A note on fractional spaces generated by the positive operator with periodic conditions and applications

Abstract

In this study, the second order differential operator $A^{x}$ defined by the formula $A^{x}u=-u_{xx}(x)+\delta u(x)$, $\delta \geq 0$, with domain $D(A^{x})= \{ u(x):u(x),u^{\prime }(x),u^{\prime \prime } (x)\in C(R^{1}), u(x)=u(x+2\pi ),x\in R^{1}, \int_{0}^{2\pi}u(x)\,dx=0 \}$ is considered. The Green function of the differential operator $A^{x}$ is constructed. The estimates for the Green function are obtained. It is proved that for any $\alpha\in(0,\frac{1}{2})$, the norms in the spaces $E_{\alpha }=E_{\alpha}(\mathring{C} (R^{1} ),A^{x})$ and $\mathring{C}^{2\alpha} (R^{1} )$ are equivalent.

The positivity of the operator $A^{x}$ in Hölder spaces of $\mathring{C}^{2\alpha} (R^{1} )$, $\alpha\in (0,\frac{1}{2})$, is proved. In the applications, theorems on well-posedness of local and nonlocal boundary value problems for elliptic equations in the Hölder spaces are obtained.

1 Introduction

Various problems for partial differential equations can be considered as an abstract boundary value problem for an ordinary differential equation in a Banach space E with a densely defined unbounded space operator A. The role played by positivity of the differential and difference operators in a Banach space in the study of various properties of boundary value problems for partial differential equations, of stability of difference schemes for partial differential equations and summation of Fourier series is well known [1–5]. The positivity of a wider class of differential and difference operators in Banach spaces has been studied by many researchers [6–25].

Let E be a Banach space and $A:D(A)\subset E\rightarrow E$ be a linear unbounded operator densely defined in E. A is called a positive operator in the Banach space if the operator $(\lambda I+A)$ has a bounded inverse in E and for any $\lambda \geq 0$, the following estimate holds:

$$\bigl\Vert (\lambda I+A)^{-1}\bigr\Vert _{E\rightarrow E}\leq \frac{M}{\lambda +1}. $$

Throughout the present paper, M denotes positive constants, which may differ in time and thus is not a subject of precision. However, we will use $M(\alpha ,\beta ,\ldots)$ to stress the fact that the constant depends only on $\alpha ,\beta ,\ldots$ .

For a positive operator A in the Banach space E, let us introduce the fractional spaces $E_{\alpha }=E_{\alpha }(E,A)$ ($0<\alpha <1$) consisting of those $v\in E$ for which the norm

$$\Vert v\Vert _{E_{\alpha }}=\sup_{\lambda >0}\lambda ^{\alpha }\bigl\Vert A(\lambda +A)^{-1}v\bigr\Vert _{E}+\Vert v\Vert _{E} $$

is finite.

Let us introduce the Banach space $C^{\beta } ( R^{1} )$, $\beta \in (0,\frac{1}{2})$, of all continuous 2π periodic functions $\varphi (x)$ defined on $R^{1}$ and satisfying a Hölder condition for which the following norm is finite:

$$\Vert \varphi \Vert _{C^{\beta } ( R^{1} ) }=\Vert \varphi \Vert _{\mathring{C} ( R^{1} ) }+\sup_{{ x,x+\tau \in [ 0,2\pi ], \tau \neq 0 }}\frac{|\varphi (x+\tau )-\varphi (x)|}{|\tau |^{\beta }}, $$

where $\mathring{C} ( R^{1} )$ is the Banach space of all continuous 2π periodic functions $\varphi (x)$ defined on $R^{1}$ with the norm

$$\Vert \varphi \Vert _{\mathring{C} ( R^{1} ) }=\max_{x\in [ 0,2\pi ] }\bigl|\varphi (x)\bigr|. $$

In [3], a new method of summations of Fourier series converging in

$$\mathring{C} \bigl( R^{1} \bigr) = \biggl\{ \varphi (x)\in C \bigl(R^{1}\bigr): \int_{0}^{2\pi }\varphi (x)\,dx=0 \biggr\} $$

is presented. It is based on the following result on the positivity of the differential operator $A^{x}$ defined by the formula

$$ A^{x}u=-u_{xx}(x)+\delta u(x),\quad \delta \geq 0, $$

(1)

with domain

$$D\bigl(A^{x}\bigr)= \biggl\{ u(x):u(x),u^{\prime }(x),u^{\prime \prime }(x) \in C\bigl(R^{1}\bigr), u(x)=u(x+2\pi ),x\in R^{1}, \int _{0}^{2\pi }u(x)\,dx=0 \biggr\} . $$

Theorem 1.1

[3]

The operator $( A^{x}+\lambda )$ has a bounded in $\mathring{C} ( R^{1} )$ inverse for $\delta =0$, $\lambda \geq 0$ and the following estimate holds:

$$ \bigl\Vert \bigl(A^{x}+\lambda \bigr)^{-1}\bigr\Vert _{\mathring{C} ( R^{1} ) \rightarrow \mathring{C} ( R^{1} ) }\leq \frac{1+16\pi ^{2}}{1+\lambda }. $$

(2)

It is easy to see that (2) is true for all $\delta >0$.

In the present study, the resolvent equation of the operator $A^{x}$

$$ A^{x}u+\lambda u=\varphi $$

(3)

or

$$ \left \{ \begin{array}{@{}l} -u^{\prime \prime }(x)+ ( \delta +\lambda ) u(x)=\varphi (x),\quad 0< x<2\pi ,\\ u(x)=u(x+2\pi ),\quad x\in R^{1}, \int_{0}^{2\pi }u(x)\,dx=0, \end{array} \right . $$

(4)

will be investigated. The Green function of $A^{x}$ is constructed. The estimates for the Green function are obtained. It is proved that for any $\alpha \in (0,\frac{1}{2})$, the norms in the spaces $E_{\alpha }=E_{\alpha }(\mathring{C} ( R^{1} ) ,A^{x})$ and $\mathring{C}^{2\alpha } ( R^{1} )$ are equivalent. Here, $\mathring{C}^{2\alpha } ( R^{1} )$ is the subspace of $C^{2\alpha } ( R^{1} )$ such that $\int_{0}^{2\pi }\varphi (x)\,dx=0$. The positivity of the operator $A^{x}$ in the Hölder spaces $\mathring{C}^{2\alpha } ( R^{1} )$, $\alpha \in (0,\frac{1}{2})$, is proved. The structure of fractional spaces generated by this operator is investigated. In the applications, theorems on well-posedness of local and nonlocal boundary value problems for elliptic equations in Hölder spaces are obtained.

2 The Green function of $A^{x}$

Lemma 2.1

Assume that $\varphi \in C ( R^{1} )$ and $\varphi (x)=\varphi (x+2\pi )$, $x\in R^{1}$, $\int_{0}^{2\pi }\varphi (x)\,dx=0$. For any $\lambda \geq 0$, problem (4) is uniquely solvable and the following formula holds:

$$ u(x)= \bigl( A^{x}+\lambda I \bigr) ^{-1}\varphi (x)=\int _{0}^{2\pi }G(x,s;\lambda )\varphi (s)\,ds, $$

(5)

where

$$\begin{aligned}& G(x,s;\lambda )=\frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \left \{ \begin{array}{@{}l@{\quad}l} e^{-\sqrt{\delta +\lambda }(x-s)}+e^{-\sqrt{\delta +\lambda }(2\pi -x+s)}, & 0\leq s\leq x,\\ e^{-\sqrt{\delta +\lambda }(s-x)}+e^{-\sqrt{\delta +\lambda }(2\pi +x-s)},& x\leq s\leq 2\pi, \end{array} \right . \\& G(2\pi +x,s;\lambda )=G(x,s;\lambda ),\quad x\in R^{1}. \end{aligned}$$

(6)

Proof

Let $x\in [ 0,2\pi ]$. From (3) there follows the problem

$$\begin{aligned}& -u^{\prime \prime }(x)+ ( \delta +\lambda ) u(x) =\varphi (x),\quad 0< x<2\pi , \\& u(0)=u(2\pi ),\quad x\in R^{1}, \int_{0}^{2\pi }u(x)\,dx=0. \end{aligned}$$

(7)

We will try to obtain a formula for (7). It is clearly

$$\begin{aligned} u(x)={}& \bigl( 1-e^{-\sqrt{\delta +\lambda }4\pi } \bigr) ^{-1} \biggl\{ \bigl( e^{-\sqrt{\delta +\lambda }x}-e^{-\sqrt{\delta +\lambda } ( 4\pi -x ) } \bigr) u(0) \\ &{} + \bigl( e^{-\sqrt{\delta +\lambda } ( 2\pi -x ) }-e^{-\sqrt{\delta +\lambda } ( 2\pi +x ) } \bigr) u(2\pi )- \bigl( e^{-\sqrt{\delta +\lambda } ( 2\pi -x ) }-e^{-\sqrt{\delta +\lambda } ( 2\pi +x ) } \bigr) \\ &{} \times ( 2\sqrt{\delta +\lambda } ) ^{-1}\int _{0}^{2\pi } \bigl( e^{-\sqrt{\delta +\lambda }(2\pi -s)}-e^{-\sqrt{\delta +\lambda }(2\pi +s)} \bigr) \varphi (s)\,ds \biggr\} \\ &{}+(2\sqrt{\delta +\lambda })^{-1}\int_{0}^{2\pi } \bigl( e^{-\sqrt{\delta +\lambda }\vert x-s\vert }-e^{-\sqrt{\delta +\lambda }(x+s)} \bigr) \varphi (s)\,ds \end{aligned}$$

(8)

is the solution of the following problem:

$$\left \{ \begin{array}{@{}l} -u^{\prime \prime }(x)+ ( \delta +\lambda ) u(x)=\varphi (x),\quad 0< x<2\pi , \\ u(0),u(2\pi) \mbox{ are given}. \end{array} \right . $$

Using (8) and the conditions $\int_{0}^{2\pi}u(x)\,dx=0$, $u(0)=u(2\pi )$, we get

$$\begin{aligned} \int_{0}^{2\pi }u(x)\,dx={}& \bigl( 1-e^{-\sqrt{\delta +\lambda }4\pi } \bigr) ^{-1}\int _{0}^{2\pi } \biggl\{ \bigl(e^{-\sqrt{\delta +\lambda }x}-e^{-\sqrt{\delta +\lambda }(4\pi -x)}\bigr)u(0) \\ &{}+\bigl(e^{-\sqrt{\delta +\lambda }(2\pi -x)}-e^{-\sqrt{\delta +\lambda }(2\pi +x)}\bigr)u(0)-\bigl(e^{-\sqrt{\delta +\lambda }(2\pi -x)}-e^{-\sqrt{\delta +\lambda }(2\pi +x)} \bigr)\\ &{} \times (2\sqrt{\delta +\lambda })^{-1}\int _{0}^{2\pi }\bigl(e^{-\sqrt{\delta +\lambda }(2\pi -s)}-e^{-\sqrt{\delta +\lambda }(2\pi +s)} \bigr)\varphi (s)\,ds \biggr\} \,dx\\ &{}+(2\sqrt{\delta +\lambda })^{-1}\int_{0}^{ 2\pi} \int_{0}^{2\pi } \bigl( e^{-\sqrt{\delta +\lambda }\vert x-s\vert }-e^{-\sqrt{\delta +\lambda }(x+s)} \bigr) \varphi (s)\,ds\,dx=0. \end{aligned}$$

Then

$$\begin{aligned} u(0)={}& \bigl( 1-e^{-\sqrt{\delta +\lambda }2\pi } \bigr) ^{-1} ( 2\sqrt{\delta +\lambda } ) ^{-1}\int _{0}^{2\pi } \bigl( e^{-\sqrt{\delta +\lambda }(2\pi -s)} + e^{-\sqrt{\delta +\lambda }s}-e^{-\sqrt{\delta +\lambda }2\pi }-1 \bigr) \varphi (s)\,ds. \end{aligned}$$

By using the assumption $\int_{0}^{2\pi } \varphi (x)\,dx=0$, we have

$$ u(0)= \bigl( 1-e^{-\sqrt{\delta +\lambda }2\pi } \bigr) ^{-1} ( 2\sqrt{\delta +\lambda } )^{-1} \int_{0}^{2\pi } \bigl( e^{-\sqrt{\delta +\lambda }(2\pi -s)}+e^{-\sqrt{\delta +\lambda }s} \bigr) \varphi (s)\,ds. $$

(9)

From (8) and (9) it follows that

$$u(x)=\int_{0}^{2\pi } \frac{1}{2\sqrt{\delta +\lambda }} \frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \bigl( e^{-\sqrt{\delta +\lambda } ( 2\pi -\vert x-s\vert ) }+e^{-\sqrt{\delta +\lambda }\vert x-s\vert } \bigr) \varphi (s)\,ds. $$

Therefore,

$$\begin{aligned} G(x,s;\lambda ) = \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \left \{ \begin{array}{@{}l@{\quad}l} e^{-\sqrt{\delta +\lambda }(x-s)}+e^{-\sqrt{\delta +\lambda }(2\pi -x+s)},& 0\leq s\leq x, \\ e^{-\sqrt{\delta +\lambda }(s-x)}+e^{-\sqrt{\delta +\lambda }(2\pi +x-s)},& x\leq s\leq 2\pi . \end{array} \right . \end{aligned}$$

Let $x\in R^{1}$. Then, from $u(x)=u(x+2\pi )$, $x\in R^{1}$ it follows that $G(2\pi +x,s;\lambda )=G(x,s;\lambda )$. Lemma 2.1 is proved. □

Note that the following pointwise estimates for $G(x,s ;\lambda )$ and its first order derivatives hold:

$$\begin{aligned}& \bigl\vert G(x,s ;\lambda )\bigr\vert \leq \frac{M(\delta)}{\sqrt{\delta +\lambda }} \left \{ \begin{array}{@{}l@{\quad}l} e^{-\sqrt{\delta +\lambda }(2\pi -x+s)}, &0\leq s\leq x-\pi ,\\ e^{-\sqrt{\delta +\lambda }(x-s)},& x-\pi \leq s\leq x, \\ e^{-\sqrt{\delta +\lambda }(s-x)},& x\leq s\leq \pi +x, \\ e^{-\sqrt{\delta +\lambda }(2\pi +x-s)}, & \pi +x\leq s\leq 2\pi , \end{array} \right . \end{aligned}$$

(10)

$$\begin{aligned}& \bigl\vert G_{x}(x,s;\lambda )\bigr\vert ,\bigl\vert G_{s}(x,s;\lambda )\bigr\vert \leq M(\delta) \left \{ \begin{array}{@{}l@{\quad}l} e^{-\sqrt{\delta +\lambda }(2\pi -x+s)}, &0< s<x-\pi , \\ e^{-\sqrt{\delta +\lambda }(x-s)},& x-\pi <s<x, \\ e^{-\sqrt{\delta +\lambda }(s-x)},& x<s<\pi +x, \\ e^{-\sqrt{\delta +\lambda }(2\pi +x-s)}, &\pi +x<s<2\pi ,\end{array} \right . \\& \bigl\vert G(2\pi +x,s;\lambda )\bigr\vert =\bigl\vert G(x,s;\lambda ) \bigr\vert ,\quad x\in R^{1}. \end{aligned}$$

(11)

Here, $M=\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}$.

3 The structure of fractional spaces generated by $A^{x}$ and positivity of $A^{x}$ in Hölder spaces

Clearly, the operator $A^{x}$ and its resolvent $( A^{x}+\lambda ) ^{-1}$commute. By the definition of the norm in the fractional space $E_{\alpha }=E_{\alpha }(\mathring{C} ( R^{1} ) ,A^{x})$, we get

$$\bigl\Vert \bigl( A^{x}+\lambda \bigr) ^{-1}\bigr\Vert _{E_{\alpha }\rightarrow E_{\alpha }}\leq \bigl\Vert \bigl(A^{x}+\lambda \bigr)^{-1}\bigr\Vert _{\mathring{C} ( R^{1} ) \rightarrow \mathring{C} ( R^{1} ) }. $$

Thus, from Theorem 1.1 it follows that $A^{x}$ is a positive operator in the fractional spaces $E_{\alpha }(\mathring{C} ( R^{1} ) ,A^{x})$. Moreover, we have the following result.

Theorem 3.1

For $\alpha \in (0,\frac{1}{2})$, the norms of the spaces $E_{\alpha }(\mathring{C} (R^{1} ),A^{x})$ and the Hölder space $\mathring{C}^{2\alpha } ( R^{1} )$ are equivalent.

Proof

For any $\lambda \geq 0$ we have the obvious equality

$$A^{x}\bigl(A^{x}+\lambda \bigr)^{-1}\varphi (x)= \varphi (x)-\lambda \bigl(A^{x}+\lambda \bigr)^{-1}\varphi (x). $$

By (5), we can write

$$\begin{aligned} A^{x}\bigl(A^{x}+\lambda \bigr)^{-1} \varphi (x)&=\varphi (x)-\lambda \int_{0}^{2\pi }G ( x,s ;\lambda ) \varphi (s)\,ds \\ &=\frac{\delta }{\delta +\lambda }\varphi (x)+\frac{\lambda }{\delta +\lambda }\varphi (x)-\lambda \int _{0}^{2\pi }G ( x,s ;\lambda ) \varphi (s)\,ds. \end{aligned}$$

(12)

By (5) and identity (12), we can write

$$ A^{x}\bigl(A^{x}+\lambda \bigr)^{-1}\varphi (x)= \frac{\delta }{\delta +\lambda }\varphi (x)+\lambda \int_{0}^{2\pi }G ( x,s;\lambda ) \bigl( \varphi (x)-\varphi (s) \bigr)\,ds. $$

(13)

Then

$$\begin{aligned} \lambda ^{\alpha }A^{x}\bigl(A^{x}+\lambda \bigr)^{-1}\varphi (x) =&\frac{\delta \lambda ^{\alpha }}{\delta +\lambda }\varphi (x)+\lambda ^{\alpha +1}\int_{0}^{2\pi }G ( x,s;\lambda ) \bigl( \varphi (x)-\varphi (s) \bigr)\,ds \\ =&P_{1}(x)+P_{2}(x), \end{aligned}$$

where

$$\begin{aligned}& P_{1}(x) =\frac{\delta \lambda ^{\alpha }}{\delta +\lambda }\varphi (x), \\& P_{2}(x) =\lambda ^{\alpha +1}\int_{0}^{2\pi }G ( x,s;\lambda ) \bigl( \varphi (x)-\varphi (s) \bigr)\,ds. \end{aligned}$$

Using the definition of the norm of space $C^{2\alpha } (R^{1} )$ and $\frac{\lambda ^{\alpha }\delta ^{1-\alpha }}{\delta +\lambda }\leq 1$, we can write

$$\bigl\vert P_{1}(x)\bigr\vert \leq \frac{\delta \lambda ^{\alpha }\delta ^{1-\alpha }}{\delta +\lambda }\bigl\vert \varphi (x)\bigr\vert \leq \delta ^{\alpha }\max_{x\in [ 0,2\pi ] } \bigl\vert \varphi (x)\bigr\vert \leq \delta ^{\alpha }\Vert \varphi \Vert _{C^{2\alpha } ( R^{1} ) } $$

for any $x\in [ 0,2\pi ]$. Then

$$\max_{x\in [ 0,2\pi ] }\bigl\vert P_{1}(x)\bigr\vert \leq \delta ^{\alpha }\Vert \varphi \Vert _{C^{2\alpha } (R^{1} ) } $$

or

$$ \Vert P_{1}\Vert _{C ( R^{1} ) }\leq \delta ^{\alpha } \Vert \varphi \Vert _{C^{2\alpha } (R^{1} ) }. $$

(14)

Then, using estimate (10), we get

$$\begin{aligned} \bigl\vert P_{2}(x)\bigr\vert \leq{}& \lambda ^{\alpha +1}\int _{0}^{2\pi }\bigl\vert G ( x,s;\lambda ) \bigr\vert \bigl\vert \varphi (x)-\varphi (s)\bigr\vert \,ds \\ \leq&{} \frac{M\lambda ^{\alpha +1}}{2\sqrt{\delta +\lambda }} \biggl( \int_{0}^{x-\pi }e^{-\sqrt{\delta +\lambda }(2\pi -x+s)} \bigl\vert \varphi (x)-\varphi (s)\bigr\vert \,ds \\ &{}+\int_{x-\pi }^{x}e^{-\sqrt{\delta +\lambda }(x-s)}\bigl\vert \varphi (x)-\varphi (s)\bigr\vert \,ds +\int_{x}^{x+\pi }e^{-\sqrt{\delta +\lambda }(s-x)}\bigl\vert \varphi (x)-\varphi (s)\bigr\vert \,ds \\ &{} +\int_{x+\pi }^{2\pi }e^{-\sqrt{\delta +\lambda }(2\pi +x-s)}\bigl\vert \varphi (x)-\varphi (s)\bigr\vert \,ds \biggr) \\ ={}&P_{21}(x)+P_{22}(x)+P_{23}(x)+P_{24}(x), \end{aligned}$$

where

$$\begin{aligned}& P_{21}(x)=\frac{M ( \delta ) \lambda ^{\alpha +1}}{2\sqrt{\delta +\lambda }}\int_{0}^{x-\pi }e^{-\sqrt{\delta +\lambda }(2\pi -x+s)} \bigl\vert \varphi (x)-\varphi (s)\bigr\vert \,ds, \\& P_{22}(x)=\frac{M ( \delta ) \lambda ^{\alpha +1}}{2\sqrt{\delta +\lambda }}\int_{x-\pi }^{x}e^{-\sqrt{\delta +\lambda }(x-s)} \bigl\vert \varphi (x)-\varphi (s)\bigr\vert \,ds, \\& P_{23}(x)=\frac{M ( \delta ) \lambda ^{\alpha +1}}{2\sqrt{\delta +\lambda }}\int_{x}^{x+\pi }e^{-\sqrt{\delta +\lambda }(s-x)} \bigl\vert \varphi (x)-\varphi (s)\bigr\vert \,ds, \\& P_{24}(x)=\frac{M ( \delta ) \lambda ^{\alpha +1}}{2\sqrt{\delta +\lambda }}\int_{x+\pi }^{2\pi }e^{-\sqrt{\delta +\lambda }(2\pi +x-s)} \bigl\vert \varphi (x)-\varphi (s)\bigr\vert \,ds. \end{aligned}$$

Here $M( \delta ) =\frac{1}{1-e^{-\sqrt{\delta }2\pi }}$.

Clearly, using the condition $\varphi (s)=\varphi (s+2\pi )$, $P_{21}(x)$ can be rewritten as

$$\begin{aligned} P_{21}(x)&=\frac{M ( \delta ) \lambda ^{\alpha +1}}{2\sqrt{\delta +\lambda }}\int_{0}^{x-\pi }e^{-\sqrt{\delta +\lambda }(2\pi -x+s)} \bigl\vert \varphi (x)-\varphi (2\pi +s)\bigr\vert \,ds \\ &\leq \Vert \varphi \Vert _{C^{2\alpha } (R^{1} ) } \frac{M ( \delta ) \lambda ^{\alpha +1}}{2 ( \delta +\lambda ) ^{\alpha +1}}\Gamma ( 2 \alpha +1 ) \end{aligned}$$

for any $x\in [ 0,2\pi ]$. Then

$$ \max_{x\in [ 0,2\pi ] }P_{21}(x)\leq \Vert \varphi \Vert _{C^{2\alpha } ( R^{1} ) }\frac{M ( \delta ) }{\alpha }\Gamma ( 2\alpha +1 ) . $$

(15)

Let us estimate $P_{22}(x)$.

$$\begin{aligned} P_{22}(x)&\leq \Vert \varphi \Vert _{C^{2\alpha } (R^{1} ) } \frac{M ( \delta ) \lambda ^{\alpha +1}}{2\sqrt{\delta +\lambda }}\int_{x-\pi }^{x}e^{-\sqrt{\delta +\lambda }(x-s)} ( x-s ) ^{2\alpha }\,ds. \\ &\leq \Vert \varphi \Vert _{C^{2\alpha } (R^{1} ) }\frac{M ( \delta ) \lambda ^{\alpha +1}}{2 ( \delta +\lambda ) ^{\alpha +1}}\Gamma ( 2\alpha +1 ) \end{aligned}$$

for any $x\in [ 0,2\pi ]$. Then

$$ \max_{x\in [ 0,2\pi ] }P_{22}(x)\leq \Vert \varphi \Vert _{C^{2\alpha } ( R^{1} ) }\frac{M ( \delta ) }{\alpha }\Gamma ( 2\alpha +1 ) . $$

(16)

Let us estimate $P_{23}(x)$.

$$\begin{aligned} P_{23}(x)&\leq \Vert \varphi \Vert _{C^{2\alpha } (R^{1} ) } \frac{M ( \delta ) \lambda ^{\alpha +1}}{2\sqrt{\delta +\lambda }}\int_{x}^{x+\pi }e^{-\sqrt{\delta +\lambda }(s-x)} ( s-x ) ^{2\alpha }\,ds \\ &\leq \Vert \varphi \Vert _{C^{2\alpha } (R^{1} ) } \frac{M ( \delta ) \lambda ^{\alpha +1}}{2 ( \delta +\lambda ) ^{\alpha +1}}\Gamma ( 2 \alpha +1 ) \end{aligned}$$

for any $x\in [ 0,2\pi ]$. Then

$$ \max_{x\in [ 0,2\pi ] }P_{23}(x)\leq \Vert \varphi \Vert _{C^{2\alpha } ( R^{1} ) }\frac{M ( \delta ) }{\alpha }\Gamma ( 2\alpha +1 ) . $$

(17)

Clearly, using the condition $\varphi (x)=\varphi (x+2\pi )$, $P_{24}(x)$ can be rewritten as

$$\begin{aligned} P_{24}(x)&\leq \Vert \varphi \Vert _{C^{2\alpha } ( R^{1} ) } \frac{M ( \delta ) \lambda ^{\alpha +1}}{2\sqrt{\delta +\lambda }}\int_{x+\pi }^{2\pi }e^{-\sqrt{\delta +\lambda }(2\pi +x-s)} ( 2 \pi +x-s ) ^{2\alpha }\,ds \\ &\leq \Vert \varphi \Vert _{C^{2\alpha } (R^{1} ) } \frac{M ( \delta ) \lambda ^{\alpha +1}}{2 ( \delta +\lambda ) ^{\alpha +1}}\Gamma ( 2 \alpha +1 ) \end{aligned}$$

for any $x\in [ 0,2\pi ]$. Then

$$ \max_{x\in [ 0,2\pi ] }P_{24}(x)\leq \Vert \varphi \Vert _{C^{2\alpha } ( R^{1} ) }\frac{M ( \delta ) }{\alpha }\Gamma ( 2\alpha +1 ) . $$

(18)

Combining estimates (15)-(18), we have

$$\begin{aligned} \max_{x\in [ 0,2\pi ] }\bigl\vert P_{2}(x)\bigr\vert &\leq \max_{x\in [ 0,2\pi ] }P_{21}(x)+\max _{x\in [ 0,2\pi ] }P_{22}(x) +\max_{x\in [ 0,2\pi ] }P_{23}(x)+\max_{x\in [ 0,2\pi ] }P_{24}(x) \\ &\leq \Vert \varphi \Vert _{C^{2\alpha } (R^{1} ) }\frac{M(\delta )}{\alpha }\Gamma ( 2 \alpha +1 ) . \end{aligned}$$

(19)

Using estimate (14) and (19), we get

$$\max_{x\in [ 0,2\pi ] }\bigl\vert \lambda ^{\alpha }A^{x} \bigl(\lambda +A^{x}\bigr)^{-1}\varphi (x)\bigr\vert \leq \frac{M(\delta )}{\alpha }\Vert \varphi \Vert _{C^{2\alpha } (R^{1} ) } + \frac{M(\delta )}{\alpha }\Gamma ( 2\alpha +1 ) \Vert \varphi \Vert _{C^{2\alpha } (R^{1} ) } $$

for any $\lambda \geq 0$. Thus,

$$ \Vert \varphi \Vert _{E_{\alpha } ( C (R^{1} ) ,A^{x} ) }\leq \frac{M(\delta )}{\alpha } \bigl[ 1+\Gamma ( 2\alpha +1 ) \bigr] \Vert \varphi \Vert _{C^{2\alpha } ( R^{1} ) }. $$

(20)

Now, let us prove the opposite inequality. For any positive operator $A^{x}$ in the Banach space, we can write

$$\varphi ( x )=\int_{0}^{\infty } A^{x} \bigl(\lambda +A^{x}\bigr)^{-2}\varphi ( x )\,d\lambda . $$

From this relation and (5), it follows that

$$\begin{aligned}[b] \varphi ( x ) &=\int_{0}^{\infty } \bigl(A^{x}+\lambda \bigr)^{-1}A^{x} \bigl(A^{x}+\lambda \bigr)^{-1}\varphi ( x )\,d\lambda \\ &=\int_{0}^{\infty }\int_{0}^{2\pi }G ( x,s;\lambda ) A^{x}\bigl(A^{x}+\lambda \bigr)^{-1} \varphi ( s ) \,ds\,d\lambda . \end{aligned} $$

Consequently,

$$\begin{aligned} \varphi ( x_{1} ) -\varphi ( x_{2} ) &=\int _{0}^{\infty }\int_{0}^{2\pi } \bigl( G ( x_{1},s;\lambda ) -G ( x_{2},s;\lambda ) \bigr) A^{x}\bigl(A^{x}+\lambda \bigr)^{-1}\varphi (s)\,ds\,d \lambda \\ &=\int_{0}^{\infty }\lambda ^{-\alpha }\int _{0}^{2\pi } \bigl( G ( x_{1},s;\lambda ) -G ( x_{2},s;\lambda ) \bigr) \lambda ^{\alpha }A^{x} \bigl(A^{x}+\lambda \bigr)^{-1}\varphi (s)\,ds\,d\lambda . \end{aligned}$$

Therefore,

$$\bigl\vert \varphi ( x_{1} ) -\varphi ( x_{2} ) \bigr\vert \leq \int_{0}^{\infty }\lambda ^{-\alpha } \int_{0}^{2\pi }\bigl\vert G ( x_{1},s; \lambda )-G ( x_{2},s;\lambda ) \bigr\vert \,ds\,d\lambda \Vert \varphi \Vert _{E_{\alpha }(C^{2\alpha } ( R^{1} ) ,A^{x})}. $$

Let

$$T=\vert x_{1}-x_{2}\vert ^{-2\alpha } \biggl( \int _{0}^{\infty }\lambda ^{-\alpha }\int _{0}^{2\pi }\bigl\vert G ( x_{1},s; \lambda ) -G ( x_{2},s;\lambda ) \bigr\vert \,ds\,d\lambda \biggr) . $$

Then, for any $x_{1},x_{2} \in [ 0,2\pi ]$ such that $x_{2}\geq x_{1}$, we have

$$\frac{\vert \varphi ( x_{1} ) -\varphi ( x_{2} ) \vert }{\vert x_{1}-x_{2}\vert ^{2\alpha }}\leq T\Vert \varphi \Vert _{E_{\alpha }(C^{2\alpha } (R^{1} ) ,A^{x})}. $$

Now, we will prove that

$$ T\leq \frac{M(\delta )}{2\alpha ( 1-2\alpha ) }. $$

(21)

Using (6), we get

$$\begin{aligned} \bigl\vert G ( x_{1},s;\lambda ) -G ( x_{2},s;\lambda ) \bigr\vert \leq{}& \biggl\vert \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\biggr\vert \\ &{}\times\! \left \{ \begin{array}{@{}l@{\quad}l} \vert e^{-\sqrt{\delta +\lambda }(2\pi -x_{1}+s)}-e^{-\sqrt{\delta +\lambda }(2\pi -x_{2}+s)}\vert ,& 0\leq s\leq x_{1}-\pi ,\\ \vert e^{-\sqrt{\delta +\lambda }(x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(2\pi -x_{2}+s)}\vert ,& x_{1}-\pi \leq s\leq x_{2}-\pi , \\ \vert e^{-\sqrt{\delta +\lambda }(x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(x_{2}-s)}\vert ,& x_{2}-\pi \leq s\leq x_{1}, \\ \vert e^{-\sqrt{\delta +\lambda }(s-x_{1})}-e^{-\sqrt{\delta +\lambda }(x_{2}-s)}\vert ,& x_{1}\leq s\leq \frac{x_{1}+x_{2}}{2}, \\ \vert e^{-\sqrt{\delta +\lambda }(s-x_{1})}-e^{-\sqrt{\delta +\lambda }(x_{2}-s)}\vert ,&\frac{x_{1}+x_{2}}{2}\leq s\leq x_{2}, \\ \vert e^{-\sqrt{\delta +\lambda }(s-x_{1})}-e^{-\sqrt{\delta +\lambda }(s-x_{2})}\vert ,& x_{2}\leq s\leq x_{1}+\pi , \\ \vert e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(s-x_{2})}\vert ,& x_{1}+\pi \leq s\leq x_{2}+\pi , \\ \vert e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(2\pi +x_{2}-s)}\vert ,& x_{2}+\pi \leq s\leq 2\pi .\end{array} \right . \end{aligned}$$

Then

$$\begin{aligned} T =&\vert x_{1}-x_{2}\vert ^{-2\alpha }\int _{0}^{\infty }\lambda ^{-\alpha } \\ &{}\times\biggl[ \int _{0}^{x_{1}-\pi }\frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \bigl\vert e^{-\sqrt{\delta +\lambda }(2\pi -x_{1}+s)}-e^{-\sqrt{\delta +\lambda }(2\pi -x_{2}+s)}\bigr\vert \,ds \\ &{}+\int_{x_{1}-\pi }^{x_{2}-\pi }\frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\bigl\vert e^{-\sqrt{\delta +\lambda }(x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(2\pi -x_{2}+s)}\bigr\vert \,ds \\ &{}+\int_{x_{2}-\pi }^{x_{1}}\frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\bigl\vert e^{-\sqrt{\delta +\lambda }(x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(x_{2}-s)}\bigr\vert \,ds \\ &{}+\int_{x_{1}}^{\frac{x_{1}+x_{2}}{2}}\frac{1}{2\sqrt{\delta +\lambda }} \frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\bigl\vert e^{-\sqrt{\delta +\lambda }(s-x_{1})}-e^{-\sqrt{\delta +\lambda }(x_{2}-s)}\bigr\vert \,ds \\ &{}+\int_{\frac{x_{1}+x_{2}}{2}}^{x_{2}}\frac{1}{2\sqrt{\delta +\lambda }} \frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\bigl\vert e^{-\sqrt{\delta +\lambda }(s-x_{1})}-e^{-\sqrt{\delta +\lambda }(x_{2}-s)}\bigr\vert \,ds \\ &{}+\int_{x_{2}}^{x_{1}+\pi }\frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\bigl\vert e^{-\sqrt{\delta +\lambda }(s-x_{1})}-e^{-\sqrt{\delta +\lambda }(s-x_{2})}\bigr\vert \,ds \\ &{}+\int_{x_{1}+\pi }^{x_{2}+\pi }\frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\bigl\vert e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(s-x_{2})}\bigr\vert \,ds \\ &{} +\int_{x_{2}+\pi }^{2\pi }\frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\bigl\vert e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(2\pi +x_{2}-s)}\bigr\vert \,ds \biggr]\,d\lambda \\ =&T_{1}+T_{2}+T_{3}+T_{4}+T_{5}+T_{6}+T_{7}+T_{8}. \end{aligned}$$

Let us estimate the expression

$$\begin{aligned} T_{1}={}&\vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\int_{0}^{x_{1}-\pi } \biggl( \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\\ &{}\times \bigl\vert e^{-\sqrt{\delta +\lambda }(2\pi -x_{1}+s)}-e^{-\sqrt{\delta +\lambda }(2\pi -x_{2}+s)}\bigr\vert \,ds \biggr)\,d\lambda . \end{aligned}$$

For $0\leq s\leq x_{1}-\pi$, using the estimates

$$\begin{aligned}& \bigl\vert e^{-\sqrt{\delta +\lambda }(2\pi -x_{1}+s)}-e^{-\sqrt{\delta +\lambda }(2\pi -x_{2}+s)}\bigr\vert \leq 2e^{-\sqrt{\delta +\lambda }(2\pi -x_{2}+s)}, \\& \bigl\vert e^{-\sqrt{\delta +\lambda }(2\pi -x_{1}+s)}-e^{-\sqrt{\delta +\lambda }(2\pi-x_{2}+s)}\bigr\vert \leq \sqrt{\delta + \lambda }2e^{-\sqrt{\delta +\lambda }(2\pi -x_{2}+s)} ( x_{2}-x_{1} ) , \end{aligned}$$

we can write

$$\begin{aligned} T_{1}\leq{}& \vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha } \biggl( \int_{0}^{x_{1}-\pi } \frac{1}{\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \\ &{} \times \bigl[ \sqrt{\delta +\lambda }e^{-\sqrt{\delta +\lambda }(2\pi -x_{2}+s)} ( x_{2}-x_{1} )\,ds \bigr] \biggr)\,d\lambda \\ \leq&{} M_{2} ( \delta ) \int_{0}^{\infty } \lambda ^{-\alpha }\frac{1}{2\sqrt{\delta +\lambda }}e^{-\sqrt{\delta +\lambda }(\pi -x_{2}+x_{1})}\,d\lambda \\ \leq&{} M_{2} ( \delta ) \int_{0}^{\infty } \lambda ^{-\alpha }\frac{1}{2\sqrt{\lambda }}e^{-\sqrt{\lambda }(\pi -x_{2}+x_{1})}\,d\lambda . \end{aligned}$$

Using the substitutions $\sqrt{\lambda }=\tau$ and $\pi -x_{2}+x_{1}=a$, we get

$$ T_{1}\leq M_{2} ( \delta ) \biggl( \int _{0}^{1}\tau ^{-2\alpha }\,d\tau +\int _{1}^{\infty }e^{-\tau a}\,d\tau \biggr) \leq M_{2} ( \delta ) \biggl( \frac{1}{1-2\alpha }+\frac{1}{a}e^{-a} \biggr) . $$

(22)

Using (22), we get

$$ T_{1}\leq \frac{M_{2}(\delta ,a)}{1-2\alpha }. $$

(23)

Now, let us estimate the expression

$$\begin{aligned} T_{2}={}&\vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\int_{x_{1}-\pi }^{x_{2}-\pi } \biggl( \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\\ &{}\times \bigl\vert e^{-\sqrt{\delta +\lambda }(x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(2\pi -x_{2}+s)}\bigr\vert \,ds \biggr)\,d\lambda . \end{aligned}$$

Using

$$\Delta _{1}=\int_{x_{1}-\pi }^{x_{2}-\pi }\bigl\vert e^{-\sqrt{\delta +\lambda }(x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(2\pi -x_{2}+s)}\bigr\vert \,ds $$

and the substitution

$$\frac{x_{1}-s-\pi }{x_{2}-x_{1}}=-y, $$

we have

$$\begin{aligned} \Delta _{1}&= ( x_{2}-x_{1} ) \int _{0}^{1} \bigl( e^{-\sqrt{\delta +\lambda }(\pi - ( x_{2}-x_{1} ) y)}-e^{-\sqrt{\delta +\lambda }(\pi - ( x_{2}-x_{1} ) (1-y))} \bigr)\,dy \\ &\leq ( x_{2}-x_{1} ) ^{2}\sqrt{\delta +\lambda }\int_{0}^{1}e^{-\sqrt{\delta +\lambda }(\pi - ( x_{2}-x_{1} ) )z}\,dz. \end{aligned}$$

Then

$$\begin{aligned} T_{2}\leq{}& \vert x_{1}-x_{2}\vert ^{-2\alpha } \biggl( \int_{0}^{\infty }\lambda ^{-\alpha }\frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \\ &{} \times ( x_{2}-x_{1} ) ^{2}\sqrt{ \delta +\lambda }\int_{0}^{1}e^{-\sqrt{\delta +\lambda }(\pi - ( x_{2}-x_{1} ) )z}\,dz \biggr)\,d\lambda \\ \leq{}& M_{3} ( \delta ) \int_{0}^{\infty } \lambda ^{-\alpha }\frac{1}{2\sqrt{\delta +\lambda }}e^{-\sqrt{\delta +\lambda }(\pi -x_{2}+x_{1})}\,d\lambda \\ \leq&{} M_{3} ( \delta ) \int_{0}^{\infty } \lambda ^{-\alpha }\frac{1}{2\sqrt{\lambda }}e^{-\sqrt{\lambda }(\pi -x_{2}+x_{1})}\,d\lambda . \end{aligned}$$

Using the substitutions $\sqrt{\lambda }=\tau$ and $\pi -x_{2}+x_{1}=a$, we get

$$\begin{aligned} T_{2}\leq M_{3} ( \delta ) \biggl( \int _{0}^{1}\tau ^{-2\alpha }e^{-\tau a}\,d\tau +\int_{1}^{\infty }\tau ^{-2\alpha }e^{-\tau a}\,d\tau \biggr) \leq M_{4} ( \delta ) \biggl( \frac{1}{{1-2\alpha }}+ \frac{1}{a}e^{-a} \biggr) . \end{aligned}$$

(24)

Using (24), we obtain

$$ T_{2}\leq \frac{M_{4}(\delta ,a)}{1-2\alpha }. $$

(25)

Now, let us estimate the expression

$$\begin{aligned}[b] T_{3}={}&\vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\int_{x_{2}-\pi }^{x_{1}} \biggl( \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\\ &{}\times \bigl\vert e^{-\sqrt{\delta +\lambda }(x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(x_{2}-s)}\bigr\vert \,ds \biggr)\,d\lambda . \end{aligned} $$

We have

$$\begin{aligned} T_{3}={}&\vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\frac{1}{2\sqrt{\delta +\lambda }} \\ &{}\times \bigl( 1-e^{-\sqrt{\delta +\lambda }(x_{2}-x_{1})} \bigr) \bigl( 1-e^{-\sqrt{\delta +\lambda }2\pi } \bigr) ^{-1}\int_{x_{2}-\pi }^{x_{1}}e^{-\sqrt{\delta +\lambda }(x_{1}-s)}\,ds\,d\lambda \\ \leq&{} M_{5} ( \delta ) \vert x_{1}-x_{2} \vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\frac{1}{2 ( \delta +\lambda ) } \bigl( 1-e^{-\sqrt{\delta +\lambda }(x_{2}-x_{1})} \bigr)\,d\lambda \\ \leq&{} M_{5} ( \delta ) \vert x_{1}-x_{2} \vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\frac{1}{2\lambda } \bigl( 1-e^{-\sqrt{\lambda }(x_{2}-x_{1})} \bigr)\,d\lambda . \end{aligned}$$

Using the substitution $\sqrt{\lambda } ( x_{2}-x_{1} ) =\tau$, we get

$$T_{3}\leq M_{5} ( \delta ) \int_{0}^{\infty } \frac{1-e^{-\tau }}{\tau ^{1+2\alpha }}\,d\tau =M_{5} ( \delta ) \int_{0}^{\infty } \frac{1}{\tau ^{1+2\alpha }}\int_{0}^{\tau }e^{-s}\,ds\,d\tau . $$

Let $0\leq s\leq \infty$ and $s\leq \tau \leq \infty$. Then

$$ T_{3}\leq \frac{M_{5} ( \delta ) }{2\alpha }\int_{0}^{\infty }e^{-s}s^{-2\alpha }\,ds= \frac{M_{5} ( \delta ) }{2\alpha }\Gamma (2\alpha ). $$

(26)

Using (26), we obtain

$$ T_{3}\leq \frac{M_{5}(\delta )}{2\alpha (1-2\alpha )}. $$

(27)

Let us estimate the expression

$$\begin{aligned} T_{4}={}&\vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\int_{x_{1}}^{\frac{x_{1}+x_{2}}{2}} \biggl( \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \\ &{}\times \bigl\vert e^{-\sqrt{\delta +\lambda }(s-x_{1})}-e^{-\sqrt{\delta +\lambda }(x_{2}-s)}\bigr\vert \,ds \biggr)\,d\lambda . \end{aligned}$$

We have

$$\begin{aligned} T_{4}&\leq \vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\frac{1}{2 ( \delta +\lambda ) } \bigl( 1-e^{-\sqrt{\delta +\lambda } ( \frac{x_{2}-x_{1}}{2} ) } \bigr)\,d\lambda \\ &\leq \vert x_{1}-x_{2}\vert ^{-2\alpha }\int _{0}^{\infty }\lambda ^{-\alpha }\frac{1}{2\lambda } \bigl( 1-e^{-\sqrt{\lambda } ( \frac{x_{2}-x_{1}}{2} ) } \bigr)\,d\lambda . \end{aligned}$$

Using the substitution $\sqrt{\lambda } ( \frac{x_{2}-x_{1}}{2} ) =\tau$, we get

$$T_{4}\leq \frac{M_{6} ( \delta ) }{4^{\alpha }}\int_{0}^{\infty } \frac{1-e^{-\tau }}{\tau ^{1+2\alpha }}\,d\tau =\frac{M_{6} ( \delta ) }{4^{\alpha }}\int_{0}^{\infty } \frac{1}{\tau ^{1+2\alpha }} \biggl( \int_{0}^{\tau }e^{-s}\,ds \biggr)\,d\tau . $$

Let $0\leq s\leq \infty$ and $s\leq \tau \leq \infty$. Then

$$ T_{4}\leq \frac{M_{6} ( \delta ) }{4^{\alpha }2\alpha }\int_{0}^{\infty }e^{-s}s^{-2\alpha }\,ds= \frac{M_{6} ( \delta ) }{4^{\alpha }\alpha }\Gamma (2\alpha )\leq \frac{M_{6} ( \delta ) }{4^{\alpha }2\alpha (1-2\alpha )}. $$

(28)

Using (28), we have

$$ T_{4}\leq \frac{M_{6}(\delta )}{2\alpha (1-2\alpha )}. $$

(29)

Let us estimate the expression

$$\begin{aligned} T_{5}={}&\vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\int_{\frac{x_{1}+x_{2}}{2}}^{x_{2}} \biggl( \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \\ &{}\times \bigl\vert e^{-\sqrt{\delta +\lambda }(s-x_{1})}-e^{-\sqrt{\delta +\lambda }(x_{2}-s)}\bigr\vert \,ds \biggr)\,d\lambda . \end{aligned}$$

We have

$$\begin{aligned} T_{5}&\leq \vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\frac{1}{2 ( \delta +\lambda ) } \bigl( 1-e^{-\sqrt{\delta +\lambda } ( \frac{x_{2}-x_{1}}{2} ) } \bigr)\,d\lambda \\ &\leq \vert x_{1}-x_{2}\vert ^{-2\alpha }\int _{0}^{\infty }\lambda ^{-\alpha }\frac{1}{2\lambda } \bigl( 1-e^{-\sqrt{\lambda } ( \frac{x_{2}-x_{1}}{2} ) } \bigr)\,d\lambda . \end{aligned}$$

Using the substitution $\sqrt{\lambda } ( \frac{x_{2}-x_{1}}{2} ) =\tau$, we get

$$T_{5}\leq \frac{M_{7} ( \delta ) }{4^{\alpha }}\int_{0}^{\infty } \frac{1-e^{-\tau }}{\tau ^{1+2\alpha }}\,d\tau =\frac{M_{7} ( \delta ) }{4^{\alpha }}\int_{0}^{\infty } \frac{1}{\tau ^{1+2\alpha }} \biggl( \int_{0}^{\tau }e^{-s}\,ds \biggr)\,d\tau . $$

Let $0\leq s\leq \infty$ and $s\leq \tau \leq \infty$. Then

$$ T_{5}\leq \frac{M_{7} ( \delta ) }{4^{\alpha }2\alpha }\int_{0}^{\infty }e^{-s}s^{-2\alpha }\,ds= \frac{M_{7} ( \delta ) }{4^{\alpha }\alpha }\Gamma (2\alpha )\leq \frac{M_{7} ( \delta ) }{4^{\alpha }2\alpha (1-2\alpha )}. $$

(30)

From (30) it follows that

$$ T_{5}\leq \frac{M_{7}(\delta )}{2\alpha (1-2\alpha )}. $$

(31)

Let us estimate the expression

$$\begin{aligned} T_{6}={}&\vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\int_{x_{2}}^{x_{1}+\pi } \biggl( \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \\ &{} \times \bigl\vert e^{-\sqrt{\delta +\lambda }(s-x_{1})}-e^{-\sqrt{\delta +\lambda }(s-x_{2})}\bigr\vert \,ds \biggr)\,d\lambda . \end{aligned}$$

We have

$$\begin{aligned} T_{6}={}&\vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\int_{x_{2}}^{x_{1}+\pi } \biggl( \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\\ &{}\times e^{-\sqrt{\delta +\lambda }(s-x_{2})} \bigl( 1-e^{-\sqrt{\delta +\lambda }(x_{2}-x_{1})} \bigr)\,ds \biggr)\,d\lambda \\ \leq{}& \vert x_{1}-x_{2}\vert ^{-2\alpha }\int _{0}^{\infty }\lambda ^{-\alpha }\frac{1}{2 ( \delta +\lambda ) } \bigl( 1-e^{-\sqrt{\delta +\lambda }(x_{2}-x_{1})} \bigr)\,d\lambda \\ \leq&{} \vert x_{1}-x_{2}\vert ^{-2\alpha }\int _{0}^{\infty }\lambda ^{-\alpha }\frac{1}{2\lambda } \bigl( 1-e^{-\sqrt{\lambda }(x_{2}-x_{1})} \bigr)\,d\lambda . \end{aligned}$$

Using the substitution $\sqrt{\lambda } ( x_{2}-x_{1} ) =\tau$, we get

$$T_{6}\leq M_{8} ( \delta ) \int_{0}^{\infty } \frac{1-e^{-\tau }}{\tau ^{1+2\alpha }}\,d\tau =M_{8} ( \delta ) \int_{0}^{\infty } \frac{1}{\tau ^{1+2\alpha }}\int_{0}^{\tau }e^{-s}\,ds\,d \tau . $$

Let $0\leq s\leq \infty$ and $s\leq \tau \leq \infty$. Then

$$ T_{6}\leq \frac{M_{9} ( \delta ) }{2\alpha }\int_{0}^{\infty }e^{-s}s^{-2\alpha }\,ds \leq \frac{M_{9} ( \delta ) }{2\alpha }\Gamma (2\alpha ). $$

(32)

Using (32), we have

$$ T_{6}\leq \frac{M_{9}(\delta )}{2\alpha (1-2\alpha )}. $$

(33)

Let us estimate the expression

$$\begin{aligned} T_{7}={}&\vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\int_{x_{1}+\pi }^{x_{2}+\pi } \biggl( \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \\ &{} \times \bigl\vert e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(s-x_{2})}\bigr\vert \,ds \biggr)\,d\lambda . \end{aligned}$$

Using

$$\Delta _{2}=\int_{x_{1}+\pi }^{x_{2}+\pi }\bigl\vert e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(s-x_{2})}\bigr\vert \,ds $$

and the substitution

$$\frac{\pi +x_{2}-s}{x_{2}-x_{1}}=y, $$

we obtain

$$\begin{aligned} \Delta _{2}&=\int_{1}^{0}- ( x_{2}-x_{1} ) \bigl( e^{-\sqrt{\delta +\lambda }(\pi - ( x_{2}-x_{1} ) (1-y))}-e^{-\sqrt{\delta +\lambda }(\pi - ( x_{2}-x_{1} ) y)} \bigr)\,dy \\ &\leq ( x_{2}-x_{1} ) ^{2}\sqrt{\delta +\lambda }\int_{0}^{1}\bigl\vert -e^{-\sqrt{\delta +\lambda }(\pi - ( x_{2}-x_{1} ) )z}\bigr\vert \,dz. \end{aligned}$$

Let us estimate the expression

$$\begin{aligned} T_{7}\leq{}& \vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty } \biggl( \lambda ^{-\alpha }\frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \\ &{}\times ( x_{2}-x_{1} ) ^{2}\sqrt{ \delta +\lambda }\int_{0}^{1}-e^{-\sqrt{\delta +\lambda }(\pi - ( x_{2}-x_{1} ) )z}\,dz \biggr)\,d\lambda \\ \leq&{} M_{10} ( \delta ) \int_{0}^{\infty } \lambda ^{-\alpha }\frac{1}{2\sqrt{\delta +\lambda }} \bigl( e^{-\sqrt{\delta +\lambda }(\pi -x_{2}+x_{1})} \bigr)\,d\lambda \\ \leq{}& M_{10} ( \delta ) \int_{0}^{\infty } \lambda ^{-\alpha }\frac{1}{2\sqrt{\lambda }} \bigl( e^{-\sqrt{\lambda }(\pi -x_{2}+x_{1})} \bigr)\,d\lambda . \end{aligned}$$

Using the substitutions $\sqrt{\lambda }=\tau$ and $(\pi -x_{2}+x_{1})=a$, we get

$$ T_{7}\leq 2M_{10} ( \delta ) \biggl( \int _{0}^{1}\tau ^{-2\alpha }\,d\tau +\int _{1}^{\infty }e^{-\tau a}\,d\tau \biggr) =M_{11} ( \delta ) \biggl( \frac{1}{{1-2\alpha }}+\frac{1}{a}e^{-a} \biggr) . $$

(34)

Using (34), we get

$$ T_{7}\leq \frac{M_{11}(\delta ,a)}{2\alpha (1-2\alpha )}. $$

(35)

Let us estimate the expression

$$\begin{aligned} T_{8}={}&\vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\int_{x_{2}+\pi }^{2\pi } \biggl( \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }}\\ &{} \times \bigl\vert e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(2\pi +x_{2}-s)}\bigr\vert \,ds \biggr)\,d\lambda . \end{aligned}$$

For $x_{2}+\pi \leq s\leq 2\pi$, using the estimates

$$\begin{aligned}& \bigl\vert e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(2\pi +x_{2}-s)}\bigr\vert \leq 2e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)}, \\& \bigl\vert e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)}-e^{-\sqrt{\delta +\lambda }(2\pi +x_{2}-s)}\bigr\vert \leq \sqrt{\delta + \lambda }2e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)} ( x_{2}-x_{1} ) , \end{aligned}$$

we can write

$$\begin{aligned} T_{8}\leq{}& \vert x_{1}-x_{2}\vert ^{-2\alpha }\int_{0}^{\infty }\lambda ^{-\alpha }\int_{x_{2}+\pi }^{2\pi } \biggl( \frac{1}{2\sqrt{\delta +\lambda }}\frac{1}{1-e^{-\sqrt{\delta +\lambda }2\pi }} \\ &{}\times \sqrt{\delta +\lambda }2e^{-\sqrt{\delta +\lambda }(2\pi +x_{1}-s)} ( x_{2}-x_{1} )\,ds \biggr)\,d\lambda \\ \leq&{} M_{12} ( \delta ) \int_{0}^{\infty } \lambda ^{-\alpha }\frac{1}{\sqrt{\delta +\lambda }}e^{-\sqrt{\delta +\lambda }x_{1}}\,d\lambda \\ \leq{}& 2M_{12} ( \delta ) \int_{0}^{\infty } \lambda ^{-\alpha }\frac{1}{2\sqrt{\lambda }}e^{-\sqrt{\lambda }x_{1}}\,d\lambda . \end{aligned}$$

Using the substitutions $\sqrt{\lambda }=\tau$ and $x_{1}=a$, we get

$$ T_{8}\leq M_{13} ( \delta ) \biggl( \int _{0}^{1}\tau ^{-2\alpha }\,d\tau +\int _{1}^{\infty }e^{-\tau a}\,d\tau \biggr) \leq M_{13} ( \delta ) \biggl( \frac{1}{1-2\alpha }+\frac{1}{a}e^{-a} \biggr) . $$

(36)

Equation (36) yields

$$ T_{8}\leq \frac{M_{13}(\delta ,a)}{2\alpha ( 1-2\alpha ) }. $$

(37)

Applying the triangle inequality and estimates (21), (23), (25), (27), (29), (31), (33), (35), and (37), we get

$$T\leq \frac{M(\delta ,a)}{2\alpha (1-2\alpha )}. $$

So, (21) is proved. Thus, for any $x_{1},x_{2}\in [ 0,2\pi ]$ we have

$$\vert x_{1}-x_{2}\vert ^{-2\alpha }\bigl\vert \varphi ( x_{1} ) -\varphi ( x_{2} ) \bigr\vert \leq \frac{M(\delta ,a)}{2\alpha (1-2\alpha )}\Vert \varphi \Vert _{E_{\alpha }(C ( R^{1} ) ,A^{x})}. $$

This means that the following inequality holds:

$$ \Vert \varphi \Vert _{\mathring{C}^{2\alpha } (R^{1} ) }\leq \frac{M(\delta ,a)}{2\alpha (1-2\alpha )}\Vert \varphi \Vert _{E_{\alpha }(C ( R^{1} ) ,A^{x}).} $$

(38)

Estimates (20) and (38) finish the proof of Theorem 3.1. □

Since $A^{x}$ is a positive operator in the fractional spaces $E_{\alpha }(\mathring{C} ( R^{1} ) ,A^{x})$, from the result of Theorem 3.1 it follows also it is positive operator in the Hölder space $\mathring{C}^{2\alpha } ( R^{1} )$. Namely, we have the following.

Theorem 3.2

The operator $(A^{x}+\lambda )$ has an inverse bounded in $\mathring{C}^{2\alpha } ( R^{1} )$ for any $\lambda \geq 0$ and the following estimate holds:

$$\bigl\Vert \bigl(A^{x}+\lambda \bigr)^{-1}\bigr\Vert _{\mathring{C}^{2\alpha } ( R^{1} ) \rightarrow \mathring{C}^{2\alpha } (R^{1} ) }\leq \frac{M(\delta ,a)}{2\alpha (1-2\alpha )}\frac{M_{14}}{\delta +\lambda }. $$

4 Applications

First, we consider the boundary value problem

$$ \left \{ \begin{array}{@{}l} -\frac{\partial ^{2}u(t,x)}{\partial t^{2}}-\frac{\partial ^{2}u(t,x)}{\partial x^{2}}+\delta u(t,x)=f(t,x),\quad 0< t<T,x\in R^{1}, \\ u(0,x)=\varphi (x),\qquad u(T,x)=\psi (x),\quad x\in R^{1}, \\ u(t,x)=u(t,x+2\pi ),\quad\int_{0}^{2\pi }u(t,s)\,ds=0, 0\leq t\leq T,x\in R^{1}.\end{array} \right . $$

(39)

Here, $\varphi (x)$, $\psi (x)$, and $f(t,x)$ are sufficiently smooth 2π-periodic functions in x and they satisfy any compatibility conditions which guarantee problem (39) has a smooth solution $u(t,x)$.

Theorem 4.1

Let $0<2\alpha <1$. Then, for the solution of the boundary value problem (39), we have the following coercive stability inequality:

$$\begin{aligned} &\Vert u_{tt}\Vert _{C ( [ 0,T ] ,\mathring{C}^{2\alpha } ( R^{1} ) ) }+\Vert u\Vert _{C ( [ 0,T ] ,\mathring{C}^{2\alpha +2} (R^{1} ) ) } \\ &\quad\leq M(\alpha ) \bigl[ \Vert \varphi \Vert _{\mathring{C}^{2\alpha +2} ( R^{1} ) }+\Vert \psi \Vert _{\mathring{C}^{2\alpha +2} ( R^{1} ) }+\Vert f\Vert _{C ( [ 0,T ] ,\mathring{C}^{2\alpha } (R^{1} ) ) } \bigr] . \end{aligned}$$

The proof of Theorem 4.1 is based on Theorem 3.1 on the structure of the fractional spaces $E_{\alpha }=E_{\alpha }(C ( R^{1} ) ,A^{x})$, Theorem 1.1 on the positivity of the operator $A^{x}$, on the following theorems on coercive stability of boundary value for the abstract elliptic equation and on the structure of the fractional space $E_{\alpha }^{\prime }=E_{\alpha }(E,A^{1/2})$ which is the Banach space consisting of those $v\in E$ for which the norm

$$\Vert v\Vert _{E_{\alpha }^{\prime }}=\sup_{\lambda >0}\lambda ^{\alpha }\bigl\Vert A^{1/2} \bigl( \lambda +A^{1/2} \bigr) ^{-1}v\bigr\Vert _{E_{\alpha }}+\Vert v\Vert _{E} $$

is finite.

Theorem 4.2

[5]

The spaces $E_{\alpha}(E,A)$ and $E_{2\alpha}^{\prime}(E,A^{1/2})$ coincide for any $0<\alpha<\frac{1}{2}$, and their norms are equivalent.

Theorem 4.3

[8]

Let A be positive operator in a Banach space E and $f\in C([0,T],E_{\alpha }^{\prime })$, $0<\alpha <1$. Then, for the solution of the nonlocal boundary value problem

$$ \left \{ \begin{array}{@{}l} -u^{\prime \prime }+Au(t)=f(t),\quad 0< t<T, \\ u(0)=\varphi,\qquad u(T)=\psi, \end{array} \right . $$

(40)

in a Banach space E with positive operator A, we have the coercive inequality

$$\begin{aligned} &\bigl\Vert u^{\prime \prime }\bigr\Vert _{C([0,T ],E_{\alpha }^{\prime })}+\Vert Au\Vert _{C([0,T ],E_{\alpha }^{\prime })} \\ &\quad\leq M \biggl[ \Vert A\varphi \Vert _{E_{\alpha }^{\prime }}+\Vert A\psi \Vert _{E_{\alpha }^{\prime }}+\frac{M}{\alpha (1-\alpha )}\Vert f\Vert _{C([0,T ],E_{\alpha }^{\prime })} \biggr] . \end{aligned}$$

Second, we consider the nonlocal boundary value problem for the elliptic equation

$$ \left \{ \begin{array}{@{}l} -\frac{\partial ^{2}u(t,x)}{\partial t^{2}}-\frac{\partial ^{2}u(t,x)}{\partial x^{2}}+\delta u(t,x)=f(t,x),\quad 0< t<T,x\in R^{1}, \\ u(0,x)=u(T,x),\qquad u_{t}(0,x)=u_{t}(T,x),\quad x\in R^{1}, \\ u(t,x)=u(t,x+2\pi ),\quad \int_{0}^{2\pi }u(t,s)\,ds=0, 0\leq t\leq T,x\in R^{1}.\end{array} \right . $$

(41)

Here, $f(t,x)$ is a sufficiently smooth 2π-periodic function in x and it satisfies any compatibility conditions which guarantee problem (41) has a smooth solution $u(t,x)$.

Theorem 4.4

Let $0<2\alpha <1$. Then, for the solution of boundary value problem (39), we have the following coercive stability inequality:

$$\Vert u_{tt}\Vert _{C ( [ 0,T ] ,\mathring{C}^{2\alpha } ( R^{1} ) ) }+\Vert u\Vert _{C ( [ 0,T ] ,\mathring{C}^{2\alpha +2} (R^{1} ) ) }\leq M(\alpha )\Vert f\Vert _{C ( [ 0,T ] ,\mathring{C}^{2\alpha } ( R^{1} ) ) }. $$

The proof of Theorem 4.4 is based on Theorem 3.1 on the structure of the fractional spaces $E_{\alpha }=E_{\alpha }(\mathring{C} ( R^{1} ) ,A^{x})$, Theorem 1.1 on the positivity of the operator $A^{x}$, Theorem 3.2 on the structure of the fractional space $E_{\alpha }^{\prime }=E_{\alpha }(E,A^{1/2})$ and on the following theorem on the coercive stability of the nonlocal boundary value for the abstract elliptic equation.

Theorem 4.5

[25]

Let A be a positive operator in a Banach space E and $f\in C([0,T],E_{\alpha }^{\prime })$, $0<\alpha <1$. Then, for the solution of the nonlocal boundary value problem

$$ \left \{ \begin{array}{@{}l} -u^{\prime \prime }+Au(t)=f(t),\quad 0< t<T, \\ u(0)=u(T),\qquad u^{\prime }(0)=u^{\prime }(T),\end{array} \right . $$

(42)

in a Banach space E with positive operator A the coercive inequality

$$\bigl\Vert u^{\prime \prime }\bigr\Vert _{C ( [ 0,T ] ,E_{\alpha }^{\prime } ) }+\Vert Au\Vert _{C ( [ 0,T ] ,E_{\alpha }^{\prime } ) }\leq \frac{M}{\alpha (1-\alpha )}\Vert f\Vert _{C ( [ 0,T ] ,E_{\alpha }^{\prime } ) } $$

holds.