Proof. Let $u$ is a $2\pi$-periodic solution of (2.4). Multiplying by $u$ and integrating the equation (2.4) we find

(2.6)\begin{equation} \int\limits_{0}^{2\pi} (-\triangle)^{s}u u - \lambda \int\limits_{0}^{2\pi} f(C + u)u' u={-} \lambda \int\limits_{0}^{2\pi} eu. \end{equation}

Using (2.1), together with the $2\pi$-periodicity of $u$ we find

\[ [u]_{H^{s}(0,2\pi)}^{2}\leq [u]_{X}^{2} = \int_{0}^{2\pi} u\mathcal{L} u =\int_{0}^{2\pi} (-\triangle)^{s}u u \]

Let $H\in C^{2}(\mathbb {R})$ such that

\[ H'(x)= \int_{0}^{x} f(t)\,\textrm{d}t. \]

Then

\begin{align*} \int_{0}^{2\pi} f(C + u(x))u'(t) u(t)\,\textrm{d}x & = \int_{0}^{2\pi} \dfrac{\textrm{d}}{\textrm{d}t}(H' (C + u(t)))u(t)\,\textrm{d}t\\ & = \int_{0}^{2\pi} \frac{\textrm{d}}{\textrm{d}t}(H'(C + u(t))u(t))\,\textrm{d}t\\ & \quad - \int_{0}^{2\pi} H'(C + u(t))u'(t)\,\textrm{d}t\\ & ={-}\int_{0}^{2\pi} \frac{\textrm{d}}{\textrm{d}t} H(C + u(t))\,\textrm{d}t=0. \end{align*}

Hence substituting in (2.6) and using Cauchy inequality we have

(2.7)\begin{align} [u]_{H^{s}(0,2\pi)}^{2} & \leq \lambda\int_{0}^{2\pi} |e(t)u(t)|\,\textrm{d}t \nonumber\\ & \leq \lambda \|e\|_{L^{2}}\|u\|_{L^{2}}. \end{align}

On the other hand, we claim that we have the following Poincaré-type inequality. There exist $K>0$ such that

(2.8)\begin{equation} \int_{0}^{2\pi}|u(t)|^{2}\,\textrm{d}t \leq K[u]_{H^{s}(0,2\pi)}^{2}. \end{equation}

Indeed, as $\bar {u}=0$ and using Cauchy inequality we get

\begin{align*} 2\pi |u(t)| & = \left| \int_{0}^{2\pi} (u(t) -u(y))\,\textrm{d}y\right| \\ & \leq \int_{0}^{2\pi} |u(t) -u(y)|\,\textrm{d}y\\ & \leq \left(\int_{0}^{2\pi} |u(t) -u(y)|^{2}\,\textrm{d}y\right)^{\dfrac{1}{2}}\left(\int_{0}^{2\pi}\,\textrm{d}y\right)^{\dfrac{1}{2}} \end{align*}

\begin{align*} |u(t)|^{2} & \leq \dfrac{1}{2\pi}\int_{0}^{2\pi} |u(t) -u(y)|^{2}\,\textrm{d}y \\ & = \dfrac{1}{2\pi}\int_{0}^{2\pi} \dfrac{|u(t) -u(y)|^{2}}{|t - y|^{1 + 2s}}|t - y|^{1 + 2s}\,\textrm{d}y, \end{align*}

as $|t - y|< 2\pi$ gives

\begin{align*} \int_{0}^{2\pi}|u(t)|^{2}\,\textrm{d}t & \leq \dfrac{(2\pi)^{1 + 2s}}{2\pi}\int_{0}^{2\pi}\int_{0}^{2\pi}\dfrac{|u(t) -u(y)|^{2}}{|t - y|^{1 + 2s}}\,\textrm{d}y\,\textrm{d}t\\ & \leq (2\pi)^{2s}[u]_{H^{s}(0,2\pi)}^{2}, \end{align*}

where $K= (2\pi )^{2s}$. Hence (2.7) and (2.8) give

\[ \|u\|_{H^{s}(0,2\pi)} \leq \lambda K\|e\|_{L^{2}([0,2\pi])} \leq K\|e\|_{L^{2}([0,2\pi])}, \]

and as $H^{s}(0,2\pi )\hookrightarrow C^{\alpha }[0,2\pi ]$ for $s>\frac {1}{2}$ we get

(2.9)\begin{equation} \|u\|_{C^{\alpha}[0,2\pi]} \leq K\|e\|_{L^{2}([0,2\pi])}. \end{equation}

Proof. Indeed, as $u$ is a periodic function, we can write its Fourier series:

\[ u(x)= a_{0} + \sum_{n=1}^{+ \infty} a_{n}\cos(nx) + b_{n}\sin(nx). \]

Then,

(2.10)\begin{align} & (-\triangle)^{s}u(x)\nonumber\\ & \quad= C_{1,s} \left(\sum_{n=1}^{+ \infty} a_{n}\int_{\mathbb{R}} \dfrac{\cos(nx) - \cos(ny)}{|x - y|^{1+ 2s}}\,\textrm{d}y + b_{n}\int_{\mathbb{R}} \dfrac{\sin(nx) - \sin(ny)}{|x - y|^{1+ 2s}}\,\textrm{d}y \right)\nonumber\\ & \quad= C_{1,s} \left(\sum_{n=1}^{+ \infty} a_{n}\int_{\mathbb{R}} \dfrac{\cos(nx) - \cos(nx - nz)}{|z|^{1+ 2s}}\,\textrm{d}z + b_{n}\int_{\mathbb{R}} \dfrac{\sin(nx) - \sin(nx -nz)}{|z|^{1+ 2s}}\,\textrm{d}z\right), \end{align}

so

\begin{align*} \int_{\mathbb{R}} \dfrac{\cos(nx) - \cos(nx - nz)}{|z|^{1+ 2s}}\,\textrm{d}z & = \int_{\mathbb{R}} \dfrac{\cos(nx) - \cos(nx)\cos(nz) - \sin(nx)\sin(nz)}{|z|^{1+ 2s}}\,\textrm{d}z \\ & = \int_{\mathbb{R}} \dfrac{1 - \cos(nz)}{|z|^{1+ 2s}}\,\textrm{d}z\cos(nx) - \int_{\mathbb{R}}\dfrac{\sin(nz)}{|z|^{1 + 2s}}\,\textrm{d}z \sin(nx), \end{align*}

where the second integral of the above is zero because we are integrating an odd function in a symmetric (with respect to zero) domain. Similarly to the previous computation, we have

\begin{align*} \int_{\mathbb{R}} \dfrac{\sin(nx) - \sin(nx - nz)}{|z|^{1+ 2s}}\,\textrm{d}z & = \int_{\mathbb{R}} \dfrac{\sin(nx) - \sin(nx)\cos(nz) + \cos(nx)\sin(nz)}{|z|^{1+ 2s}}\,\textrm{d}z \\ & = \int_{\mathbb{R}} \dfrac{1 - \cos(nz)}{|z|^{1+ 2s}}\,\textrm{d}z\sin(nx) - \int_{\mathbb{R}} \dfrac{\sin(nz)}{|z|^{1 + 2s}}\,\textrm{d}z \cos(nx). \end{align*}

Hence substituting this in (2.10) gives

\begin{align*} (-\triangle)^{s}u(x) & = C_{1,s} \left(\sum_{n=1}^{+ \infty} a_{n}\int_{\mathbb{R}} \dfrac{1 - \cos(nz)}{|z|^{1+ 2s}}\,\textrm{d}z\cos(nx) + b_{n}\int_{\mathbb{R}} \dfrac{1 - \cos(nz)}{|z|^{1+ 2s}}\,\textrm{d}z\sin(nx)\right) \\ & = \left(\int_{\mathbb{R}} \dfrac{1 - \cos(\xi_{1})}{|\xi|^{n + 2s}}\,\textrm{d}\xi\right)^{{-}1} \left(\sum_{n=1}^{+ \infty} a_{n}\int_{\mathbb{R}} \dfrac{1 - \cos(nz)}{|z|^{1+ 2s}}\,\textrm{d}z\cos(nx)\right.\\ & \quad\left.+ b_{n}\int_{\mathbb{R}} \dfrac{1 - \cos(nz)}{|z|^{1+ 2s}}\,\textrm{d}z\sin(nx)\right)\\ & = \sum_{n=1}^{+ \infty} a_{n}n^{2s}\cos(nx) + b_{n}n^{2s}\sin(nx). \end{align*}

Hence,

\[ \int_{0}^{2\pi} (-\triangle)^{s}u(x)\,\textrm{d}x = \sum_{n=1}^{+ \infty} a_{n}n^{2s} \int_{0}^{2\pi} \cos(nx)\,\textrm{d}x + b_{n}n^{2s} \int_{0}^{2\pi} \sin(nx)\,\textrm{d}x= 0. \]

Proof. Let $u\in C^{1,\alpha }_{2\pi }(\mathbb {R})$, we define for each $x\in \mathbb {R}$

\[ H(u(x),u'(x))= \begin{cases} -\beta(x) + e(x) - f(\beta(x))\beta'(x) - g(\beta(x)) , & \text{if}\ u(x)>\beta(x),\\ - u(x) + e(x) - f(u(x))u'(x) - g(u(x)), & \text{if}\ \eta(x)\leq u(x)\leq \beta(x),\\ - \eta (x) + e(x) - f(\eta(x))\eta'(x) -g(\eta(x)), & \text{if}\ u(x)<\eta(x), \end{cases} \]

then $H\in C^{\alpha }_{2\pi }(\mathbb {R})$.

Thus we can define

\begin{align*} N: C^{1,\alpha}_{2\pi} (\mathbb{R})& \to C^{\alpha}_{2\pi}(\mathbb{R}) \\ u & \longmapsto H(u,u') \end{align*}

and

\begin{align*} K: C^{\alpha}_{2\pi} (\mathbb{R})& \to C^{1,\alpha}_{2\pi}(\mathbb{R}) \\ z & \longmapsto v \end{align*}

where $v$ is the unique solution of

\[ (\triangle)^{s}v(x) - v(x)=z(x). \]

Then, we have $K \circ N: C^{1,\alpha }_{2\pi } (\mathbb {R})\to C^{1,\alpha }_{2\pi } (\mathbb {R})$ is continuous and compact, as proved in proposition (1.2). Notice that $N(C^{1,\alpha }_{2\pi } (\mathbb {R}))$ is bounded. Hence, by Schauder's fixed point theorem, $K\circ N$ has a fixed point $u$, i.e., $u$ is a solution of

\[ (\triangle)^{s}u(x) - u(x)= H(u,u') \quad in\ \mathbb{R}. \]

Now we prove that $\eta (x)\leq u(x)\leq \beta (x)$ for all $x\in \mathbb {R}$, we first show that $u(x)\leq \beta (x)$. The other inequality is similar. We assume by contradiction that $\max (u(x) - \beta (x))= u(\bar t) - \beta (\bar t)> 0$, here $\bar t$ is the point where the maximum is attained. So we have $(\triangle )^{s}u(\bar t) - (\triangle )^{s}\beta (\bar t) \leq 0$ and also

\begin{align*} (\triangle)^{s}u(\bar t) -(\triangle)^{s}\beta (\bar t) & \geq u(\bar t) + H(u(\bar t))+ f(\beta(\bar t))\beta'(\bar t) + g(\beta(\bar t)) - e(\bar t)\\ & \geq u(\bar t) -\beta(\bar t) + e(\bar t)- f(\beta(\bar t))\beta'(\bar t)- g(\beta(\bar t))\\& \quad + f(\beta(\bar t))\beta'(\bar t)+ g(\beta(\bar t)) - e(\bar t)\\ & = u(\bar t) - \beta(\bar t)>0, \end{align*}

this gives a contradict. Therefore, $u(x)\leq \beta (x)$.

Proof. Follows directly from theorem 1.1.

Proof. Fix $0< a< b$ and choose $\delta$ and $M$ with the restrictions $0<\delta < a$ and $M>b+\delta$. We consider the integral

(4.3)\begin{equation} I_{\delta,M}= \int_{a}^{b} u'(x) \int^{M}_{{- M} \atop {|y-x|\ge \delta}} \frac{u(x)-u(y)}{|x-y|^{1+2s}}\,\textrm{d}y\,\textrm{d}x= \iint_{A_{\delta,M}}u'(x) \frac{u(x)-u(y)}{|x-y|^{1+2s}}\,\textrm{d}y\,\textrm{d}x, \end{equation}

where $A_{\delta,M}=([a,b]\times [-M,M]) \cap \{(x,y)\in \mathbb {R}^{2}:\ |y-x|\ge \delta \}$.

It is not hard to see that

\begin{align*} I_{\delta,M} & = \frac{1}{2} \iint_{A_{\delta,M}} \ \left(\dfrac{(u(x)-u(y))^{2}}{|x-y|^{1+2s}}\right)_x \,\textrm{d}y\,\textrm{d}x\\ & \quad+ \frac{1+2s}{2} \ \iint_{A_{\delta,M}}\ \frac{(x-y) (u(x)-u(y))^{2}}{|x-y|^{3+2s}}\,\textrm{d}y\,\textrm{d}x. \end{align*}

We now split $A_{\delta,M}=A^{1} \cup A^{2} \cup A^{3} \cup A^{4}$, where

\begin{align*} A^{1}= \{(x,y) \in A_{\delta,M}: y \ge b\}\\ A^{2}= \{(x,y) \in A_{\delta,M}: x + \delta \le y \le b\}\\ A^{3}= \{(x,y) \in A_{\delta,M}: a \le y \le x - \delta\}\\ A^{4}= \{(x,y) \in A_{\delta,M}: y\le a\}. \end{align*}

Since the region $A^{2}$ is the reflection of $A^{3}$, with respect to the line $y=x$ and the integrand in the last integral above is antisymmetric, we get

\begin{align*} I_{\delta,M} & = \frac{1}{2}\iint\limits_{A_{\delta,M}} \left(\frac{(u(x)-u(y))^{2}}{|x-y|^{1+2s}}\right)_x\,\textrm{d}y\,\textrm{d}x\\ & \quad- \frac{1+2s}{2}\left( \iint\limits_{A^{1}} \frac{(u(x)-u(y))^{2}}{(x-y)^{2+2s}}\,\textrm{d}y\,\textrm{d}x - \iint\limits_{A^{4}}\frac{(u(x)-u(y))^{2}}{(x-y)^{2+2s}}\,\textrm{d}y\,\textrm{d}x \right)\\ & =\frac{1}{2}\oint\limits_{\partial A_{\delta,M}} \frac{(u(x)-u(y))^{2}}{|x-y|^{1+2s}}\,\textrm{d}y\\ & \quad - \frac{1+2s}{2}\left( \iint\limits_{A^{1}} \frac{(u(x)-u(y))^{2}}{(x-y)^{2+2s}}\,\textrm{d}y\,\textrm{d}x - \iint\limits_{A^{4}}\frac{(u(x)-u(y))^{2}}{(x-y)^{2+2s}}\,\textrm{d}y\,\textrm{d}x\right). \end{align*}

Now, using Green's formula, where the line integral is to be taken in the positive sense. Parameterizing the line integral we have

(4.4)\begin{align} I_{\delta,M} & ={-}\frac{1}{2}\int^{M} \limits_{{-M} \atop |y-a|\ge \delta} \frac{(u(a)-u(y))^{2}}{|a-y|^{1+2s}}\,\textrm{d}y+\frac{1}{2}\int^{M} \limits_{{-}M \atop |y-b|\ge \delta} \frac{(u(b)-u(y))^{2}}{|b-y|^{1+2s}}\,\textrm{d}y\nonumber\\ & \quad+ \frac{1}{2}\int^{b-\delta}\limits_{a} \frac{(u(x)-u(x+\delta))^{2}}{\delta^{1+2s}}\,\textrm{d}x-\frac{1}{2}\int_{a}^{b} \frac{(u(x)-u(x-\delta))^{2}}{\delta^{1+2s}}\,\textrm{d}x\nonumber\\ & \quad- \frac{1+2s}{2} \iint_{A^{1}} \frac{(u(x)-u(y))^{2}}{(x-y)^{2+2s}}\,\textrm{d}y\,\textrm{d}x + \frac{1+2s}{2} \iint_{A^{4}} \frac{(u(x)-u(y))^{2}}{(x-y)^{2+2s}}\,\textrm{d}y\,\textrm{d}x\nonumber\\ & ={-}\frac{1}{2}\int^{M} \limits_{{-}M \atop |y-a|\ge \delta} \frac{(u(a)-u(y))^{2}}{|a-y|^{1+2s}}\,\textrm{d}y +\frac{1}{2}\int^{M} \limits_{{-}M \atop |y-b|\ge \delta} \frac{(u(b)-u(y))^{2}}{|b-y|^{1+2s}}\,\textrm{d}y \nonumber\\ & \quad- \frac{1}{2}\int\limits_{a-\delta}^{a} \frac{(u(x+\delta)-u(x))^{2}}{\delta^{1+2s}}\,\textrm{d}x- \frac{1+2s}{2} \iint_{A^{1}} \frac{(u(x)-u(y))^{2}}{(x-y)^{2+2s}}\,\textrm{d}y\,\textrm{d}x\nonumber\\ & \quad+ \frac{1+2s}{2} \iint_{A^{4}} \frac{(u(x)-u(y))^{2}}{(x-y)^{2+2s}}\,\textrm{d}y\,\textrm{d}x. \end{align}

Now we can pass to the limit in $I_{\delta, M}$ as $M\to +\infty$ using dominated convergence and get

(4.5)\begin{align} \int_{a}^{b} u'(x) \int_{|y-x|\ge \delta}\frac{u(x)-u(y)}{|x-y|^{1+2s}}\,\textrm{d}y\,\textrm{d}x & = \frac{1}{2}\int_{|y-b|\ge \delta} \frac{(u(b)-u(y))^{2}}{|b-y|^{1+2s}}\,\textrm{d}y\nonumber\\ & \quad-\frac{1}{2}\int_{|y-a|\ge \delta}\frac{(u(a)-u(y))^{2}}{|a-y|^{1+2s}}\,\textrm{d}y\nonumber\\ & \quad- \frac{1}{2}\int_{a-\delta}^{a}\frac{(u(x+\delta)-u(x))^{2}}{\delta^{1+2s}}\,\textrm{d}x\nonumber\\ & \quad- \frac{1+2s}{2} \iint_{A_\delta^{1}}\frac{(u(x)-u(y))^{2}}{(x-y)^{2+2s}}\,\textrm{d}y\,\textrm{d}x,\nonumber\\ & \quad+ \frac{1+2s}{2} \iint_{A_\delta^{2}}\frac{(u(x)-u(y))^{2}}{(x-y)^{2+2s}}\,\textrm{d}y\,\textrm{d}x, \end{align}

where $A_\delta ^{1}=([a, b]\times (b, +\infty ) \cap \{(x,y)\in \mathbb {R}^{2}: y \geq x + \delta \}$ and $A_\delta ^{2}=([a, b]\times (- \infty, a) \cap \{(x,y)\in \mathbb {R}^{2}: y \le x-\delta \}$.

Finally, will want to pass to the limit as $\delta \to 0$ in (4.5). Observe that, since $u\in C^{1}_{2\pi } (\mathbb {R})$, we have that $u'$ is bounded so $u$ is Lipschitz, hence for $y$ close to $b$ gives

\[ \frac{(u(b)-u(y))^{2}}{|b-y|^{1+2s}} \le C |b-y|^{1-2s}\in L^{1}_\textrm{loc}(\mathbb{R}), \]

and we also have for $y$ close to $a$

\[ \frac{(u(a)-u(y))^{2}}{|a-y|^{1+2s}} \le C |a-y|^{1-2s}\in L^{1}_\textrm{loc}(\mathbb{R}). \]

So the passing to the limit is justified in the first and second integral in the right-hand side of (4.5) by dominated convergence. As for the third integral, for being $u$ Lipschitz gives

\[ \frac{(u(x+\delta)-u(x))^{2}}{\delta^{1+2s}} \le C\delta^{1-2s} \]

so that,

\[ \int_{t-\delta}^{t} \frac{(u(x+\delta)-u(x))^{2}}{\delta^{1+2s}}\,\textrm{d}x \le C\delta^{2-2s}\to 0 \]

as $\delta \to 0^{+}$. As for the double integral, we also have that

\[ \frac{(u(x)-u(y))^{2}}{|x-y|^{2+2s}} \le C |x-y|^{{-}2s}\in L^{1}_\textrm{loc}(\mathbb{R}^{2}), \]

for $x$ and $y$ close. Therefore, we can pass to the limit in the right-hand side of (4.5).

Now, on the left-hand side of (4.5) using the regularity of $u$ and dominated convergence, it follows that we can pass the limit and get our identity.

Proof. Now we write

\[ u(t)= a_{0} + \sum_{n=1}^{+ \infty} a_{n}\cos(nt) + b_{n}\sin(nt), \]

so,

\[ u'(t)= \sum_{k=1}^{+ \infty} b_{k}k\cos(kt) - a_{k}k\sin(kt), \]

and

\[ (\triangle)^{s}u(t)=\sum_{n=1}^{+ \infty} a_{n}n^{2s}\cos(nt) + b_{n}n^{2s}\sin(nt), \]

by orthogonality we have

\begin{align*} & <\sin (nt),\sin(kt)>{=} 0,\quad n \neq k\\ & <\cos (nt),\cos(kt)>{=} 0 ,\quad n \neq k\\ & <\cos (nt),\sin(kt)>{=} 0. \end{align*}

Hence,

(4.6)\begin{align} & \int\limits_{0}^{2\pi} (\triangle)^{s}u(t)u'(t)\,\textrm{d}t\nonumber\\ & \quad=\sum_{n=1}^{+ \infty}\int_{0}^{2\pi} (a_{n}n^{2s}\cos(nt) + b_{n}n^{2s}\sin(nt))( b_{n}n\cos(nt) - a_{n}n\sin(nt))\nonumber\\ & \quad= \sum_{n=1}^{+ \infty}\left(\int_{0}^{2\pi} a_{n}b_{n}n^{2s + 1}\cos^{2}(nt) + \int_{0}^{2\pi}(b_{n}^{2}- a_{n}^{2})n^{2s + 1}\cos(nt)\sin(nt)\right.\nonumber\\ & \qquad\left.- \int\limits_{0}^{2\pi} a_{n}b_{n}n^{2s + 1}\sin^{2}(nt)\right)\nonumber\\ & \quad= \sum_{n=1}^{+ \infty} \int_{0}^{2\pi} a_{n}b_{n}n^{2s + 1}(\cos^{2}(nt) - \sin^{2}(nt))\nonumber\\ & \quad= \sum_{n=1}^{+ \infty} \int_{0}^{2\pi} a_{n}b_{n}n^{2s + 1} \cos(2nt) =0. \end{align}

Proof. Let $w$ be a solution of

\[ w''= \eta f \quad in\ \mathbb{R}, \]

where $\eta \in C^{\infty }(\mathbb {R})$ such that $\eta \equiv 0$ outside $(x_{0} - \delta, x_{0} + \delta )$ and $\eta \equiv 1$ in $[x_{0} - \frac {3\delta }{4}, x_{0} + \frac {3\delta }{4}]$. So we get

(4.7)\begin{align} \|w\|_{C^{2,\alpha}(\mathbb{R})} & \leq C\| \eta f \|_{C^{\alpha}(\mathbb{R})}\nonumber\\ & \leq C\|f\|_{C^{\alpha}(x_{0} - \delta, x_{0} + \delta)}. \end{align}

Then, since $(\triangle )^{s}((\triangle )^{1 - s}w)= w''$ we have

\[ (\triangle)^{s}(u - (\triangle)^{1 - s}w)=0 \quad in \ [x_{0} - 3\delta/4, x_{0} + 3\delta/4], \]

we can use theorem 1.1 of [Reference Silvestre and Caffarelli31], to obtain that there exists $\gamma$ such that

\begin{align*} \|u - (\triangle)^{1 - s}w\|_{C^{2s+ \gamma}[x_{0} - \frac{3\delta}{4}, x_{0} + \frac{3\delta}{4}]}& \leq C\|u - (\triangle)^{1 - s}w\|_{L^{\infty}(\mathbb{R})}\\ & \leq C (\|u\|_{L^{\infty}(\mathbb{R})} + \|(\triangle)^{1 - s}w\|_{L^{\infty}(\mathbb{R})}). \end{align*}

Hence, using lemma 4.3 and (4.7) we get

\begin{align*} \|u\|_{C^{2s+ \gamma}[x_{0} - \frac{3\delta}{4}, x_{0} + \frac{3\delta}{4}]} & \leq C (\|(\triangle)^{1 - s}w\|_{C^{2s+\gamma}(\mathbb{R})} + \|u\|_{L^{\infty}(\mathbb{R})}) \\ & \leq C(\|f\|_{C^{\gamma}(x_{0} - \delta, x_{0} + \delta)} + \|u\|_{L^{\infty}(\mathbb{R})}). \end{align*}

Proof. Now, by the lemma 4.4,

(4.8)\begin{equation} \|u\|_{C^{2s + \gamma} [x_{0} - \frac{\delta}{2}, x_{0} + \frac{\delta}{2}]} \leq C^{*}(\|f\|_{C^{\gamma}(x_{0} - \delta, x_{0} + \delta)} + \|u\|_{C^{1,\gamma}(x_{0} - \delta, x_{0} + \delta)} + \|u\|_{L^{\infty}(\mathbb{R})}). \end{equation}

Using now the Interpolation inequalities ([Reference Krylov17], theorems 3.2.1) together with the fact that $2s > 1$, we obtain that, for any $\epsilon > 0$, there is a positive constant $C = C(\gamma, \epsilon, s)$, such that

(4.9)\begin{equation} \|u\|_{C^{1, \gamma}(x_{0} - \delta, x_{0} + \delta)}\leq \epsilon \|u\|_{C^{2s+\gamma}(x_{0} - \delta, x_{0} + \delta)} + C\|u\|_{C(x_{0} - \delta, x_{0} + \delta)}. \end{equation}

Hence, choosing $\epsilon = \frac {1}{2C^{*}}$ we get out results from (4.9) and (4.8).

Proof. We shall assume that $u$ is a solution of (4.1) for some fixed $\lambda \in (0,1)$ then

\[ \int\limits_{0}^{2\pi} (\triangle) ^{s}u(t)\,\textrm{d}t + c\int\limits_{0}^{2\pi} u'(t)\,\textrm{d}t - \lambda \int\limits_{0}^{2\pi} g(u(t))\,\textrm{d}t = \lambda \int\limits_{0}^{2\pi} e(t)\,\textrm{d}t. \]

But, for lemma 2.2 and using the $2\pi$-periodicity of $u'$ gives

(4.10)\begin{equation} \frac{1}{2\pi} \int\limits_{0}^{2\pi} g(u(t))\,\textrm{d}t + \bar{e}=0. \end{equation}

We now notice that with assumption $(G2)$, we get that there exists $R_{0}>0$ such that

\[ g(x)>0 \quad \text{and}\quad g(x) + \bar{e} >0, \]

whenever $0< x\leq R_{0}$. Therefore, if $0< u(t)\leq R_{0}$ for all $t\in [0, 2\pi ]$, we obtain $g(u(t)) + \bar {e} >$ for those $t$ and hence

\[ \frac{1}{2\pi} \int\limits_{0}^{2\pi} g(u(t))\,\textrm{d}t + \bar{e} > 0 \]

a contradiction to (4.10), thus there exists $t_{0}$ such that $u(t_{0}) > R_{0}$. On the other hand, assumption $(G1)$ implies the existence of some $R_{1} > R_{0}$ such that

\[ g(u) + \bar{e} < 0, \]

whenever $u \geq R_{1}$. Then, if $u(t) \geq R_{1}$ for all $t\in [0,2\pi ]$, gives $g(u(t)) + \bar {e} < 0$ for those $t$ and

\[ \frac{1}{2\pi} \int\limits_{0}^{2\pi} g(u(t))\,\textrm{d}t + \bar{e} < 0 \]

a contradiction to (4.10), thus there exists $t_{1}$ such that $u(t_{1}) < R_{1}$.

Proof. By lemma 4.7

(4.12)\begin{align} u(t)& \leq u(t_{1}) + \int\limits_{0}^{2\pi} |u'(t)|\,\textrm{d}t\nonumber\\ & \leq R_{1} + \sqrt{2\pi} \|u'\|_{L^{2}(0, 2\pi)}. \end{align}

Now, multiplying (4.1) for $u'$ and integrating the equation

(4.13)\begin{equation} \int_{0}^{2\pi} (\triangle) ^{s}u(t)u'(t)\,\textrm{d}t + c \int_{0}^{2\pi} |u'(t)|^{2}\,\textrm{d}t = \lambda \int_{0}^{2\pi} g(u(t))u'(t) + \lambda \int_{0}^{2\pi} e(t)u'(t)\,\textrm{d}t, \end{equation}

since

\[ \lambda \int_{0}^{2\pi} g(u(t))u'(t)\,\textrm{d}t = \lambda \int_{u(0)}^{u(2\pi)} g(w)\,\textrm{d}w= 0, \]

together with lemma 4.2 and using Hölder inequality gives

\begin{align*} c\|u'\|_{L^{2}(0,2\pi)}^{2} & = \lambda \int_{0}^{2\pi} e(t)u'(t)\\ & \leq C \|e\|_{L^{2}(0,2\pi)}\|u'\|_{L^{2}(0,2\pi)}, \end{align*}

so that,

\[ \|u'\|_{L^{2}(0,2\pi)} \leq C\|e\|_{L^{2}(0,2\pi)} \]

which, together with (4.12) implies

\[ u(t) \leq R_{0} + C\|e\|_{L^{2}(0,2\pi)}= R, \]

therefore the result follows.

Proof. First we find a $L^{1}$ bound for $g$. Notice that $(G4)$ implies that

(4.15)\begin{equation} |g(u(t))|\leq g(u(t)) + 2au(t) + 2b, \end{equation}

integrating over $[0,2\pi ]$

\[ \int_{0}^{2\pi} |g(u(t))|\,\textrm{d}t \leq \int_{0}^{2\pi} g(u(t))\,\textrm{d}t +2a\int_{0}^{2\pi} u(t)\,\textrm{d}t + 4b\pi, \]

by lemma 4.8 and (4.10) we have

(4.16)\begin{equation} \int_{0}^{2\pi} |g(u(t))|\,\textrm{d}t \leq 4\pi(aR + b). \end{equation}

Secondly, we now consider the following problem

(4.17)\begin{equation} (\Delta)^{s} v(t) =g(v(t)) + h(t)\quad \text{in}\ \mathbb{R}, \end{equation}

where $h\in L^{2}$. Let $w\in C^{2}(\mathbb {R})$ such that

(4.18)\begin{equation} w'(t)= \int_{{-}2\pi}^{t} g(v(x)) + h(x)\,\textrm{d}x, \quad t\in[{-}2\pi,4\pi]. \end{equation}

Then, $w$ is a solution of

\[ w''(t)= g(v(t)) + h(t), \]

and since $(\triangle )^{s}((\triangle )^{1 - s}w)= w''$ we have

\[ (\triangle)^{s}(v - (\triangle)^{1 - s}w)=0 \quad in \ [0,2\pi]. \]

Using [Reference Ambrosio1, theorem 2.7] we have

\begin{align*} \|v - (\triangle)^{1 - s}w\|_{C^{s + \epsilon/2}(0,2\pi)}& \leq C\|v - (\triangle)^{1 - s}w\|_{L^{\infty}(0,2\pi)}\\ & \leq C (\|v\|_{L^{\infty}(\mathbb{R})} + \|(\triangle)^{1 - s}w\|_{L^{\infty}(0,2\pi)}), \end{align*}

and as we know that for $\mu =1 - s + \epsilon /2$ (see [Reference Tan, Felmer and Quaas32, theorem 3.1])

(4.19)\begin{equation} \| (\triangle)^{1-s}w\|_{C^{s + \epsilon/2}[0,2\pi]} \leq C \|w\|_{C^{1,\mu}({-}2\pi,4\pi)}, \end{equation}

we obtain

(4.20)\begin{align} \|v\|_{C^{s+\epsilon/2}[0,2\pi]} & \leq C (\|(\triangle)^{1 - s}w\|_{C^{s+ \epsilon/2}[0,2\pi]} + \|v\|_{L^{\infty}(\mathbb{R})}) \nonumber\\ & \leq C(\|w\|_{C^{1,\mu}({-}2\pi,4\pi)} + \|v\|_{L^{\infty}(\mathbb{R})}). \end{align}

Now, we want to bound $\|w\|_{C^{1,\mu }(-2\pi,4\pi )}$, it follows from (4.18) and (4.16) we need to estimate $[w']_{C^{\mu }}$. Let $x,y\in (-2\pi,4\pi )$ and using Hölder inequality we get

(4.21)\begin{align} & |w'(x) -w'(y)|\nonumber\\ & \quad= \left|\int_{x}^{y} g(v(x)) + h(x)\,\textrm{d}x\right| \nonumber\\ & \quad\leq \left(\left(\int_{x}^{y}|g(v(x))|^{1/(s-\epsilon/2)}\,\textrm{d}x\right)^{s-\epsilon/2}+ \left(\int_{x}^{y}|h(x)|^{1/(s-\epsilon/2)}\,\textrm{d}x\right)^{s-\epsilon/2}\right)|x-y|^{1-s+\epsilon/2}\nonumber\\ & \quad\leq \left(\left(\int_{{-}2\pi}^{4\pi} |g(v(x))|^{1/(s-\epsilon/2)}\,\textrm{d}x\right)^{s-\epsilon/2} + \left(\int_{{-}2\pi}^{4\pi} |h(x)|^{1/(s-\epsilon/2)}\,\textrm{d}x\right)^{s-\epsilon/2}\right)|x-y|^{1-s+\epsilon/2}, \end{align}

so that (4.20) implies,

(4.22)\begin{equation} \|v\|_{C^{s + \epsilon/2}[0,2\pi]} \leq C\left(\left(\int_{{-}2\pi}^{4\pi} |g(v(x))|^{1/(s-\epsilon/2)}\,\textrm{d}x\right)^{s-\epsilon/2} + \|h\|_{L^{2}[{-}2\pi,4\pi]}\right). \end{equation}

Finally, let $u$ a solution of (1.5) by the above with $h(t)= cu'(t) + e(t)$ we have

\begin{align*} & \|u\|_{C^{s + \epsilon/2}[0,2\pi]}\\ & \quad\leq C\left(\left(\int_{{-}2\pi}^{4\pi} |g(u(x))|^{1/(s-\epsilon/2)}\,\textrm{d}x\right)^{s-\epsilon/2}+ \|u'\|_{L^{2}[{-}2\pi,4\pi]} + \|e\|_{L^{2}[{-}2\pi,4\pi]}\right). \end{align*}

Proof. Let $t^{n}_{3}$, $t^{n}_{4}$ be the minimum point and the maximum point of $u_n$ in $[0,2\pi ]$, notice that by lemma 4.7 we have $u_n(t^{n}_4)>R_0$.

Now we assume by contradiction that $u_n(t^{n}_3) \to 0$ as $n\to \infty$. First we suppose that $t^{n}_{4}< t^{n}_{3}$,

and multiplying (4.1) by $u'_n$, then

\begin{align*} & \int_{t^{n}_{4}}^{t^{n}_{3}} (\triangle) ^{s}u_n(t)u_n'(t)\,\textrm{d}t + c \int_{t^{n}_{4}}^{t^{n}_{3}} |u_n'(t)|^{2}\,\textrm{d}t\\ & \quad - \lambda\int_{t^{n}_{4}}^{t^{n}_{3}} g(u_n(t))u_n'(t) = \lambda \int_{t^{n}_4}^{t^{n}_{3}} e(t)u_n'(t)\,\textrm{d}t, \end{align*}

so that,

(4.23)\begin{align} \lambda\int_{u_n(t^{n}_{3})}^{u_n(t^{n}_{4})} g(w)\,\textrm{d}w & = \int_{t^{n}_{4}}^{t^{n}_{3}} (-\triangle) ^{s}u_n(t)u_n'(t)\,\textrm{d}t\nonumber\\ & \quad - c \int_{t^{n}_{4}}^{t^{n}_{3}} |u_n'(t)|^{2}\,\textrm{d}t + \lambda\int_{t^{n}_{4}}^{t^{n}_{3}} e(t)u_n'(t)\,\textrm{d}t. \end{align}

We need to bound the first term in the right-hand side, since by (4.11) the order terms in the right hand are bounded, to get a contradiction with $(G3)$ by the fact that $u_n(t^{n}_4)>R_0$ and $u_n(t^{n}_3) \to 0$.

Let us now define

\[ \delta_n=\sup\{\delta>0\ | u_n (t)>\frac{R_0}{4C_0}\ \text{for}\ t \in (t_{4}^{n}-\delta, t_{4}^{n}+\delta)\}. \]

Notice that for $n$ large $u_n(t_{3}^{n})<\frac {R_0}{4C_0}$ therefore $\delta _n$ is finite and

\[ \inf_{(t_{4}^{n}-\delta_n, t_{4}^{n}+\delta_n)}u_n=\frac{R_0}{4C_0}. \]

Now we claim that there exists $\delta _{0}>0$ such that $\delta _n>\delta _{0}$. Suppose the contrary, so there exists a sub-sequence (still denote by $n$) such that $\delta _{n} \to 0$. Define $v_{n}(t)=u_{n}(\delta _{n}t + t_{4}^{n})$, we have that

\[ (\triangle)^{s}v_{n}(t) + \delta^{2s-1}_{n}cv_n'(t)= \delta^{2s}_{n}h \quad in \ ({-}1,1), \]

where $h:=g(u_{n}(\delta _{n}t + t_{4}^{n})) + e(\delta _{n}t + t_{4}^{n})\in L^{\infty }(-1,1)$ by the definition of $\delta _n$. By lemma 4.6,

\[ \sup\limits_{\left(-\frac{1}{2},\frac{1}{2}\right)} v_{n} \leq C_{0}(\inf\limits_{({-}1,1)} v_{n} + \delta^{2s}_{n}\|h\|_{L^{\infty}({-}1,1)}), \]

as $\inf \limits _{(-1,1)} v_{n} = \dfrac {R_{0}}{4C_{0}}$, we have

\[ R_{0} < \sup\limits_{\left(-\frac{1}{2},\frac{1}{2}\right)} v_{n} \leq C_{0}\left(\frac{R_{0}}{4C_{0}} + \delta^{2s}_{n}\|h\|_{L^{\infty}({-}1,1)}\right). \]

Taking $\delta _{n}$ small, we obtain a contradiction and the claim follows.

Now we can use theorem 4.5, and with $\delta =\delta _0 >0$ independent of $u_n$ there exists $\bar C>0$ (independent of $n$) such that

\[ \|u_n\|_{C^{2s + \beta} \left[t_{4}^{n} - \frac{\delta}{2}, t_{4}^{n} + \frac{\delta}{2}\right]}\leq \bar C, \]

with $\delta > 0$, this implies that

(4.24)\begin{equation} |u_n'(x)| \leq \bar C\quad\text{in}\ [t_{4}^{n} - \delta/2, t_{4}^{n} + \delta/2]. \end{equation}

Now we simplify the notation and drop the $n$ index to estimate $\int \limits _{t^{n}_{4}}^{t^{n}_{3}} u_n'(x) (-\Delta )^{s} u_n(x)\,\textrm {d}x$.

By lemma 4.1 we have

(4.25)\begin{align} \int\limits_{t_{4}}^{t_{3}} u'(x) (-\Delta)^{s} u(x)\,\textrm{d}x & \leq\frac{c(1,s)}{2} \left( \int_{-\infty}^{+\infty} \frac{(u(t_{3})-u(y))^{2}}{|t_{3}-y|^{1+2s}}\,\textrm{d}y \right.\nonumber\\ & \quad\left. + (1+2s)\int_{t_{4}}^{t_{3}} \int_{-\infty}^{t_{4}} \frac{(u(x)-u(y))^{2}}{|x-y|^{2+2s}}\,\textrm{d}y\,\textrm{d}x\right). \end{align}

Let $\rho >0$ we have

\begin{align*} \int_{-\infty}^{+\infty} \frac{(u(t_{3})-u(y))^{2}}{|t_{3}-y|^{1+2s}}\,\textrm{d}y & = \int_{|t_{3} - y|< \rho} \frac{(u(t_{3})-u(y))^{2}}{|t_{3}-y|^{1+2s}}\,\textrm{d}y\\ & \quad + \int_{|t_{3} - y|\geq \rho} \frac{(u(t_{3})-u(y))^{2}}{|t_{3}-y|^{1+2s}}\,\textrm{d}y, \end{align*}

by lemma 4.9 together with $(G2)$,$(G4)$ and (4.15) we obtain

(4.26)\begin{align} & \int_{|t_{3} - y|< \rho} \frac{(u(t_{3})-u(y))^{2}}{|t_{3}-y|^{1+2s}}\,\textrm{d}y\nonumber\\ & \quad\leq C\left(\left(\int_{{-}2\pi}^{4\pi} |g(u(x))|^{1/(s-\epsilon/2)}\,\textrm{d}x\right)^{2s-\epsilon} + C^{*}\right)\int_{|t_{3} - y|< \rho} \frac{|t_{3}-y|^{2s+\epsilon}}{|t_{3}-y|^{1+2s}}\,\textrm{d}y\nonumber\\ & \quad\leq C \left(\left(\int_{{-}2\pi}^{4\pi} |g(u(x))|^{1 + (1/(s-\epsilon 2) -1)}\,\textrm{d}x\right)^{2s-\epsilon} + C^{*}\right)\int_{|t_{3} - y|< \rho} |t_{3}-y|^{\epsilon -1}\,\textrm{d}y\nonumber\\ & \quad\leq Cg(u(t_{3}))^{(1/(s-\epsilon/2) -1)(2s-\epsilon)} \left(\left(\int_{{-}2\pi}^{4\pi} g(u(x))\,\textrm{d}x\right)^{2s-\epsilon} +{+} C^{*}\right)\nonumber\\ & \quad\leq C(g(u(t_{3}))^{(1/(s-\epsilon/2) -1)(2s-\epsilon)} + C^{*})=C(g(u(t_{3}))^{2-2s +\epsilon} + C^{*}), \end{align}

where $C^{*}= 4a\pi R + 4b\pi + \|u'\|_{L^{2}[-2\pi,4\pi ]} + \|e\|_{L^{2}[-2\pi,4\pi ]}$, and by (4.11) we have $C^{*}< + \infty$. Now, the second integral we know is bounded

(4.27)\begin{align} \int_{|t_{3} - y|\geq \rho} \frac{(u(y) -u(t_{3}))^{2}}{|t_{3}-y|^{1+2s}}\,\textrm{d}y & \leq 4R^{2} \int_{|t_{3} - y|\geq \rho} \frac{\textrm{d}y}{|t_{3}-y|^{1+2s}}\nonumber\\ & \leq C. \end{align}

Now from (4.24) we have

\[ u(x) - u(y) = \int_{y}^{x} u'(\xi)\,\textrm{d}\xi\leq \bar C (x-y), \]

so that

(4.28)\begin{align} & \int_{t_{4}}^{t_{3}} \int_{t_{4}- \frac{\delta}{2}}^{t_{4}} \frac{(u(x)-u(y))^{2}}{|x-y|^{2+2s}}\,\textrm{d}y\,\textrm{d}x\nonumber\\ & \quad\leq \bar C^{2} \int_{t_{4}}^{t_{3}} \int_{t_{4}- \frac{\delta}{2}}^{t_{4}} \frac{(x -y)^{2}}{|x-y|^{2+2s}}\,\textrm{d}y\,\textrm{d}x \nonumber\\ & = \dfrac{\bar C^{2}}{2s -1} \int_{t_{4}}^{t_{3}} -(x-t_{4})^{{-}2s +1} + \left(x-t_{4}+ \frac{\delta}{2}\right)^{{-}2s +1}\,\textrm{d}x \nonumber\\ & \quad= \dfrac{\bar C^{2}}{(2s -1)(2 -2s)} \left(-(t_{3}-t_{4})^{{-}2s +2} + \left(t_{3}-t_{4} + \frac{\delta}{2}\right)^{{-}2s +2}- \frac{\delta}{2}^{{-}2s +2}\right)\nonumber\\ & \quad\leq \dfrac{\bar C^{2}}{(2s -1)(2 -2s)} \left(t_{3}-t_{4} + \frac{\delta}{2}\right)^{{-}2s +2}. \end{align}

To analyse the same integral when $y < t_{4}- \delta$ and $t_{4} < x< t_{3}$ we have

(4.29)\begin{align} \int_{t_{4}}^{t_{3}} \int_{- \infty}^{t_{4} - \frac{\delta}{2}} \frac{(u(x)-u(y))^{2}}{|x-y|^{2+2s}}\,\textrm{d}y\,\textrm{d}x & \leq 4\|u\|^{2}_{L_{\infty}}\int_{t_{4}}^{t_{3}} \int_{- \infty}^{t_{4} - \frac{\delta}{2}} \frac{\textrm{d}y}{|x-y|^{2+2s}}\,\textrm{d}x \nonumber\\ & \leq \frac{4R^{2}}{1+2s}\int_{t_{4}}^{t_{3}}- \left(x - t_{4} + \frac{\delta}{2}\right)^{{-}1 -2s}\,\textrm{d}x \nonumber\\ & = \frac{4R^{2}}{(1 +2s)(2s)}\left(\left(t_{3}-t_{4} + \frac{\delta}{2}\right)^{{-}2s} - \frac{\delta}{2}^{{-}2s}\right)\nonumber\\ & \leq \frac{4R^{2}}{(1 +2s)(2s)}\left(t_{3}-t_{4} + \frac{\delta}{2}\right)^{{-}2s}. \end{align}

Hence, (4.25), (4.26), (4.27), (4.28) and (4.29) give that there exists $M_0>0$ independent of $n$ such that

\[ g(u_n(t^{n}_{3}))^{2s-2-\epsilon}\int_{t^{n}_{4}}^{t^{n}_{3}} u_n'(x) (-\Delta)^{s} u_n(x)\,\textrm{d}x \leq M_{0}. \]

Thus, as mentioned above, this inequality implies a contradiction and the result follows.

Proof. We now take $g(u)= G(u) - \mu u$, so (5.1) is of the form (1.5), and satisfying the following conditions:

1. (H1’) $\lim _{t \to 0^{+}} g(t) = + \infty$

2. (H2’) $\limsup _{t \to + \infty } [g(t) + \bar {e}]< 0$ for $\mu \geq 0$

3. (H3’) $\int _{0}^{1} g(t)\,\textrm {d}t= + \infty$.

Therefore the results of theorem 1.4 are valid for equation (5.1) when $\mu \geq 0$. Then, by the continuity of degree defined in theorem 1.4 there exists $\eta >0$ such that for $-\eta \leq \mu < 0$ that degree is not trivial. So, there exists $u$ a solution for (5.1) with $-\eta \leq \mu < 0$. Now, by $(H2)$ $N(u)=o(\|u\|)$ at $u= + \infty$ then, the fundamental theorem on bifurcation from infinity from a simple eigenvalue implies the existence of a continuum $\mathcal {C}_{\infty }$ of positive solution $(\mu, u)$ bifurcating from infinity at $\mu =0$, since the solutions for $\mu \geq 0$ are bounded, the bifurcation is for the left side.