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Existence of solutions to fractional boundary value problems at resonance in Hilbert spaces
Boundary Value Problems volume 2017, Article number: 105 (2017)
Abstract
We study the existence of solutions to a nonlinear fractional differential equation in Hilbert spaces associated with threepoint boundary conditions at resonance
by using Mawhin’s continuation theorem. We propose a new technique to improve the conditions on A which have been used previously. In addition, a necessary and sufficient condition for that the fractional differential operator is Fredholm with zeroindex is established, especially for the first time when the fractional differential operator takes values in an arbitrary Hilbert space.
Introduction
Fractional differential equations arise in various areas of physics and applied mathematics and have recently become a powerful tool in modeling of many physical phenomena (for instance, see [1–7] and the references therein). In this paper, we are concerned with the existence of solutions to the following fractional threepoint boundary value problem at resonance:
where \(D^{\alpha }\) is the RiemannLiouville differential operator of order α; θ is zero element in a Hilbert space \(\mathcal{H}\); \(\eta \in (0, 1)\) is a given constant; \(A\in \mathcal{L(H)}\) is a bounded linear operator on \(\mathcal{H}\) such that
 (A1):

\(IA\) is a Fredholm operator with zeroindex;
 (A2):

\(f(\cdot, x,y)\) is Lebesgue measurable on \([0,1]\) for every \((x,y)\in \mathcal{H}\times \mathcal{H}\);
 (A3):

\(f(t,\cdot,\cdot)\) is continuous on \(\mathcal{H}\times \mathcal{H}\) for almost every \(t \in [0, 1]\);
 (A4):

for each bounded set \(\Omega \subset \mathcal{H}\times \mathcal{H}\), the function \(t\mapsto \varphi_{\Omega }(t) = \sup \{ \Vert f(t, x, y)\Vert _{\mathcal{H}} : (x, y) \in \Omega \} \) is Lebesgue integrable on \([0,1]\), and the set \(\{ f(t, x, y):(x, y)\in \Omega \} \) is relatively compact in \(\mathcal{H}\), here \(\Vert \cdot \Vert _{\mathcal{H}}\) stands for the norm induced by the scalar product \(\langle \cdot,\cdot \rangle \) in \(\mathcal{H}\).
where L and N are operators from a Banach space X to another Banach space Y (here L is linear). If L is invertible, or \(\ker L=\{0\}\), (1.2) is called nonresonant problem. Otherwise, if kerL is not a trivial space, then it is called resonant problem. For this problem, we focus on an important class of resonant problems when L is a Fredholm operator with zeroindex, as a prerequisite for applying coincidence degree theory [8].
Different from nonresonant problems being studied for a long time, the solvability of resonant problems has been extensively studied for the last decade. The reason is that resonant problems are rather complicated due to the noninvertibility of L. Noninvertibility leads to the difficulty of constructing a suitable continuous projection on a complement of ImL. Due to the fact that constructing that projection becomes more difficult when the dimension of kerL is large, most of works are investigated mainly for \(\dim \ker L=1\). We refer the reader to [9–23] and to the references therein.
For the case that \(\dim \ker L\) can be arbitrary, recently in [24], we use the Mawhin continuation theorem to investigate the existence of solutions for boundary value problems (briefly, BVPs) at resonance. We consider the following threepoint BVP:
where \(A:\mathbb {R}^{n} \rightarrow \mathbb {R}^{n}\) is a continuous linear operator satisfying
It seems that [24] is one of premier papers on the case that \(\dim \ker L\) is arbitrary. Later, problem (1.3) in [24] was extended to fractional BVP (1.1) in [25]. Most recently, the result in [25] was generalized to an infinitedimensional space \(l^{2}\) in [26]. We emphasize that condition (1.4) is an essential condition for the techniques in [24–26], with the note that in [26] the condition is slightly different in order to adapt the boundary conditions.
In our new research [27] condition (1.4) can be omitted completely thanks to the fruitful features of a continuous linear operator on \(\mathbb {R}^{n}\) which can be regarded as matrices. The breakthrough results in [27] allow us to ask the first question: For extending problem ( 1.1 ) from a finitedimensional space \(\mathbb {R}^{n}\) to a Hilbert space \(\mathcal{H}\) , is omitting condition ( 1.4 ) still possible or not? The answer is no, for example, let A be the difference of an identity operator and a finiterange operator, problem (1.1) is unsolvable via Mawhin’s continuation theorem.
Our next question is: What is the condition for a Hilbert space? In order to find the suitable condition for a Hilbert space, we observe that there is no change in the proofs in [24–26] if we replace (1.4) by the following more generalized assumption:
or equivalently,
Clearly, condition (1.4) can be derived from condition (1.5) by taking \(\kappa =1\) or \(\kappa =\frac{1}{2}\). We remark that condition (1.5) leads to some useful qualitative properties: \(\operatorname {Im}(IA)\) is closed and \(\ker (IA)\) is isomorphic to the complement of \(\operatorname {Im}(IA)\). It suggests us to generalize (1.5) to the condition for a Hilbert space as follows.
which is actually assumption (A1) for operator A. We discover that in this paper condition (1.6) is the best generalization on a Hilbert space in the following senses:

(1)
Condition (1.6) is not only a sufficient condition but also a necessary condition for that \(L=D^{\alpha }\) is the Fredholm operator with zeroindex.

(2)
When \(\mathcal{H}\) is finitedimensional, (1.6) automatically holds, therefore the result in [27] is a special case of our result in this paper.

(3)
When \(\mathcal{H}\) is infinitedimensional, it is well known that the ranknullity theorem no longer holds for an operator \(A\in \mathcal{L(H)}\). Condition (1.6) helps us to overcome this difficulty by providing another characterization of dimension in a Hilbert space, as to be seen in (2.10).

(4)
Condition (1.6) gives a unified approach and viewpoint for solving this kind of BVPs, for \(\dim \mathcal{H}\leq \infty \) and \(L=D^{\alpha }\) with \(1<\alpha \leq 2\).
The structure of the paper is organized as follows. In Section 2, we introduce some necessary background of fractional calculus as well as coincidence degree theory. In addition, we establish some essential lemmas needed for our main result later. Section 3 is devoted to presenting the statement and the proof of the main result. Finally, we give a specific example to illustrate.
Preliminaries
Fractional calculus
In this subsection, we recall some definitions and results of the fractional calculus. See [2, 6, 7] for more details.
Definition 2.1
Given \(f: [0, 1] \rightarrow \mathcal{H}\) and \(\alpha >0\). Then
 ⋄:

The fractional integral of order α of the function f is given by
$$I^{\alpha }f(t):= \frac{1}{\Gamma (\alpha)} \int_{0}^{t}(ts)^{ \alpha 1}f(s)\,ds\quad \mbox{for } t>0. $$  ⋄:

The RiemannLiouville fractional derivative of order α of f is given by
$$D^{\alpha }f(t):= \frac{d^{m}}{dt^{n}} \bigl( I^{m\alpha }f \bigr) (t) = \frac{1}{\Gamma (m  \alpha)} \frac{d^{m}}{dt^{m}} \int_{0}^{t} (ts)^{m  \alpha 1} f(s)\,ds\quad \mbox{for } t>0, $$where \(m = [\alpha ] + 1\),
For \(k \in \mathbb {N}\), we denote by \(AC^{k} ( [0,1]; \mathcal{H} ) \) the space of all functions \(f: [0, 1] \rightarrow \mathcal{H}\) which have absolutely continuous derivative up to order \(k1\). Set
The following lemma concerning basic properties of fractional calculus is needed afterward. The proof can be found in [2], Lemmas 2.4, 2.5, or [7], Theorem 2.4.
Lemma 2.2
Let \(\varphi \in L^{1} ( [0, 1]; \mathcal{H} ) \) and \(\alpha >0\), \(m=[\alpha ]+1\).

(i)
The equality \(( D^{\alpha }I^{\alpha }\varphi) (t)= \varphi (t)\) holds for almost every \(t\in [0,1]\).

(ii)
If \(D^{\alpha m}\varphi \in AC^{m} ( [0,1]; \mathcal{H} ) \), then
$$\bigl( I^{\alpha }D^{\alpha }\varphi \bigr) (t)=\varphi (t)\sum _{j=1} ^{m}\frac{D^{\alpha j}\varphi (0)}{\Gamma (\alpha j+1)}t^{\alpha j} $$for almost every \(t\in [0,1]\).
Mawhin’s continuation theorem
Let X and Y be two Banach spaces.
Definition 2.3
A linear operator \(L: \operatorname {dom}(L) \subset X \rightarrow Y\) is called Fredholm operator with zeroindex if ImL is closed in Y and
It is known that when L is a Fredholm operator with zeroindex, there exist a projection P on X and a projection Q on Y such that
Moreover, the operator \(L_{P}\) defined as \(L_{P}:=L\vert _{\operatorname {dom}(L) \cap \ker P}\) is invertible. Let \(K_{P}:=L_{P}^{1}\) and set
called the generalized inverse of L. On the other hand, for isomorphism \(J: \operatorname{Im}Q\rightarrow \ker L\), the operator \(K_{P,Q}+JQ: Y \rightarrow \operatorname {dom}(L)\) is isomorphic. Moreover,
for \(x\in \operatorname {dom}(L)\).
Definition 2.4
Suppose that L is a Fredholm operator with zeroindex and Ω is a bounded subset of X such that \(\operatorname {dom}(L) \cap \Omega \neq \emptyset \). An operator \(N : X \rightarrow Y\) is called Lcompact on Ω̅ if the following two conditions hold:

(i)
\(QN: \overline{\Omega } \rightarrow Y\) is continuous and \(QN(\overline{\Omega })\) is bounded in Y;

(ii)
\(K_{P,Q}N : \overline{\Omega } \rightarrow X\) is completely continuous.
Let L be a Fredholm operator with zeroindex and N be Lcompact on Ω̅. It follows from Mawhin’s equivalent theorem [28] that the equation
is equivalently converted into the fixed point equation
where \(\Phi =P+(JQ+K_{P,Q}N)\) is a completely continuous operator. This can be solvable thanks to the following theorem, called Mawhin’s continuation theorem.
Theorem 2.5
Assume that Ω is a bounded and open subset in X, and L is a Fredholm operator with zeroindex, and N is an Lcompact operator on Ω̅. Additionally, suppose that the following three assumptions hold:

(i)
\(Lx\neq \lambda Nx\) for \(x \in \partial \Omega \cap ( \operatorname {dom}(L) \backslash \ker L ) \) and for \(\lambda \in (0,1)\);

(ii)
\(QNx \neq 0\) for \(x \in \partial \Omega \cap \ker L\);

(iii)
for any isomorphism J from ImQ to kerL,
$$\deg_{B} ( \mathrm{JQN}\vert _{\ker L},\Omega \cap \ker L, 0 ) \neq 0, $$where \(Q:Y\rightarrow Y\) is a projection given as above.
Then the equation \(Lx=Nx\) has a solution in \(\overline{\Omega }\cap \operatorname {dom}(L)\).
For a comprehensive treatment on the subject of the coincidence degree theory and Mawhin’s continuation theorem as well, we refer the reader to [8, 28, 29].
Finally, we restate the following compactness criterion on the space of continuous functions \(C(X,Y)\) which is regarded as a generalization of the ArzelaAscoli theorem [30], Section 1.1.3.
Lemma 2.6
Let X be a compact topological space and Y be a metric space. A subset of \(C(X,Y)\) is relatively compact if and only if it is pointwise relatively compact and equicontinuous.
Key lemmas
Let
be the Banach space with the norm
where \(\Vert \cdot \Vert _{\infty }\) stands for the usual supnorm on \(C ( [0,1]; \mathcal{H} ) \). Also, let \(Y=L^{1} ( [0, 1]; \mathcal{H} ) \) be the Banach space with the Lebesgue norm
Next we define a linear operator \(L: X \rightarrow Y\) by
in which
Moreover, we define a (nonlinear) operator \(N : X \rightarrow Y\) by
Clearly, (1.1) is equivalent to
in which the operators L and N are defined by (2.1) and (2.2), respectively.
Fredholm property of the fractional differential operator L
We first note that if \(x \in \operatorname {dom}(L)\), then \(D^{\alpha }x \in Y\) and \(I^{2\alpha }x \in AC^{2} ( [0,1]; \mathcal{H} ) \). From Lemma 2.2 and \(x(0) = \theta \), it follows that
for every \(t\in [0,1]\) and some constant element \(\mathbf{c}\in \mathcal{H}\). Moreover, in view of
we deduce that
where

\(T = I  A\),

\(\phi: Y \rightarrow H\) is the following bounded linear operator:
$$ \phi ( y) = \frac{1}{\Gamma (\alpha)} \biggl( A \int_{0}^{ \eta }y(t)\,dt  \int_{0}^{1} y(t)\,dt \biggr) \quad \mbox{for } y \in Y. $$(2.3)
Thus, for convenience, a function \(x\in \operatorname {dom}(L)\) will be represented equivalently as
Note that \(D^{\alpha }x=y\).
Remark 2.7
Note that ϕ is a surjective map. Indeed, for each \(\mathbf{c} \in \mathcal{H}\), setting
we have \(\phi (y)=\mathbf{c}\), by a straightforward computation.
Due to the definition of L and (2.4), by some simple calculations, we can indicate the kernel and the image of L as follows:
and
Now let us recall the definition of the MoorePenrose inverse of linear operators. Suppose that T is a linear operator on a Hilbert space. Then we call some linear operator \(T^{\dagger }\) the MoorePenrose inverse of T if

(i)
\(T^{\dagger }TT^{\dagger }= T^{\dagger }\);

(ii)
\(TT^{\dagger }T =T\);

(iii)
\(TT^{\dagger }\) and \(T^{\dagger }T\) are selfadjoint operators.
It is well known that if T is bounded and has closed range, then \(T^{\dagger }\) uniquely exists and it is continuous. Moreover, \(TT^{\dagger }\) (resp. \(T^{\dagger }T\)) is an orthogonal projection on ImT (resp. \(\operatorname {Im}T^{\dagger }\)). For more details, one can see [31].
Lemma 2.8
Assume that T is a continuous operator with closed range. Then the following two assertions are true.

(i)
There exists a continuous projection \(Q:Y\rightarrow Y\) such that
$$\ker Q=\operatorname {Im}L \quad \textit{and}\quad Y=\operatorname {Im}L \oplus \operatorname {Im}Q. $$ 
(ii)
L is a Fredholm operator with zeroindex if and only if so is T.
Proof
(i) Define a map \(Q : Y \rightarrow Y\) by setting, for \(y \in Y\), that
where \(q=\frac{\Gamma (\alpha +1)}{\eta^{\alpha }1}\) is a real constant. It is clear that Q is a continuous linear operator. Moreover, Q is also a projection. Indeed, we observe that
since
Then, for every \(r \in \mathbb {R}\), we have
From (2.3) and (2.6)(2.8), it follows that
for every \(t\in [0,1]\). Hence Q is a projection as asserted. On the other hand, we have
due to the fact that \(TT^{\dagger }\) and \(( I  TT^{\dagger } ) \) are projections and \(\operatorname {Im}( TT^{\dagger } ) =\operatorname {Im}T\). This means that
and consequently, \(Y=\operatorname {Im}L\oplus \operatorname {Im}Q\).
(ii) To prove this part, we start with the observation that
Indeed, since \(TT^{\dagger }\) is a projection and
the ‘⊆’ inclusion of (2.9) can be obtained easily. For the converse inclusion, let \(y(t)=\mathbf{c}t^{\alpha 1}\) for \(\mathbf{c} \in \ker T^{\dagger }\). Then, by the surjectivity of ϕ, there exists \(z\in Y\) such that \(q\phi (z)=\mathbf{c}\). It follows that
for every \(t\in [0,1]\). This shows that \(y\in \operatorname {Im}Q\).
Now, by (2.9), we can deduce that ImQ, a complement of ImL in Y, is isomorphic to \(\ker T^{\dagger }\). Besides, since ImT is closed, so is \(\operatorname {Im}L=\phi^{1}(\operatorname {Im}T)\), due to the continuity of ϕ. We are to prove the main claim of this part. Using the fact that
we can see that T is a Fredholm operator with zeroindex if and only if
This is clear to be a necessary and sufficient condition for which L is a Fredholm operator with zeroindex. Then the proof of lemma is complete. □
Remark 2.9
If (1.5) holds, we can give another expression of Q quite simpler than (2.7). Specifically, we set
This also claims that the projection Q as such is in general not unique.
Remark 2.10
Lemma 2.8 will be no longer true in the general case that \(\mathcal{H}\) is a Banach space. If \(\mathcal{H}\) is a Banach space rather than a Hilbert space, for this lemma to remain true, we need to add one more assumption to this lemma. Additional assumption is that T has the MoorePenrose inverse \(T^{\dagger }\in \mathcal{L(H)}\).
Now, we define a linear operator \(P : X \rightarrow X\) by
Note that P is the continuous operator on X. Moreover, the following lemma provides some properties of P as well as \(K_{P}\).
Lemma 2.11
The following statements are true.

(i)
The operator P in (2.11) is projection and satisfies that
$$\ker L=\operatorname {Im}P,\qquad X = \ker P \oplus \ker L. $$ 
(ii)
The map \(K_{P} : \operatorname {Im}L \rightarrow \operatorname {dom}(L) \cap \ker P\) is defined by
$$ K_{P}y(t) = T^{\dagger } \phi (y)t^{\alpha 1} + I^{\alpha }y(t)\quad \textit{for } t \in [0, 1], $$(2.12)satisfying
$$K_{P} = L_{P}^{1}\quad \textit{and} \quad \Vert K_{P}y\Vert \leq C\Vert y\Vert _{1}, $$where \(C =\frac{1}{\Gamma (\alpha)} ( 1 + \Vert T^{\dagger }\Vert _{\mathcal{L(H)}} ( 1 + \Vert A\Vert _{\mathcal{L(H)}} )) \).
Proof
(i) We first notice that
for every \(t\in [0,1]\). Since \(( IT^{\dagger }T ) \) is a projection, then so is P, by a straightforward computation. Furthermore, P is onto kerL, that is, \(\operatorname {Im}P = \ker L\). Indeed, because the inclusion \(\operatorname {Im}P \subset \ker L\) can be proved simply, we need only prove the converse one: \(\ker L \subset \operatorname {Im}P\). Suppose that \(x(t)=\mathbf{c} t^{\alpha 1}\in \ker L\), where
or
Then, by again using \(D^{\alpha 1} ( t^{\alpha 1} ) (t)= \Gamma (\alpha)\), we have
This implies \(x \in \operatorname {Im}P\). As a consequence, we obtain \(X = \ker P \oplus \ker L\).
(ii) Assume that \(y \in \operatorname {Im}L\). Then \(\phi (y) = T\mathbf{c} \) for some \(\mathbf{c} \in H\). From (2.11)(2.12), we have
that means \(K_{P}y\in \ker P\). Moreover, we get \(K_{P}y\in \operatorname {dom}(L)\) due to the fact that
combined with (2.4). Hence \(K_{P}\) is well defined.
We next show that \(K_{P}\) is the inverse operator of \(L_{P}\). Clearly, it suffices to show that \(K_{P}\) is a leftinverse of \(L_{P}\). Let \(x \in \operatorname {dom}(L) \cap \ker P\), then
where
Thus
The rest of the proof regarding the continuity of \(K_{P}\) is quite simple. Noting from (2.12) that
and combining (2.3), (2.12)(2.13) together, we immediately establish the following estimates:

\(\Vert \phi (y)\Vert _{\mathcal{H}} \leq \frac{1}{\Gamma (\alpha)} ( 1 + \Vert A\Vert _{\mathcal{L(H)}} ) \Vert y\Vert _{1}\),

\(\Vert K_{P}y\Vert _{\infty }\leq \frac{1}{\Gamma (\alpha)} ( 1 + \Vert T^{\dagger }\Vert _{\mathcal{L(H)}} + \Vert A\Vert _{ \mathcal{L(H)}}\Vert T^{\dagger }\Vert _{\mathcal{L(H)}} ) \Vert y\Vert _{1}\),

\(\Vert D^{\alpha 1}(K_{P}y)\Vert _{\infty }\leq ( 1 + \Vert T^{\dagger }\Vert _{\mathcal{L(H)}} + \Vert A\Vert _{\mathcal{L(H)}} \Vert T^{\dagger }\Vert _{\mathcal{L(H)}} ) \Vert y\Vert _{1}\).
These inequalities lead to \(\Vert K_{P}y\Vert \leq C\Vert y\Vert _{1}\), where \(C =\frac{1}{\Gamma (\alpha)} ( 1 + \Vert T^{\dagger} \Vert _{\mathcal{L(H)}} ( 1 + \Vert A\Vert _{\mathcal{L(H)}} )) \) and thus finish the proof of the lemma. □
Lcomplete continuity of the nonlinear operator N
Lemma 2.12
The operator N is (2.2) is Lcompact.
Proof
Assume that \(\Omega \subset X\) is bounded. By assumption (A4) on f, there is a function \(\varphi_{\Omega } \in Y\) such that
for almost every \(t \in [0, 1]\) and for all \(x \in \Omega \). It follows from the equality
and (2.3) and (2.14) that \(QN(\overline{\Omega })\) is bounded. In addition, the continuity of QN is deduced thanks to the Lebesgue dominated convergence theorem.
Next we prove that \(K_{P,Q}N\) is completely continuous. Again using the Lebesgue dominated convergence theorem, we can show that \(K_{P,Q}N\) is continuous. It suffices to prove that \(K_{P,Q}N\) is compact. For this purpose, we first observe that, for every \(x \in \Omega \) and for every \(0\leq t_{1}< t_{2} \leq 1\),
Using the inequality that
we imply from (2.15) that
In addition, due to (2.13) we have
These two estimates (2.16), (2.17) show that the families \(K_{P,Q}N(\Omega)\) and \(D^{\alpha 1}K_{P,Q}N(\Omega)\) are equicontinuous in \(C([0,1]; \mathcal{H})\).
On the other hand, in view of (A4), the set \(\{ Nx(t): x\in \Omega \} \) is relatively compact in \(\mathcal{H}\) almost all \(t\in [0,1]\). Then, due to the Lebesgue dominated convergence theorem, we can prove that \(\{ K_{P,Q}Nx(t): x\in \Omega \} \) and \(\{ D^{\alpha 1}K_{P,Q}Nx(t): x\in \Omega \} \) are relatively compact in \(\mathcal{H}\) for every \(t\in [0,1]\). Therefore, Lemma 2.6 guarantees that \(K_{P,Q}N(\Omega)\) is relatively compact in X. This means that the operator \(K_{P,Q}N\) is compact. The lemma is then proved. □
Main result
The main result of this paper is the following theorem.
Theorem 3.1
Let (A1)(A4) hold, and let the following assumptions (B1)(B3) be satisfied.

(B1)
There exist positive real functions \(a, b, c \in L^{1}[0,1]\) with \(( \frac{1}{\Gamma (\alpha)} +C ) ( \Vert a\Vert _{L^{1}} + \Vert b\Vert _{L^{1}} ) < 1 \) such that
$$ \bigl\Vert f(t, x, y)\bigr\Vert _{\mathcal{H}} \leq a(t)\Vert x\Vert _{ \mathcal{H}} + b(t)\Vert y\Vert _{\mathcal{H}} + c(t) $$(3.1)for almost every \(t\in [0,1]\), for every \((x, y) \in H\times H\), with constant C given in Lemma 2.11.

(B2)
There is a constant \(\Lambda_{1}>0\) such that if \(x \in \operatorname {dom}(L)\) and \(\min_{t \in [0, 1]}\Vert D^{\alpha 1}x(t)\Vert _{\mathcal{H}} > \Lambda_{1}\), then
$$ \int_{\eta }^{1}f \bigl( t, x(t), D^{\alpha 1}x(t) \bigr)\,dt \notin \operatorname {Im}T. $$(3.2) 
(B3)
There is a constant \(\Lambda_{2}>0\) together with an isomorphism \(J: \operatorname {Im}Q \rightarrow \ker L \) satisfying either
$$ \bigl\langle \mathbf{c}, \mathrm{JQN}\bigl( \mathbf{c} t^{\alpha 1} \bigr) \bigr\rangle \geq 0, $$(3.3)or
$$ \bigl\langle \mathbf{c}, \mathrm{JQN}\bigl( \mathbf{c} t^{\alpha 1} \bigr) \bigr\rangle \leq 0 $$(3.4)for every \(\mathbf{c} \in \ker T\) satisfying \(\Vert \mathbf{c}\Vert _{\mathcal{H}} > \Lambda_{2}\), and for some \(t\in (0,1]\).
Then problem (1.1) has at least one solution in X.
The proof of the theorem needs several auxiliary results. We present them first, in the next three lemmas.
Lemma 3.2
Let \(\Omega_{1} = \{ x \in \operatorname {dom}(L) \setminus \ker L : Lx = \lambda Nx, 0<\lambda \leq 1 \} \). Then \(\Omega_{1}\) is a bounded subset in X.
Proof
Assume \(x \in \Omega_{1}\), so \(Lx = \lambda Nx\) for some \(0< \lambda \leq 1\). We have
then, by (2.6), we have \(\phi (Nx) \in \operatorname {Im}T\). Therefore
It follows from (B2) that \(\Vert D^{\alpha 1}x(s)\Vert _{ \mathcal{H}} \leq \Lambda_{1}\) for some \(s\in [0,1]\). We thus obtain
and hence
Furthermore, since P is the projection on X, \(( \mathrm{Id}_{X}P ) x \in \operatorname {dom}(L) \cap \ker P\), we have
where constant C in Lemma 2.11 and \(\mathrm{Id}_{X}\) stands for the identity mapping on X. Combining (3.6)(3.7) yields
Besides, (B1) gives us
Thus
that is, \(\Omega_{1}\) is bounded. The proof is complete. □
Lemma 3.3
Let \(\Omega_{2} = \{ x \in \ker L : Nx \in \operatorname {Im}L \} \). Then \(\Omega_{2}\) is a bounded subset in X.
Proof
Assume \(x \in \Omega_{2}\). Then \(Nx \in \operatorname {Im}L\) and there is some \(\mathbf{c} \in \ker T\) for which \(x(t) =\mathbf{c} t^{\alpha 1}\). Similarly, arguing as in the proof of Lemma 3.2, we also have \(\Vert D^{\alpha 1}x(t_{0})\Vert _{\mathcal{H}} \leq \Lambda_{1}\) for some \(s\in [0,1]\). This leads to \(\Vert \mathbf{c}\Vert _{\mathcal{H}} \leq \frac{\Lambda_{1}}{\Gamma (\alpha)}\), and hence
that is, \(\Omega_{2}\) is bounded, or the proof is complete. □
Lemma 3.4
Setting
and
we have

(i)
The set \(\Omega^{+}\) is bounded in X if (3.3) holds.

(ii)
The set \(\Omega^{}\) is bounded in X if (3.4) holds.
Proof
(i) Assume \(x \in \Omega_{3}^{+}\). Then \(x(t) = \mathbf{c} t^{\alpha 1}\) for some \(\mathbf{c} \in \ker T\) and
This deduces for the case \(\lambda = 0\) that \(Nx \in \ker (JQ) = \ker Q = \operatorname{Im}L\), then \(x \in \Omega_{2}\). Due to Lemma 3.3, \(\Vert x\Vert \) is bounded.
For the case \(\lambda >0\), we argue by contradiction. Suppose that \(\Vert \mathbf{c}\Vert _{\mathcal{H}} > \Lambda_{2}\). Then, by (3.3), there exists \(t>0\) for which
this cannot happen. Hence, we have \(\Vert x\Vert =\Vert x\Vert _{\infty } = \Vert \mathbf{c} \Vert _{\mathcal{H}} \leq \Lambda _{2}\), or \(\Omega_{3}^{}\) is bounded.
(ii) Similarly, the boundedness of \(\Omega_{3}^{+}\) is established. The lemma is proved. □
Now we focus on proving Theorem 3.1.
Proof of Theorem 3.1
Our purpose is to apply Theorem 2.5 by verifying all its conditions (i)(iii). Firstly, we set Ω is a bounded and open subset in X for which \(\bigcup_{i=1}^{3}\overline{\Omega }_{i} \subset \Omega \), where
Simply, assumption (i) and assumption (ii) in Theorem 2.5 are satisfied thanks to Lemma 3.2 and Lemma 3.3. It is sufficient to check if assumption (iii) also holds. To do this, we use the Brouwer degree property of the invariance with a continuous homotopy. Namely, consider a homotopy \({H_{\lambda }: X \rightarrow X}\) defined by
in which \(J : \operatorname {Im}Q \rightarrow \ker L\) is isomorphic assumed in (B3). Lemma 3.4 gives us that
Therefore,
Hence, (iii) holds as claimed. The proof of the theorem is complete. □
Remark 3.5
If \(\mathcal{H}\) is a Banach space, assumption (B3) needs to be slightly modified. Precisely, two inequalities (3.3), (3.4) are respectively generalized as
and
in which \(\tilde{\mathbf{c}}\in \mathcal{H}'\) (the dual space of \(\mathcal{H}\)) with \(\langle \tilde{\mathbf{c}}, \mathbf{c} \rangle = \Vert \mathbf{c}\Vert ^{2}_{\mathcal{H}}\), and \(\langle \cdot,\cdot \rangle \) now stands for the scalar product for the duality \(\mathcal{H}'\), \(\mathcal{H}\). Then the proof of Lemma 3.4 is modified correspondingly; the result of Theorem 3.1 still holds.
Example and discussion
This section is to provide one illustrative example of Theorem 3.1. Before presenting the example, we should note that giving such a significant example is a challenging task in the case that \(\mathcal{H}\) is infinite dimensional. It is caused by two major reasons. The first one is that it is not trivial to give a suitable nonlinear function f in an infinite dimensional space. The second one, which is more difficult to overcome, is that so far there has been no effective method to find the MoorePenrose inverse of a general bounded linear operator on Hilbert spaces or of a Fredholm operator with zeroindex. For instance, some recent works [26, 32] for the first time give the examples in an infinite dimensional space. However, to the best of our knowledge, these examples are still controversial.
Indeed, in [26], Section 4, and also in [32], Section 4, for the case \(\mathcal{H}=l^{2}\), there is a gap in setting the expression of f. Namely, the function \(f_{1}(t, u, v)\) in [26], Section 4, is not continuous with respect to the third variable at points which \(\Vert v\Vert _{l^{2}}=\Vert (y_{i})_{i=1}^{\infty }\Vert _{l^{2}}\geq 1\) such that \(y_{1}=0\). This means that f is not a Carathéodory function, one assumption must be satisfied. Moreover, in [32], Section 4, finding the MoorePenrose inverse of \(\mathcal{M}\) is not much helpful because the given operator \(\mathcal{M}\) does not need to be Fredholm with zeroindex as a prerequisite. This is derived from another gap in [32], Lemma 2.6, that the condition on A is not sufficient to conclude that \(\operatorname {Im}Q=\ker L\).
In what follows, we give an example for the case that \(\mathcal{H}\) is finite dimensional. The example in an infinite dimensional space would be presented in our future research with more careful consideration.
Example 4.1
Consider the existence of solutions to the following fractional differential equations:
subject to
Set \(\alpha =\frac{3}{2}\), \(\eta =\frac{1}{2}\),
and define \(f: [0, 1] \times \mathbb{R}^{2}\times \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) by setting
where \(f_{1}, f_{2}: [0, 1] \times \mathbb{R}^{2}\times \mathbb{R} ^{2} \rightarrow \mathbb{R}\) are functions given by
and
for \(t \in [0, 1]\) and \(x=(x_{1}, x_{2})\), \(y=(y_{1}, y_{2}) \in \mathbb{R}^{2}\). Then problem (4.1)(4.2) becomes problem (1.1). Our purpose is to apply Theorem 3.1 in order to establish the existent result of (4.1)(4.2).
It is clear that (A1)(A4) are satisfied. We need to verify (B1)(B3). From (4.3)(4.5) we have
for every \(0\leq t\leq 1\) and for every \(x, y \in \mathbb{R}^{2}\), in which
Moreover, some direct calculations give us
It follows that
Hence, (B1) holds. On the other hand, observe that
for all \(x \in \operatorname {dom}(L)\). This implies that
due to the fact that \(\operatorname {Im}T = \langle (1,1) \rangle = \{ (c, c): c \in \mathbb{R} \} \); that is, (B2) holds. Finally, to check (B3) we see that
and
and \(J:\operatorname {Im}Q\rightarrow \ker L\) isomorphism defined as
Thus, we get
for all \(z \in L^{1}([0, 1], \mathbb{R}^{2})\). By \(\ker T=\langle (9,2) \rangle \), suppose that \(\gamma = (9a, 2a) \in \ker T\) for some \(a\in \mathbb {R}\). Then
and hence
It follows from (2.3) and (4.7) that
Combining (4.6) and (4.8), we obtain
Therefore,
where \(\theta (t)=\frac{85\sqrt{t}}{1664(\sqrt{2}4)}\leq 0\) for every \(0\leq t \leq 1\). This deduces that \(\langle \gamma, \mathrm{JQN}( \gamma t^{\alpha 1} ) \rangle \leq 0\) for all \(t\in [0,1]\), for \(\vert a\vert \) large enough. Hence, (B3) holds. Problem (4.1)(4.2) thus has at least one solution.
Remark 4.2
We should comment on the computation of the MoorePenrose inverse matrix \(T^{\dagger }\) in the above example since this step is crucial. Basically, the common method for computing the MoorePenrose inverse of matrix is the singular value decomposition (SVD) method (see [31]). This method is accurate but timeintensive since it requires a large amount of computational resources, especially in the case of large scale matrix. Another approach for computing the MoorePenrose inverse of matrix has been proposed recently, i.e., the method based on Tensorproduct matrix (see [33]). The new method can be applied for rectangular matrix with fullrank and square matrix with rankdeficient only. However, the method works well compared to SVD method in some aspects: larger dimension matrix and less timeconsuming.
Conclusions
In this work, we have established an existence result for the class of fractional boundary value problems with a general resonant condition in Hilbert spaces. We have provided a sufficient and necessary condition for which some RiemannLiouville fractional differential operators associated with threepoint boundary condition are Fredholm with zeroindex. This means that we have handled the problem with a wide range of resonant conditions in which Mawhin’s continuation theorem can be applied. Our result is a natural generalization of some recent ones [24–27].
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We are thankful to the editor and the anonymous reviewers for many valuable suggestions to improve this paper.
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MSC
 34A08
 34B10
 34B15
Keywords
 coincidence degree
 threepoint boundary value problem
 fractional differential equation
 resonance