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
$$I^{\alpha }f(t):= \frac{1}{\Gamma (\alpha)} \int_{0}^{t}(t-s)^{ \alpha -1}f(s)\,ds\quad \mbox{for } t>0. $$
- ⋄:
-
The Riemann-Liouville fractional derivative of order α of f is given by
$$D^{\alpha }f(t):= \frac{d^{m}}{dt^{n}} \bigl( I^{m-\alpha }f \bigr) (t) = \frac{1}{\Gamma (m - \alpha)} \frac{d^{m}}{dt^{m}} \int_{0}^{t} (t-s)^{m - \alpha -1} f(s)\,ds\quad \mbox{for } t>0, $$
where \(m = [\alpha ] + 1\),
provided that the terms on the right-hand side of the above equalities are pointwise defined on \([0, 1]\). In the above expressions, the sign ‘∫’ denotes the Bochner integral.
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 \(k-1\). Set
$$AC \bigl( [0,1]; \mathcal{H} \bigr):= AC^{1} \bigl( [0,1]; \mathcal{H} \bigr). $$
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 zero-index if ImL is closed in Y and
$$\operatorname {codim}\operatorname {Im}L=\dim \ker L< \infty. $$
It is known that when L is a Fredholm operator with zero-index, there exist a projection P on X and a projection Q on Y such that
$$\ker L=\operatorname {Im}P ,\qquad \operatorname {Im}L= \ker Q . $$
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
$$K_{P,Q} := K_{P}(I-Q), $$
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,
$$( K_{P,Q}+JQ ) ^{-1}x = \bigl( L+J^{-1}P \bigr) x $$
for \(x\in \operatorname {dom}(L)\).
Definition 2.4
Suppose that L is a Fredholm operator with zero-index 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 L-compact 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 zero-index and N be L-compact on Ω̅. It follows from Mawhin’s equivalent theorem [28] that the equation
$$Lx=Nx \quad \mbox{for } x \in \overline{\Omega } $$
is equivalently converted into the fixed point equation
$$x=\Phi x \quad \mbox{for } x \in \overline{\Omega }, $$
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 zero-index, and
N
is an
L-compact 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 Arzela-Ascoli 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
$$X = \bigl\{ x\in C \bigl( [0,1]; \mathcal{H} \bigr): x(t)= I^{\alpha -1} _{0^{+}}\varphi (t), \varphi \in C \bigl( [0,1]; \mathcal{H} \bigr) \bigr\} $$
be the Banach space with the norm
$$\Vert x\Vert = \max \bigl\{ \Vert x\Vert _{\infty }, \bigl\Vert D ^{\alpha -1}x\bigr\Vert _{\infty } \bigr\} , $$
where \(\Vert \cdot \Vert _{\infty }\) stands for the usual sup-norm on \(C ( [0,1]; \mathcal{H} ) \). Also, let \(Y=L^{1} ( [0, 1]; \mathcal{H} ) \) be the Banach space with the Lebesgue norm
$$\Vert y\Vert _{1}= \int_{0}^{1} \bigl\Vert y(t)\bigr\Vert _{\mathcal{H}}\,dt. $$
Next we define a linear operator \(L: X \rightarrow Y\) by
$$ Lx=D^{\alpha }x, $$
(2.1)
in which
$$x\in \operatorname {dom}(L) = \bigl\{ x = I_{0^{+}}^{\alpha -1}\varphi \in X: \varphi \in AC \bigl( [0, 1]; \mathcal{H} \bigr), x(0) = \theta, D^{\alpha -1}x(1) = AD^{\alpha -1}x(\eta) \bigr\} . $$
Moreover, we define a (nonlinear) operator \(N : X \rightarrow Y\) by
$$ Nx(t) = f \bigl( t, x(t), D^{\alpha -1}x(t) \bigr)\quad \mbox{for almost every } t \in [0, 1]. $$
(2.2)
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
$$x(t) = \mathbf{c}t^{\alpha -1} + \bigl( I^{\alpha }D^{\alpha }x \bigr) (t) $$
for every \(t\in [0,1]\) and some constant element \(\mathbf{c}\in \mathcal{H}\). Moreover, in view of
$$D^{\alpha -1}x(1) = AD^{\alpha -1}x(\eta), $$
we deduce that
$$T(\mathbf{c}) = \phi \bigl( D^{\alpha }x \bigr), $$
where
Thus, for convenience, a function \(x\in \operatorname {dom}(L)\) will be represented equivalently as
$$ x(t) = \mathbf{c} t^{\alpha -1} + I^{\alpha }y(t) \quad \mbox{with } T(\mathbf{c}) = \phi ( y). $$
(2.4)
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
$$y(t)=\frac{\Gamma (\alpha)(\eta -2t)}{1-\eta } \mathbf{c} \quad \mbox{for } t \in [0,1], $$
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:
$$ \ker L = \bigl\{ x \in X : x(t) = \mathbf{c} t^{\alpha -1} \mbox{ for } \mathbf{c}\in \ker T \bigr\} \cong \ker T, $$
(2.5)
and
$$ \operatorname{Im}L = \bigl\{ y \in Y : \phi (y) \in \operatorname{Im}T \bigr\} = \phi^{-1}(\operatorname {Im}T). $$
(2.6)
Now let us recall the definition of the Moore-Penrose inverse of linear operators. Suppose that T is a linear operator on a Hilbert space. Then we call some linear operator \(T^{\dagger }\) the Moore-Penrose inverse of T if
-
(i)
\(T^{\dagger }TT^{\dagger }= T^{\dagger }\);
-
(ii)
\(TT^{\dagger }T =T\);
-
(iii)
\(TT^{\dagger }\) and \(T^{\dagger }T\) are self-adjoint 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 zero-index if and only if so is
T.
Proof
(i) Define a map \(Q : Y \rightarrow Y\) by setting, for \(y \in Y\), that
$$ Qy(t)=q \bigl( I - TT^{\dagger } \bigr) \phi (y)t^{\alpha -1}\quad \mbox{for } t \in [0, 1], $$
(2.7)
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
$$\bigl( I - TT^{\dagger } \bigr) A = \bigl( I - TT^{\dagger } \bigr) $$
since
$$\bigl( I - TT^{\dagger } \bigr) (I-A) = \bigl( I - TT^{\dagger } \bigr) T = T-TT^{\dagger }T=0. $$
Then, for every \(r \in \mathbb {R}\), we have
$$\begin{aligned} \bigl( I - TT^{\dagger } \bigr) (r A-I) =& r \bigl( I - TT^{\dagger } \bigr) A - \bigl( I - TT^{\dagger } \bigr) \\ =& r \bigl( I - TT ^{\dagger } \bigr) - \bigl( I - TT^{\dagger } \bigr) \\ =& (r-1) \bigl( I - TT^{\dagger } \bigr). \end{aligned}$$
(2.8)
From (2.3) and (2.6)-(2.8), it follows that
$$\begin{aligned} Q(Qy) (t) =& q \bigl( I - TT^{\dagger } \bigr) \phi (Qy)t^{\alpha -1} \\ =&q \bigl( I - TT^{\dagger } \bigr) \biggl( q \bigl( I - TT^{\dagger } \bigr) \frac{( \eta^{\alpha }A-I)}{\alpha \Gamma (\alpha)}\phi (y) \biggr) t^{\alpha -1} \\ =& q \bigl( I - TT^{\dagger } \bigr) \phi (y) t^{\alpha -1} \\ =& Qy(t) \end{aligned}$$
for every \(t\in [0,1]\). Hence Q is a projection as asserted. On the other hand, we have
$$\begin{aligned} y \in \ker Q \quad \Leftrightarrow &\quad \phi (y)\in \ker \bigl( I - TT^{ \dagger } \bigr) \\ \quad \Leftrightarrow &\quad \phi (y)\in \operatorname {Im}\bigl( TT^{\dagger } \bigr) \\ \quad \Leftrightarrow &\quad \phi (y) \in \operatorname {Im}T \\ \quad \Leftrightarrow &\quad y \in \operatorname {Im}L, \end{aligned}$$
due to the fact that \(TT^{\dagger }\) and \(( I - TT^{\dagger } ) \) are projections and \(\operatorname {Im}( TT^{\dagger } ) =\operatorname {Im}T\). This means that
$$\ker Q = \operatorname {Im}L, $$
and consequently, \(Y=\operatorname {Im}L\oplus \operatorname {Im}Q\).
(ii) To prove this part, we start with the observation that
$$ \operatorname {Im}Q = \bigl\{ y\in Y: y(t)=\mathbf{c}t^{\alpha -1}, t\in [0,1] \mbox{ for } \mathbf{c}\in \ker T^{\dagger } \bigr\} . $$
(2.9)
Indeed, since \(TT^{\dagger }\) is a projection and
$$\ker \bigl( TT^{\dagger } \bigr) =\ker T^{\dagger }, $$
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
$$\begin{aligned} Qz(t) = q \bigl( I - TT^{\dagger } \bigr) \phi (z) t^{\alpha -1}= \bigl( I - TT^{\dagger } \bigr) \mathbf{c} t^{\alpha -1}= \mathbf{c} t^{\alpha -1} = y(t) \end{aligned}$$
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
$$H=\operatorname {Im}\bigl( TT^{\dagger } \bigr) \oplus \ker \bigl( TT^{\dagger } \bigr) =\operatorname {Im}T \oplus \ker T^{\dagger }, $$
we can see that T is a Fredholm operator with zero-index if and only if
$$ \dim \ker T^{\dagger }= \dim \ker T < \infty. $$
(2.10)
This is clear to be a necessary and sufficient condition for which L is a Fredholm operator with zero-index. 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
$$Qy(t)=q(I-\kappa T)\phi (y)t^{\alpha -1}. $$
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 Moore-Penrose inverse \(T^{\dagger }\in \mathcal{L(H)}\).
Now, we define a linear operator \(P : X \rightarrow X\) by
$$ Px(t) = \frac{1}{\Gamma (\alpha)} \bigl( I - T^{\dagger }T \bigr) D^{ \alpha -1}x(0)t^{\alpha -1},\quad t \in [0, 1], x\in X. $$
(2.11)
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
$$\textstyle\begin{cases} D^{\alpha -1} ( t^{\alpha -1} ) (t)=\Gamma (\alpha), \\ D^{\alpha } ( t^{\alpha -1} ) (t)=0 \end{cases} $$
for every \(t\in [0,1]\). Since \(( I-T^{\dagger }T ) \) 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
$$\mathbf{c}\in \ker T = \operatorname {Im}\bigl( I-T^{\dagger }T \bigr), $$
or
$$\mathbf{c}= \bigl( I-T^{\dagger }T \bigr) \mathbf{c}_{1} \quad \mbox{for some } \mathbf{c}_{1}\in \mathcal{H}. $$
Then, by again using \(D^{\alpha -1} ( t^{\alpha -1} ) (t)= \Gamma (\alpha)\), we have
$$Px(t)= \bigl( I-T^{\dagger }T \bigr) \mathbf{c} t^{\alpha -1}= \bigl( I-T ^{\dagger }T \bigr) ^{2} \mathbf{c}_{1}t^{\alpha -1}= \bigl( I-T^{ \dagger }T \bigr) \mathbf{c}_{1}t^{\alpha -1}=x(t). $$
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
$$PK_{P}y(t) = \bigl( I-T^{\dagger }T \bigr) T^{\dagger } \phi (y)= \bigl( I-T ^{\dagger }T \bigr) T^{\dagger }T\mathbf{c} = 0\quad \mbox{for all } t \in [0, 1], $$
that means \(K_{P}y\in \ker P\). Moreover, we get \(K_{P}y\in \operatorname {dom}(L)\) due to the fact that
$$T \bigl( T^{\dagger }\phi (y) \bigr) = \bigl( TT^{\dagger } \bigr) (T \mathbf{c}) = T\mathbf{c}=\phi (y) $$
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 left-inverse of \(L_{P}\). Let \(x \in \operatorname {dom}(L) \cap \ker P\), then
$$x(t)=\mathbf{c} t^{\alpha -1}+I^{\alpha }Lx(t), $$
where
$$T(\mathbf{c})=\phi (Lx)\quad \mbox{for } \mathbf{c} \in \ker \bigl( I-T^{\dagger }T \bigr). $$
Thus
$$\begin{aligned} K_{P}L_{P}x(t) = & T^{\dagger }\phi (Lx)t^{\alpha -1} + I^{\alpha }Lx(t) \\ = & T^{\dagger } ( T \mathbf{c} ) t^{\alpha -1} + I^{\alpha }Lx(t) \\ =& \mathbf{c} t^{\alpha -1} + I^{\alpha }Lx(t) \\ =& x(t),\quad \forall t\in [0,1]. \end{aligned}$$
The rest of the proof regarding the continuity of \(K_{P}\) is quite simple. Noting from (2.12) that
$$ \bigl( D^{\alpha -1}K_{P}y \bigr) (t) =\Gamma (\alpha)T^{\dagger } \phi (y)+ \int_{0}^{t} y(s)\,ds\quad \mbox{for all } t\in [0, 1] $$
(2.13)
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. □
L-complete continuity of the nonlinear operator N
Lemma 2.12
The operator
N
is (2.2) is
L-compact.
Proof
Assume that \(\Omega \subset X\) is bounded. By assumption (A4) on f, there is a function \(\varphi_{\Omega } \in Y\) such that
$$ \bigl\Vert Nx(t)\bigr\Vert _{\mathcal{H}} = \bigl\Vert f\bigl(t, x(t), D^{\alpha -1}x(t)\bigr)\bigr\Vert _{\mathcal{H}} \leq \varphi_{\Omega }(t) $$
(2.14)
for almost every \(t \in [0, 1]\) and for all \(x \in \Omega \). It follows from the equality
$$QNx(t) = q \bigl( I - TT^{\dagger } \bigr) \phi (Nx) t^{\alpha -1} $$
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 \(0\leq t_{1}< t_{2} \leq 1\),
$$\begin{aligned}& \bigl\Vert K_{P,Q}Nx(t_{2}) - K_{P,Q}Nx(t_{1})\bigr\Vert _{\mathcal{H}} \\& \quad \leq \frac{1}{\Gamma (\alpha)} \int_{0}^{t_{1}}\bigl\vert (t_{2}-s)^{ \alpha -1}-(t_{1}-s)^{\alpha -1} \bigr\vert \bigl\Vert Nx(s)\bigr\Vert _{ \mathcal{H}}\,ds \\& \qquad {} + \frac{1}{\Gamma (\alpha)} \int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha -1}\bigl\Vert Nx(s)\bigr\Vert _{\mathcal{H}}\,ds \\& \qquad {} + \bigl\Vert T^{\dagger }\phi (Nx)\bigr\Vert _{\mathcal{H}} \vert t_{2} - t_{1} \vert ^{\alpha -1} \\& \qquad {} +\frac{\vert q\vert \Gamma (\alpha)}{\Gamma (2\alpha)}\bigl\Vert \phi (Nx)\bigr\Vert _{ \mathcal{H}}\bigl\vert t_{2}^{2\alpha -1}-t_{1}^{2\alpha -1}\bigr\vert . \end{aligned}$$
(2.15)
Using the inequality that
$$\bigl\vert a^{p}-b^{p}\bigr\vert \leq \vert a-b \vert ^{p}\quad \mbox{for all } a,b\geq 0 \mbox{ and } 0< p\leq 1, $$
we imply from (2.15) that
$$\begin{aligned}& \bigl\Vert K_{P,Q}Nx(t_{2}) - K_{P,Q}Nx(t_{1})\bigr\Vert _{\mathcal{H}} \\& \quad \leq \frac{2}{\Gamma (\alpha)}\vert t_{2}-t_{1}\vert ^{\alpha -1} \Vert \varphi_{\Omega }\Vert _{1} + \frac{1}{\Gamma (\alpha)} \bigl( 1 + \Vert A\Vert _{ \mathcal{L(H)}} \bigr) \bigl\Vert T^{\dagger }\bigr\Vert _{\mathcal{L(H)}} \vert t_{2} - t_{1}\vert ^{\alpha -1}\Vert \varphi_{\Omega }\Vert _{1} \\& \qquad {}+ \frac{\vert q\vert }{\Gamma (2\alpha)} \bigl( 1 + \Vert A\Vert _{ \mathcal{L(H)}} \bigr) \bigl\Vert T^{\dagger }\bigr\Vert _{\mathcal{L(H)}} \bigl\vert t_{2}^{2\alpha -1}-t_{1}^{2\alpha -1}\bigr\vert \Vert \varphi_{\Omega }\Vert _{1}. \end{aligned}$$
(2.16)
In addition, due to (2.13) we have
$$ \bigl\Vert D^{\alpha -1} ( K_{P,Q}Nx) (t_{2}) - D^{\alpha -1} ( K_{P,Q}Nx) (t_{1}) \bigr\Vert _{\mathcal{H}} \leq \int_{t_{1}} ^{t_{2}}\varphi_{\Omega }(s)\,ds + C_{3}\Vert \varphi_{\Omega }\Vert _{1}\bigl\vert t_{2}^{\alpha } - t_{1}^{\alpha }\bigr\vert . $$
(2.17)
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. □