  Let $H=-D^2+V$ be a Schr\"odinger operator on $ L^2(\mathbb{R})$, or on $
L^2(0,\infty)$. Suppose the potential satisfies $\limsup_{x\to
\infty}|xV(x)|=a<\infty$. We prove that $H$ admits no eigenvalue larger than $
\frac{4a^2}{\pi^2}$. For any positive $a$ and $\lambda$ with $0<\lambda<
\frac{4a^2}{\pi^2}$, we construct potentials $V$ such that $\limsup_{x\to
\infty}|xV(x)|=a $ and the associated Sch\"rodinger operator $H=-D^2+V$ has
eigenvalue $\lambda$.
