A question on nontrivial solution of ODE

Question

It is well known that there exists no non-trivial bounded solution of $-u''+u=0$ in $\mathbb R.$ Is this result even true, the problem $$ \bigg(-\frac{d^2}{dx^2}\bigg)^{s} u+u=0 $$ has no bounded solution in $\mathbb R$ where $s\in (0, 1).$ Here $$\bigg(-\frac{d^2}{dx^2}\bigg)^{s} u(x)=c\int_{\mathbb R}\frac{u(x)-u(y)}{|x-y|^{1+2s}}dy$$ where $c$ is positive constant depending on $s.$

Answer 1

Yes, the result is true, in arbitrary dimension $d$. Once I needed this and I could not find a reference; the following is the relevant part of the proof of Proposition 2.1 in my paper Ten equivalent definitions of the fractional Laplace operator (DOI:10.1515/fca-2017-0002).

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Suppose that $u$ is a bounded, non-zero solution of $$(-\Delta)^s u + u = 0.$$ With no loss of generality we may assume that $\|u\|_\infty = 1$, and that in fact $\sup \{f(x) : x \in \mathbb{R}^d\} = 1$. Fix $x_1$ such that $f(x_1) > \tfrac{1}{2}$, and choose a smooth, compactly supported $v$ such that $$v(x_1) = \tfrac{1}{2}, \qquad \|v\|_\infty = \tfrac{1}{2} , \qquad \|(-\Delta)^s v\|_\infty \leqslant \tfrac{1}{2}.$$ Then $u + v$ is a continuous function such that $$\limsup_{|x| \to \infty} |u(x) + v(x)| = \limsup_{|x| \to \infty} |u(x) + v(x)| \leqslant 1$$ and $u(x_1) + v(x_1) > 1$. This means that $u + v$ attains a global maximum at some point $x_2$, and $u(x_2) + v(x_2) > 1$. By the positive maximum principle, we have $$ \begin{aligned} 0 & \leqslant (-\Delta)^s (u + v)(x_2) \\ & = (-\Delta)^s u (x_2) + (-\Delta)^s v(x_2) \\ & = -u(x_2) + (-\Delta)^s v(x_2) \\ & \leqslant -u(x_2) + \tfrac{1}{2} \\ & < (v(x_2) - 1) + \tfrac{1}{2} \\ & \leqslant (\tfrac{1}{2} - 1) + \tfrac{1}{2} = 0 ,\end{aligned} $$ a contradiction.