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Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$ with smooth boundary $\partial \Omega$ and let $\nu$ denote the outer unit normal. Let $u$ be an eigenfunction of $-\Delta$ in $\Omega$ satisfying $\frac{\partial u}{\partial \nu} - u = 0$ on $\partial \Omega$. Must the associated eigenvalue for $u$ be negative?

Since $u$ is an eigenfunction of $-\Delta$ and satisfies the above boundary condition on $\partial \Omega$, integration by parts gives that $$\lambda \int_{\Omega}u^{2}\, dx = \int_{\Omega}|\nabla u|^{2}\, dx - \int_{\partial \Omega}u^{2}\, d\sigma.$$ However, it is not immediate that the right hand side is always $< 0$.

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In one dimension, if we set $u(x)=\sin(x)$ on $(a_-,a_+)$, we have $-\Delta u=u$ (eigenvalue $+1$) and $$ \partial_\nu u(a_\pm)-u(a_\pm) = \pm u'(a_\pm)-u(a_\pm) = \pm\cos(a_\pm)-\sin(a_\pm) = -\sqrt2\sin(a_\pm\mp\pi/4). $$ We can now choose the endpoints so that this vanishes: for example $a_\pm=\pm\pi/4$.

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