Erratum to: Existence of solutions of scalar field equations with fractional operator

1 Erratum to: J. Fixed Point Theory Appl. (2017) 19:649–690 DOI 10.1007/s11784-016-0369-x

In the latter part of [2, Proposition 3.6 (i)], it is claimed that any weak solution of

$$\begin{aligned} (1-\Delta )^\alpha u = f(u) \quad \hbox {in} \ \mathbf {R}^N,\qquad u \in H^\alpha (\mathbf {R}^N)\end{aligned}$$

decays faster than any polynomial. However, the proof there is valid only for positive solutions since the function $g(x) :=( f(u(x)) - (1-\delta _0) u(x))_+$ may not be compactly supported.

We shall modify the proof and show the decay estimate of any weak solution of

$$\begin{aligned} (1-\Delta )^\alpha u = f(x,u) \quad \hbox {in} \ \mathbf {R}^N, \qquad u \in H^\alpha (\mathbf {R}^N). \end{aligned}$$

(1)

Proposition 0.1

Let $f \in C(\mathbf {R}^N\times \mathbf {R}, \mathbf {R})$ satisfy $|f(x,s)| \le C(|s| + |s|^{2^*_\alpha -1} )$ for each $(x,s) \in \mathbf {R}^N\times \mathbf {R}$ and

$$\begin{aligned} \limsup _{s\rightarrow 0} \sup _{x \in \mathbf {R}^N} \frac{f(x,s)}{s} < 1. \end{aligned}$$

(2)

Then for every weak solution u of (1) and every $k \in \mathbf {N},$ there exists a $C_{k,u}$ such that $|u(x)| \le C_{k,u}(1+|x|)^{-k}$ for all $x \in \mathbf {R}^N$.

We first introduce the extension problem (see [1]):

$$\begin{aligned} \left\{ \begin{array}{ll} t^{1-2\alpha } (-\Delta _x + 1) w - (t^{1-2\alpha } w_t)_t=0&{}\quad \hbox {in} \ \mathbf {R}^{N+1}_+,\\ w=u&{}\quad \hbox {on} \ \partial \mathbf {R}^{N+1}_+\end{array} \right. \end{aligned}$$

(3)

where $\Delta _x = \sum _{i=1}^N \partial _{x_i}^2$. We also set

$$\begin{aligned}&X^\alpha := \{w : \mathbf {R}^{N+1}_+\rightarrow \mathbf {R}\mid \Vert w \Vert _{X^\alpha } < \infty \},\\&\quad \Vert w \Vert _{X^\alpha }^2 := \int _{\mathbf {R}^{N+1}_+} t^{1-2 \alpha } \left( |\nabla w|^2 + w^2 \right) \mathrm {d}X, \quad X := (x,t) \end{aligned}$$

where $\nabla = (\nabla _x, \partial _t)$. Recall [2, Proposition 5.2]:

Proposition 0.2

• (i) There exists the trace operator $\mathrm{Tr}: X^\alpha \rightarrow H^\alpha (\mathbf {R}^N)$.

• (ii) For each $u \in H^\alpha (\mathbf {R}^N),$ there is a unique solution $Eu \in X^\alpha$ of (3). In addition,  there is a constant $\kappa _\alpha > 0$ so that

  $$\begin{aligned} \kappa _\alpha \left\langle u, \mathrm{Tr\,}w \right\rangle _\alpha = \int _{\mathbf {R}^{N+1}_+} t^{1-2\alpha } (\nabla (Eu) \cdot \nabla w + (Eu) w)\,\mathrm {d}X \end{aligned}$$

  for every $u \in H^\alpha (\mathbf {R}^N)$ and $w \in X^\alpha$.

• (iii) For each $u \in H^\alpha (\mathbf {R}^N)$ and $w \in X^\alpha$ with $\mathrm{Tr\,}w = u,$ $\kappa _\alpha \Vert u \Vert _\alpha ^2 = \Vert Eu \Vert _{X^\alpha }^2 \le \Vert w \Vert _{X^\alpha }^2$ hold.

Now we prove Proposition 0.1:

Proof of Proposition 0.1

Let u be a weak solution of (1). By (2), we may choose $s_0>0$ and $\delta _0>0$ such that

$$\begin{aligned} \frac{f(x,s)}{s} < 1 - 2\delta _0 \quad \hbox {for any }(x,s) \in \mathbf {R}^N\times ([-s_0,s_0] \setminus \{0\}). \end{aligned}$$

From the former part of the proof of [2, Proposition 3.6 (i)], we see that $u \in C^{\beta }_\mathrm{b}(\mathbf {R}^N)$ for any $\beta \in (0,2\alpha )$ and $u(x) \rightarrow 0$ as $|x| \rightarrow \infty$. Hence, we may choose an $R_0>0$ such that

$$\begin{aligned} |x| \ge R_0, \ u(x) \ne 0 \quad \Rightarrow \quad \frac{f(x,u(x))}{u(x)} \le 1 - \delta _0. \end{aligned}$$

(4)

Denote by $\chi _{R_0}(x)$ the characteristic function of $B_{R_0}(0)$. Notice that $\chi _{R_0}(x) | f(x,u(x)) | \in L^2(\mathbf {R}^N) \cap L^\infty (\mathbf {R}^N)$ is compactly supported. Let v be a unique solution of

$$\begin{aligned} (1-\Delta )^\alpha v - (1-\delta _0) v = \chi _{R_0}(x) | f(x,u(x)) | \quad \hbox {in} \ \mathbf {R}^N, \quad v \in H^\alpha (\mathbf {R}^N). \end{aligned}$$

Then [2, Proposition 5.1] asserts that for any $k \in \mathbf {N}$ there is a $c_k>0$ so that

$$\begin{aligned} 0< v(x) \le c_k (1 + |x|)^{-k} \quad \hbox {for all }x \in \mathbf {R}^N. \end{aligned}$$

Therefore, it suffices to prove $-v(x) \le u(x) \le v(x)$ for all $x \in \mathbf {R}^N$.

To this end, remark that u satisfies

$$\begin{aligned} (1-\Delta )^\alpha u = \chi _{R_0}(x) f(x,u(x)) + (1-\chi _{R_0}(x)) f(x,u(x)) \quad \hbox {in }\mathbf {R}^N. \end{aligned}$$

Setting $U(x) := v(x) - u(x)$, one finds that

$$\begin{aligned} (1-\Delta )^\alpha U= & {} \chi _{R_0}(x) \{|f(x,u(x))| - f(x,u(x))\}\\&+\,\{-(1-\chi _{R_0}(x)) f(x,u(x)) + (1-\delta _0) v(x)\}. \end{aligned}$$

Now consider EU and $(EU)_-$ where $w_-(X) := \max \{ 0, -w(X) \}$. It is easily seen that $(EU)_- \in X^\alpha$ and $\mathrm{Tr\,}(EU)_- = U_-$. Applying Proposition 0.2, we get

$$\begin{aligned}&\kappa _\alpha \left\langle U, U_- \right\rangle _\alpha \\&\quad = \int _{\mathbf {R}^N} t^{1-2\alpha } ( \nabla EU \cdot \nabla (EU)_- + (EU) (EU)_-)\,\mathrm {d}X = - \Vert (EU)_- \Vert _{X^\alpha }^2. \end{aligned}$$

On the other hand,

$$\begin{aligned} \kappa _\alpha \left\langle U, U_- \right\rangle _\alpha= & {} \kappa _\alpha \int _{\mathbf {R}^N} \left[ \chi _{R_0} \left\{ |f(x,u)| - f(x,u) \right\} \right. \nonumber \\&\left. +\left\{ -(1-\chi _{R_0}) f(x,u) + (1-\delta _0) v \right\} \right] U_- \mathrm {d}x \nonumber \\\ge & {} \kappa _\alpha \int _{\mathbf {R}^N} (1-\chi _{R_0}) \left[ - f(x,u) + (1-\delta _0) v \right] U_- \mathrm {d}x. \end{aligned}$$

(5)

If $U_-(x) > 0$ and $|x| \ge R_0$, then from $u(x)> v(x) > 0$ and (4) it follows that

$$\begin{aligned}&(1-\chi _{R_0}(x)) \left[ - f(x,u(x)) + (1-\delta _0) v(x) \right] \nonumber \\&\quad \ge (1-\chi _{R_0}(x)) \left[ - (1-\delta _0) u(x) + (1-\delta _0) v(x) \right] \\&\quad = (1-\delta _0) (1-\chi _{R_0}(x)) U(x). \end{aligned}$$

Hence,

$$\begin{aligned} \kappa _\alpha \int _{\mathbf {R}^N} (1-\chi _{R_0}) \left[ - f(x,u) + (1-\delta _0) v \right] U_- \mathrm {d}x \ge - (1-\delta _0) \kappa _\alpha \Vert U_- \Vert _{L^2}^2. \end{aligned}$$

(6)

Next, by the Plancherel theorem and Proposition 0.2, one sees

$$\begin{aligned} \kappa _\alpha \Vert U_- \Vert ^2_{L^2} = \kappa _\alpha \int _{\mathbf {R}^N} |\widehat{U_-} |^2 \mathrm {d}\xi \le \kappa _\alpha \Vert U_- \Vert _\alpha ^2 \le \Vert (EU)_- \Vert _{X^\alpha }^2. \end{aligned}$$

Combining this with (4)–(6), we finally obtain

$$\begin{aligned} - \Vert (EU)_- \Vert _{X^\alpha }^2 \ge - (1-\delta _0) \Vert (EU)_- \Vert _{X^\alpha }^2, \end{aligned}$$

which implies $(EU)_- \equiv 0$, hence, $U_- \equiv 0$ and $u(x) \le v(x)$.

For the opposite inequality $-v(x) \le u(x)$, we proceed similarly. Set $V(x) := v(x) + u(x)$. Then we have

$$\begin{aligned} - \Vert (EV)_- \Vert _{X^\alpha }^2= & {} \kappa _\alpha \left\langle V, V_- \right\rangle _\alpha \\\ge & {} \kappa _\alpha \int _{\mathbf {R}^N} (1-\chi _{R_0} (x) ) \left[ f(x,u(x)) + (1-\delta _0) v(x) \right] V_-(x) \mathrm {d}x. \end{aligned}$$

If $V_-(x) > 0$ and $|x| \ge R_0$, then $u(x)< -v (x) < 0$. Thus, by (4), one gets

$$\begin{aligned}&(1-\chi _{R_0} (x) ) \left[ f(x,u(x)) + (1-\delta _0) v(x) \right] \\&\quad \ge (1-\chi _{R_0} (x) ) \left[ (1-\delta _0) u(x) + (1-\delta _0) v(x) \right] \\&\quad = (1-\delta _0) (1-\chi _{R_0}(x))V(x). \end{aligned}$$

The rest of the argument is same as the above and we obtain $V_- \equiv 0$. Thus, $- v(x) \le u(x)$ holds and we complete the proof. $\square$