Suppose that $u\in C^2(\mathbb R^3)$ is a bounded solution of the eigenvalue problem below:$$\Delta u=u,\ x\in\mathbb R^3.$$ Here $\Delta$ denotes the Laplacian. Please show that $u$ is nothing but zero.
In the Sobolev space $H^1(\mathbb R^3),$ we consider the functional $I:H^1(\mathbb R^3)\to\mathbb R$ which is given by $$I(v)=\frac{1}{2}\int_{\mathbb R^3}|\nabla v|^2+v^2dx=\frac{1}{2}\|v\|_{H^1(\mathbb R^3)}^2.$$ Then the functional $I$ possesses a unique critical point $u_0=0$ in $H^1(\mathbb R^3).$ So if the solution $u$ of the eigenvalue problem above belongs to $H^1(\mathbb R^3),$ then $u=0.$ I tried a lot but failed on this.
Actually, once $u\in L^2(\mathbb R^3),$ then we can also obtain that $u=0.$ Consider the function $w=u^2:\mathbb R^3\to\mathbb R,$ it is clear that $w$ is subharmonic in $\mathbb R^3.$ So we can get the desired result by using the mean-value ineqaulity to $w.$ But I also failed to verify that $u\in L^2(\mathbb R^3).$
Hope someone could give me some hints. Thanks!
At present, I believe I have obtained the desired result by the comparison principle.
Let $v:\mathbb R^3\setminus\{0\}\to\mathbb R$ be defined by $$v(x)=|x|^4+10^{2017}|x|^{-1},x\in\mathbb R^3\setminus\{0\},$$then $v\to\infty,~\hbox{as $x\to0$ or $x\to\infty$.} $ For any $\epsilon>0,$ since $u$ is bounded in $\mathbb R^3,$ a direct calculation yields that there exists some $R_\epsilon>0$ such that \begin{align*}Lu&=-\Delta u+u=0\leq L(\epsilon v),~\hbox{in}~\Omega_\epsilon,\\u&\leq\epsilon v,~\hbox{on the boundary of}~\Omega_\epsilon,\end{align*} here we denote by $\Omega_\epsilon$ the set $$\{x\in\mathbb R^3;|x|\in(1/R_\epsilon,R_\epsilon)\}.$$ Thanks to the comparison principle, we have $$u\leq\epsilon v,\ \hbox{on the closure of}~\Omega_\epsilon.$$ Let $\epsilon\to0+,$ we note that $R_\epsilon\to\infty,$ so we obtain $u(x)\leq 0$ for any $x\in\mathbb R^3\setminus\{0\}.$ Thus $u$ is non-positive in $\mathbb R^3.$ So is $-u$, thus $u$ is zero in the whole $\mathbb R^3.$