scarce; as to our knowledge, the works [27, 24, 1] are of relevance in connection with our model problem (1) (for Semigroup theory for related Volterra equations see [23]). These works have in common that (i) their functional analytic setting is not based exclusively on classical Sobolev regularity in time, α but rather involves the operator ∂t , and that (ii) the initial value g is taken from L2(Ω). The goal of the present work is to derive the well-posedness of variational formulations set up in classical Sobolev-Bochner spaces and to clarify the question of regularity needed for the initial data. Now, as our functional analytic setting is based only on Sobolev regularity, a result of this kind is specifically interesting for numerical analysis of the equation (1). Indeed, approximation results for functions with certain Sobolev regularity are well known and ubiquitous in numerical analysis. The property (i) is owed to the fact that there is no rigorous definition of time-fractional on fractional Sobolev-Bochner spaces available. It sure is true that operators defined between real valued Sobolev spaces L (J) L (J) do extend to vector-valued counterparts 2 → 2 L (J; X) L (J; X) (for X a , this is a classical result of Marcienkiwicz and 2 → 2 Zygmund [16]), but the fact that we are dealing with Sobolev regularity Hα(J; X) in time needs some care and additional analysis. To that end, we will show first that the fractional Caputo derivative is a linear and on a time-fractional Sobolev-Bochner space. This way, we can consider a variational formulation of (1) based exclusively on Sobolev regularity, which resembles classical variational formulations for parabolic equations. Regarding the point (ii), the choice of g L (Ω) as initial value is indeed admissible, but one has to bear in mind ∈ 2 the following: While the space L (J; H1(Ω)) L (J; H−1(Ω)), used in variational formulations 2 ∩ 2 of parabolic equations, is continuously embedded in C(J; L2(Ω)), this is no longer true for the equation (1). We will show that the spacee of solutions of our variational formulation of (1) is con- tinuously embedded in C(J; H1−1/α−ε(Ω)) for all ε> 0. Our main result is then well-posedness of the variational formulation, cf. Theorem 2.

2 2 Mathematical setting and main results

2.1 Sobolev and Bochner spaces We denote by Ω Rd a (spatial) Lipschitz domain, and by J = (0, T ) for T > 0 a temporal ⊂ 1 interval. We use Lebesgue and Sobolev spaces L2(Ω) and H (Ω), the tilde denoting vanishing trace on the boundary ∂Ω. The fractional Sobolev spaces Hs(Ω) for s (0, 1) are defined by the s 1 ∈ K-method of interpolation as H (Ω) := [L2(Ω), H (Ω)]s,2, cf.e [26], where e ∞ [B ,B ] := u B u e < , e u 2 := t−2s−1K (t, u)2 dt 0 1 s,2 0 [B0,B1]s,2 [B0,B1] 2 [B0,B1] ∈ | k k ∞ k k s, 0 n o Z with the K-functional

2 2 2 2 K[B0,B1](t, u) := inf u w B0 + t w B1 . w∈B1 k − k k k

The topological dual of a X is denoted by X′, and we define H−s(Ω) := Hs(Ω)′ as the topological duals with respect to the extended L (Ω) scalar product ( , ), and duality 2 · · will be denoted by , . We set H0(Ω) := L (Ω). In time, we additionally use Lebesguee and h· ·i 2 Sobolev spaces L (J) and Hα(J), for α (0, 1]. The L (J) scalar product will also be denoted 2 ∈ 2 by ( , ), but it will always be cleare which scalar product we are using. We use the notation Du · · to denote the distributional derivative of a function u given on J. For α (0, 1) the on ∈ the space Hα(J) is given by

2 2 2 2 2 f(s) f(t) f α := f + f α < , where f α := | − | ds dt. k kH (J) k kL2(J) | |H (J) ∞ | |H (J) s t 2α+1 ZJ ZJ | − | α 1 −s s ′ We mention that H (J) = [L2(J),H (J)]α,2 with equivalent norms. We set H (J) := H (J) and H−s(J) := Hs(J) for s (0, 1). For a Banach space X, we use the Bochner space L (J; X) ∈ 2 of functions f : J X which are strongly measurable with respect to the Lebesgue measuree ds → on J eand

f 2 := f(s) 2 ds < . k kL2(J;X) k kX ∞ ZJ

For w a measurable, positive function on J, we denote by L2(J, w; X) the w-weighted Lebesgue space of strongly measurable functions with norm

f 2 := w(s)2 f(s) 2 ds < . k kL2(J,w;X) k kX ∞ ZJ We say that a function f L (J; X) has a weak time-derivative ∂ f L (J; X), if ∈ 2 t ∈ 2 ∂ fϕ = f∂ ϕ for all ϕ C∞(J). (2) t − t ∈ 0 ZJ ZJ

3 Note that this last integral has to be understand in the sense of Bochner. We define the space H1(J; X) as the space of functions with

2 2 2 f 1 := f + ∂ f . k kH (J;X) k kL2(J;X) k t kL2(J;X) For 0 <α< 1, we also use the fractional Sobolev-Bochner space Hα(J; X) of ds-strongly measurable functions f : J X with → 2 2 2 2 2 f(s) f(t) X f α := f + f α < , where f α := k − k ds dt. k kH (J;X) k kL2(J;X) | |H (J;X) ∞ | |H (J;X) s t 2α+1 ZJ ZJ | − | We will also use these Bochner spaces on R instead of J. For a recent introduction to Bochner spaces, we refer to [11].

2.2 Fractional time derivative on Bochner spaces For 0 <β< 1, we define the left and right-sided Riemann-Liouville fractional integral operators

1 t 1 T D−βu(t) := (t s)β−1u(s) ds and D−βu(t) := (s t)β−1u(s) ds, 0 Γ(β) − T Γ(β) − Z0 Zt where Γ is Euler’s Gamma function. For sufficiently smooth functions u, the left-sided Caputo fractional derivative ∂α for α (0, 1) is defined as ∂αu := Dα−1Du. We will show below in ∈ 0 Lemma 10 that the tensorised version D−β I defined by ( D−β I)(u x) := ( D−βu) x 0 ⊗ 0 ⊗ ⊗ 0 ⊗ can be extended uniquely to a linear and bounded operator D−β : L (J; X) Hβ(J; X) for 0 2 → a Hilbert space X. This allows us to prove the following result. The proof will be carried out below in Section 3.3.

Theorem 1. Let ∂ be the weak time derivative defined in (2). Then, for α (1/2, 1), the t ∈ operator ∂α := Dα−1 ∂ is linear and bounded as ∂α : Hα(J; H−1(Ω)) L (J; H−1(Ω)). t 0 ◦ t t → 2 2.3 Variational formulation and main result Our variational formulation of (1) is the following: Given f L (J; H−1(Ω)) and g H1−1/α+δ(Ω) ∈ 2 ∈ for some δ > 0, find u L (J; H1(Ω)) with u Hα(J; H−1(Ω)) such that ∈ 2 ∈ α 1 ∂t u , v e+ ( u , v) = (f , v) for all v H (Ω), (3) h i ∇ ∇ ∈ α in J, and u(0, ) = g( ). The duality ∂t u , v ine (3) makes sense due to the α · · h i mapping properties of ∂t from Theorem 1, and the initial condition makes makes sense as we will show in Corollary 9 below that L (J; H1(Ω)) Hα(J; H−1(Ω)) is continuously embedded in 2 ∩ C(J; H1−1/α−ε(Ω)) for all ε > 0. The following theorem is our main result and will be proven below in Section 3.3. e

4 Theorem 2. The variational formulation (3) is well posed: there exists a unique solution u, and

u e 1 + u α −1 Cδ g 1−1/α+δ + f −1 . k kL2(J;H (Ω)) k kH (J;H (Ω)) ≤ k kH (Ω) k kL2(J;H (Ω))   The constant Cδ > 0 depends only on δ. Remark 3. It is textbook knowledge that there holds the continuous embedding

.((L (J; H1(Ω)) H1(J; H−1(Ω)) ֒ C(J; L (Ω 2 ∩ → 2 In the present case, we have thee embedding ((L (J; H1(Ω)) Hα(J; H−1(Ω)) ֒ C(J; H1−1/α−ε(Ω 2 ∩ → for all ε > 0, cf. Lemma 8e below. The reason for the missing power of ε is that we use the embedding result H1/2+ε(J; X) ֒ C(J; X). Furthermore, note that the stability estimate of 1−1/α+δ→ Theorem 2 involves the H (Ω) norm of the initial data g, and the constant Cδ is expected to blow up for δ 0. → 3 Technical results

3.1 Fractional integral and differential operators We have the following results. The first point is an extension of a recent result in [12] and can be found in [7, Lemma 5], while the second point is part of the proof of [7, Lemma 6]. The third part can be found in [15, Lem. 2.7].

Lemma 4. (i) For every s R with β s and β > 0, the operators D−β and D−β can be ∈ − ≤ 0 T extended to bounded linear operators Hs(J) Hs+β(J). → −β −β/2 (ii) For 0 <β< 1, the operator 0D is elliptice on H (J). (iii) For 0 <β< 1 and u, v L (J) it holds (D−βu , v) = (u , D−βv). ∈ 2 0 T

The Mittag-Leffler function arises naturally in the study of fractional differential equations. We refer to [6, Section 18.1] for an overview. It is defined as

∞ zk En1,n2 (z) := . Γ(kn1 + n2) Xk=0 According to [22, Thm. 1.6], for z R, ∈ 1 E (z) . , (4) n1,n2 1+ z | |

5 and due to [5, Thm. 4.3],

α−1 α α 0D DEα,1(λt )= λEα,1(λt ). (5)

Furthermore, by [25], E ( z) is completely monotone for 0 < α 1 and positive z, in partic- α,1 − ≤ ular,

E′ ( z) 0 for positive z. (6) α,1 − ≥ We will need the following result on fractional seminorms, which combines the Hs norm and the dual norm of the distributional derivative. Lemma 5. Let s (0, 1) be fixed. There holds ∈ s u s . Du s−1 for all u H (J), | |H (J) k kH (J) ∈ where Du is the distributional derivative of u.

−1 Proof. As u L2(J), it holds Du H (J). We can write u = Dψ + c with c R, where 1 ∈ ∈ 1 ∈ ψ H (J) is the unique solution of (Dψ,Dϕ) = (u,Dϕ) for all ϕ H (J). Then, ψ e 1 . ∈ ∈ k kH (J) u L2(J), and due to the definition of the distributional derivative we see k k e e

(u,Dψ) = (Du,ψ) . Du −1 ψ e1 . Du −1 u . | | | | k kH (J)k kH (J) k kH (J)k kL2(J) We conclude that for u L (J), it holds ∈ 2 2 u = (u,Dψ) + (u , c) . Du −1 u + (u , c). k kL2(J) k kH (J)k kL2(J) Now we apply this estimate to u u, where u denotes the mean value of u, and obtain − u u . Du −1 . (7) k − kL2(J) k kH (J) The standard Poincaré inequality states that

u u 1 . Du . (8) k − kH (J) k kL2(J) The Hs(J) norm can equivalently be obtained by the K-method of interpolation via

∞ 2 2 −2s 2 2 2 dt 1 1 u u Hs(J) u u [L2(J),H (J)] 2 = t inf u u v L2(J) + t v H (J) . k − k ≃ k − k s, v∈H1(J) k − − k k k t Z0   Using (7) and (8), we obtain

2 2 2 2 2 2 1 1 inf u u v L2(J) + t v H (J) inf u u v L2(J) + t v H (J) v∈H1(J) k − − k k k ≤ v∈H1(J) k − − k k k v=0 2 2 2 −1 . inf Du Dv H (J) + t Dv L2(J) v∈H1(J) k − k k k v=0

6 Next we use that for w L (J) there is a ψ H1(J) with ψ = 0 such that Dψ = w. We ∈ 2 ∈ conclude ∞ 2 −2s 2 2 2 dt −1 u u Hs(J) . t inf Du w H (J) + t w L2(J) . k − k w∈L2(J) k − k k k t Z0   2 2 By definition, the right-hand side is Du −1 , which is equivalent to Du s−1 . k k[H (J),L2(J)]s,2 k kH (J) This concludes the proof.

The next lemma establishes a norm equivalence on a fractional .

Lemma 6. Let 1/2

s−1 D Du . Du e −1 . Du s−1 . u s . k0 kL2(J) k kHs (J) k kH (J) k kH (J) Here, the first estimate follows from Lemma 4, and the second one can be found in [10, Lem. 5]. To see the third estimate, recall that D is the distributional derivative, and hence Du −1 k kH (J) ≤ u as well as Du u 1 . The third estimate now follows from an interpolation k kL2(J) k kL2(J) ≤ k kH (J) argument. The fact that Du = D(u u) for the mean value u of u and Poincare’s inequality − show

s−1 u s & D Du . | |H (J) k0 kL2(J) To show the converse estimate, we take u C∞(J) and estimate with Lemmas 5 and 4 ∈ 2 2 2(s−1) u s . Du s−1 . ( D Du , Du) | |H (J) k kH (J) 0 = ( Ds−1Du,Ds−1Du) Ds−1Du Ds−1Du . 0 T ≤ k0 kL2(J)k T kL2(J) Here, the identity follows from Lemma 4, (iii). Due to Lemma 4, it also holds Ds−1Du . k T kL2(J) Du e −1 . u s , where the second estimate wa already shown at the beginning of this k kHs (J) k kH (J) proof. Applying the whole argument to u u and using Poincare’s inequality finally shows the − statement.

3.2 Sobolev and Bochner spaces For s ( 1, 0] we have the interpolation estimate ∈ − (1−s)/2 (1+s)/2 u s C(s) u −1 u e , (9) k kH (Ω) ≤ k kH (Ω) · k kH1(Ω) with C(s) > 0 a constant depending only on s. This estimate follows for s = 0 by duality, and for s ( 1, 0) using additionally [26, 1.3.3 (g)] and the fact that duality and interpolation commute, ∈ −

7 cf. [26, 1.11.2]. For a measurable set M R, we denote by 1 the characteristic function on ⊂ M M, and for a function φ : J R and x X, we define φ x := J X as (φ x) (s) := φ(s)x. → ∈ ⊗ → ⊗ We denote by (J) the σ-algebra of all ds-measurable sets on J, and by S(J) the set of simple L functions. It is known that the following subsets are dense for J bounded or J = R,

n S(J; X) := 1 x A (J),x X for all i = 1,...,n,n N L (J; X), Ai ⊗ i | i ∈L i ∈ ∈ ⊂ 2 ( i=1 ) Xn ∞ ∞ R N 1 S (J; X) := ϕi xi ϕi Cc ( ),xi X for all i = 1,...,n,n H (J; X). ( ⊗ | ∈ ∈ ∈ ) ⊂ Xi=1 We assume from now on that the Banach spaces X are reflexive; this implies that they have the so-called Radon-Nikodým property, cf. [11, Thm. 1.95], which is sufficient and necessary in ′ ′ order to have that L2(J; X) is isometrically isomorphic to L2(J; X ), cf. [11, Thm. 1.84]. We can extend ∂ for u L (J; X) by defining ∂ u H1(J; X′)′ as t ∈ 2 t ∈ 0 u(s) , ∂ ϕ(s) ds for ϕ H1(J; X′). − h t i ∈ 0 ZJ Then, we have that ∂ : L (J; X) H1(J; X′)′ is bounded. Furthermore, ∂ : H1(J; X) t 2 → 0 t → L (J; X)= L (J; X′)′ is bounded, and by interpolation, we have that for s (0, 1) 2 2 ∈ ∂ : L (J; X),H1(J; X) H1(J; X′)′,L (J; X′)′ (10) t 2 s → 0 2 s is bounded. We will need the following results on interpolat ion of Sobolev-Bochner spaces. Lemma 7. There holds R 1 R s R L2( ; X),H ( ; X) s = H ( ; X) and   ′ ′ 1 ′ ′ ′ ′ 1 ′ ′ ′ 1 ′ 1−s ′ ′ H (J; X ) ,L2(J; X ) = H (J; X ),L2(J; X ) = L2(J; X ), H (J; X ) = H (J; X ) . s s 1−s h i h i h i Proof.e The first identity is duee to [11, Thm. 2.91], and the seconde and third identitiese are well- known results in interpolation theory, cf. [26, 1.11.2]. The last identity is a variant of the first one with bounded interval and zero traces. Using extension theorems, its proof can in fact be reduced to the first identity. In the case of scalar-valued Sobolev spaces, we refer to [3, Thm. 14.2.3] for details.

Next, we will establish continuous embeddings for the function space of our variational formula- tion. Lemma 8. Suppose that α (0, 1), s ( 1, 0] and 0 < r are such that r < α(1 s)/2. Then, ∈ ∈ − − we have the continuous embedding

.((L (J; H1(Ω)) Hα(J; H−1(Ω)) ֒ Hr(J; Hs(Ω 2 ∩ →

e 8 r Proof. It is clear that u s u e 1 . To bound the H -seminorm, we write for k kL2(J;H (Ω)) ≤ k kL2(J;H (Ω)) r < α(1 s)/2 − 1 s 1+ s 2r + 1 = (2α + 1) − + (1 ε) 2 − 2 for some ε > 0. The interpolation estimate (9) and the inequalities of Cauchy-Schwarz and Young then yield

2 1−s 1+s u(s) u(t) e u(s) u(t) −1 u(s) u(t) e Hs(Ω) H (Ω) H1(Ω) k − k dtds . k − k k − k dtds s t 2r+1 s t 2r+1 ZJ ZJ | − | ZJ ZJ | − | 2 (1−s)/2 u(s) u(t) H−1(Ω) . k − k dt s t 2α+1 ZJ ZJ | − | ! 2 (1+s)/2 u(s) u(t) e H1(Ω) k − k dt ds  s t 1−ε  ZJ | − |  2  u(s) u(t) 2 u(s) u(t) H−1(Ω) He 1(Ω) . k − k dtds + k − k dtds s t 2α+1 s t 1−ε ZJ ZJ | − | ZJ ZJ | − |

and as ε> 0, the last integral can be bounded by u e 1 . k kL2(J;H (Ω)) Corollary 9. Suppose that α (0, 1) and s ( 1, 0] are such that s< 1 1/α. Then, we have ∈ ∈ − − the continuous embedding

.((L (J; H1(Ω)) Hα(J; H−1(Ω)) ֒ C(J; Hs(Ω 2 ∩ → Proof. If s< 1 1/α, then 1/2 < α(1 s)/2, and according to Lemma 8 there holds the continuous − e − .embedding L (J; H1(Ω)) Hα(J; H−1(Ω)) ֒ H1/2+ε(J; Hs(Ω)) for a sufficiently small ε > 0 2 ∩ → ֒ ((According to [11, Thm. 2.95] there also holds the continuous embedding H1/2+ε(J; Hs(Ω → C(J; Hs(Ω)), ande this proves the statement.

The next lemma shows that the Riemann-Liouville fractional integral operators can be extended in the canonical way (i.e., by tensorisation) to Sobolev-Bochner spaces.

Lemma 10. Suppose that X is a Hilbert space and 0 <β< 1/2. Then, the operator

S(J; X) Hβ(J; X) D−β I := → 0 n 1 n −β1 ⊗ ( xi Ai (0D Ai )xi i=1 7→ i=1 can be extended uniquely to a linear andP bounded operatorP D−β : L (J; X) Hβ(J; X). The 0 2 → same statement is true for the operator D−β I. T ⊗

9 Proof. It follow from Lemma 4 (i) that the operator D−β : L (J) L (J) is bounded. Fur- 0 2 → 2 thermore, it is a positive operator, i.e., D−β(1 ) 0 on J. It is then easy to see, cf. [11, 0 Ai ≥ Thm. 2.3], that D−β Iu D−β u for u S(J; X), (11) k0 ⊗ kL2(J;X) ≤ k0 kL2(J)→L2(J)k kL2(J;X) ∈ and as S(J; X) is dense in L (J; X), we obtain boundedness D−β : L (J; X) L (J; X). 2 0 2 → 2 Next, we will follow the ideas developed in [12, Thm. 3.1]. Denoting by f L (R) the extension ∈ 2 of f by zero, it holds D−βf(x) = D−βf(x). Denote by : L (R) L (R) the Fourier 0 −∞ F 2 → 2 transformation. Then, the operator I extends to an isometry : L (eR; X) L (R; X): For F⊗ F 2 → 2 general operators, this is a classical result by Marcienkiwe icz and Zygmund [16], cf. [11, Thm. 2.9], but in the present case of the Fourier transformation it can be seen readily by using density of simple functions S(R; X) in L2(R; X) and the Plancherel theorem for the scalar-valued Fourier transformation. Furthermore, for u L (R; X) we have u = u with u(x) = u( x) the ∈ 2 FF P P − parity operator. For a function ϕ = n ϕ x S∞(R; X), we conclude i=1 i ⊗ i ∈ n n 2 P 2 2 ϕ 1 R = ϕ x R + ∂ ϕ x R k kH ( ;X) k i ⊗ ikL2( ;X) k t i ⊗ ikL2( ;X) i=1 i=1 Xn Xn 2 2 = ϕ x R + (∂ ϕ ) x R k F i ⊗ ikL2( ;X) k F t i ⊗ ikL2( ;X) i=1 i=1 X 2 X = g ϕ R , k F kL2 ( ;X) with weight function g(ω) := √1+ ω2. By density, this shows that I can be extended to 1 F ⊗ an isometry : H (R, X) L2(R, g; X). By interpolation and Lemma 7, we conclude that s F s → : H (R; X) L2(R, g ; X) is bounded. To show that this operator is an isometry, consider F → R R a decomposition u0 + u1 = u with u0 L2( ; X), u1 L2( , g; X). From u1 H1(R;X) = F ∈ 1 ∈ kF k u R = u R we conclude u H (R; X) and due to u + u = u = kFF 1kL2( ,g;X) k 1kL2( ,g;X) F 1 ∈ F 0 F 1 FF u we have that u + u = u is a decomposition of u. Hence, P ePF 0e PF 1 e e e e 2 e 2 e 2 2 e e u R + t u R = u R + t u 1 R k 0kL2( ;X) k 1kL2( ,g;X) kF 0kL2( ;X) kF 1kH ( ;X) e e 2 2 1 = u0 L2(R;X) + t u1 H (R;X), e e kPFe k kPFe k 2 s which implies K R 1 R (t, u) K R R (t, u). This shows that : H (R; X) [L2( ;X),H ( ;X)] ≤ [L2( ;X),L2( ,g;X)] F F → L (R, gs; X) is an isometry. Next, for a simple functione u S(R; X), e 2 ∈

−β 2 2s β 2 −∞D Iu Hs(R;X) = g(ω) −∞D u(ω) X dω k ⊗ k R kF k Z . Dβ u(ω) 2 dω + (ω−2 + 1)s u(ω) 2 dω kF−∞ kX kF kX Z|ω|≤1 Z|ω|>1 β 2 2 −∞D u(ω) X dω + u(ω) X dω ≤ R kF k kF k Z Z|ω|>1 β 2 2 2 . −∞D u(s) X ds + u(s) X ds . u L2(R;X), R k k R k k k k Z Z

10 −β and by density we get the desired result. The proof for DT follows along the same lines. Lemma 11. The operator D−βu I has a unique extension as bounded and linear operator 0 ⊗ D−β I : Hβ(J; H1(Ω))′ L (J; H−1(Ω)). 0 ⊗ → 2 n −1 n 1 Proof. For u = 1A ui S(J; H (Ω))e , v = 1A vi S(J; H (Ω)), we compute i=1 i ⊗ ∈ i=1 i ⊗ ∈ P −β −βP e (u , DT Iv)= u(s) , DT Iv(s) ds ⊗ J h ⊗ i Z n 1 −β1 = ui , vi Ai (s)DT Aj (s) ds h i J i,jX=1 Z (12) n −β1 1 = ui , vi 0D Ai (s) Aj (s) ds h i J i,jX=1 Z = ( D−β Iu , v). 0 ⊗ β 1 1 −1 1 ′ β 1 ′ As H (J; H (Ω)) is dense in L2(J; H (Ω)), L2(J; H (Ω)) = L2(J; H (Ω)) is dense in H (J; H (Ω)) . −β 1 β 1 According to Lemma 10, DT : L2(J; H (Ω)) H (J; H (Ω)) is bounded, and hence the equal- −β → ity (12) showse that 0D can be extendede as stipulated. This finishese the proof. e e e 3.3 Proof of the main theorems

Proof of Theorem 1. Using the boundedness (10) of ∂t and Lemmas 7 and 11, we conclude that

Dα−1 ∂ : Hα(J; H−1(Ω)) H1−α(J; H1(Ω))′ = H1−α(J; H1(Ω))′ L (J; H−1(Ω)) 0 ◦ t → → 2 is bounded. e e e

Proof of Theorem 2. We mimic the proof for parabolic PDE. Take (wk)k≥1 the L2(Ω)-orthonormal basis of eigenfunctions and (λ ) the eigenvalues of ∆. We make the ansatz k k≥1 − m k um(t) := dm(t)wk. Xk=1 Now, we are looking for dk : J R such that m → α k k ∂ d (t)+ λkd (t)= f(t) , wk , k = 1,...,m, t m m h i (13) dk (0) = g , w , k = 1,...,m. m h ki According to [2, Thm. 2.1], cf. [5, Thm. 7.2] and [15, Chapter 3.1], the solutions to these equations are given uniquely by

dk (t)= g , w E ( λ tα)+ ψ (t), k = 1,...,m, m h ki α − k k

11 where ψ (t) := α t f(t s) , w sα−1E′ ( λ sα) ds. In order to obtain energy estimates for the k 0 h − ki α − k u , we can extend the calculations carried out in [24]. However, as we aim at weaker initial m R values, we need a finer analysis. First, using the bound (4), we have for ε [0, 1] ∈ 1 E (z) . z −(1−ε). | α,1 | 1+ z ≤ | | | | Furthermore, α tα−1E′ ( λ tα) dt = λ−1(1 E ( λ T α)), and tα−1E′ ( λ tα) 0 due J α,1 − k k − α,1 − k α,1 − k ≥ to (6). Hence, we see R tα−1E′ ( λ tα) dt . λ−1. (14) | α,1 − k | k ZJ We conclude that for 2α(1 ε) < 1, it holds − E ( λ ( )α) 2 C λ−2(1−ε). (15) k α,1 − k · kL2(J) ≤ ε k Furthermore, due to Lemma 6, the identity (5) and the previous estimate, we also have

α 2 2 α 2 2ε E ( λ ( ) ) α λ E ( λ ( ) ) C λ . (16) | α − k · |H (J) ∼ kk α,1 − k · kL2(J) ≤ ε k By Young’s inequality and (14),

2 ψ 2 . f(t) , w 2 dt tα−1E′ ( λ tα) dt . λ−2 f(t) , w 2 dt. (17) k kkL2(J) h ki · | α − k | k h ki ZJ  ZJ  ZJ According to [22, pp. 140], it holds ∂αψ (t) = f(t) , w λ ψ (t). Hence, using Lemma 6 t k −h ki− k k and (17), we also see

2 2 ψ α . f(t) , w dt. (18) | |H (J) h ki ZJ Now choose ε := (2 1/α+δ)/2 and observe that 2α(1 ε) = 1 αδ < 1 and 1+2ε = 1 1/α+δ. − − − − − Using (15) and (17), we estimate

m m m 2 k 2 −1+2ε 2 −1 2 um e 1 = λk dm . λ g , wk + λ f(t) , wk dt k kL2(J;H (Ω)) k kL2(J) k h i k h i k=0 k=0 k=0 ZJ X X2 2X . g 1−1 + + f −1 . k kH /α δ (Ω) k kL2(J;H (Ω)) Using (16) and (18), we can analogously estimate

2 2 2 u α −1 . g 1−1 + + f −1 . k mkH (J;H (Ω)) k kH /α δ (Ω) k kL2(J;H (Ω)) 1 α −1 Therefore, (um)m∈N is a bounded sequence in L2(J; H (Ω)) and in H (J; H (Ω)), and we conclude that there is a subsequence (u ) N which converges weakly to some u L (J; H1(Ω)) mk k∈ ∈ 2 e 12 e and to some u Hα(J; H−1(Ω)). It follows that u also converges weakly in L (J; H−1(Ω)) to u ∈ 2 as well as to u, which yields u = u. Taking into account the construction of the um and invoking the weak limit, we obtain for all v L (J; H1(Ω)) e ∈ 2 e e ∂αu , v + eu , v dt = f , v dt. h t i h∇ ∇ i h i ZJ ZJ 1−1/α+δ Note that due to Corollary 9, umk also converges weakly to u in C([0, T ]; H (Ω)), hence g = u (0) u(0). This yields u(0) = g, and we conclude that u is a weak solution. As for mk → uniqueness, if u is a weak solution with vanishing data, then the functions uk(t) := (u(t) , wk) solve the equations (13) with vanishing right-hand side, and hence uk(t) = 0. Acknowledgement: The author would like to thank Vincent J. Ervin for his valuable com- ments.

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