On the hyperbolic space $\mathbb{H}^n$, it is known that the spectrum of the Laplacian satisfies $\text{Spec}(\Delta) \subset [\frac{(n  1)^2}{4}, \infty)$. Consider the operator $P = \Delta + a$, where $a >  \frac{(n  1)^2}{4}$. Consider the norm $\Vert .\Vert$ defined by $\Vert u\Vert^2 = (Pu, u)$, where $(u, v)$ is the usual $L^2(\mathbb{H}^n)$ inner product. Is $\Vert u\Vert \simeq \Vert u\Vert_{H^1}$?
closed as offtopic by Deane Yang, Joonas Ilmavirta, Willie Wong, Alex Degtyarev, Stefan Kohl Jun 3 '15 at 8:37
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3$\begingroup$ Am I missing something, or isn't this just integration by parts? $\endgroup$ – Nate Eldredge Jun 3 '15 at 1:27

$\begingroup$ @NateEldredge Suppose $a = 1/8$. Then $(Pu, u) = (\Delta u  1/8 u, u) = \Vert \nabla u\Vert^2  1/8\Vert u\Vert^2$. That negative $1/8$ part is throwing me off. $\endgroup$ – rook Jun 3 '15 at 1:54
Let's call your new norm $\\cdot\_a$ and reserve $\\cdot\$ for the usual $L^2$ norm. Integration by parts shows us $\u\_a^2 = \\nabla u\^2 + a\u\^2$. If $a > 0$ then this is easy, so let $\lambda = \frac{(n1)^2}{4}$ be the bottom of the spectrum of $\Delta$ and suppose $\lambda < a \le 0$. In this case $\u\^2_a \le \u\^2_{H^1}$ is obvious.
For the other inequality, the operator $\Delta  \lambda$ is positive definite, so integrating by parts gives us the inequality $\\nabla u\^2 \ge \lambda \u\^2$. Set $r = 1 + \frac{a}{\lambda} > 0$. Then $$\begin{align*} \u\_a^2 &= \\nabla u\^2 + a\u\^2 \\ &= r \\nabla u\^2  \frac{a}{\lambda}(\\nabla u\^2  \lambda \u\^2) \\ &\ge r \\nabla u\^2 \\ &\ge \frac{r}{2} \\nabla u\^2 +\frac{r \lambda}{2} \u\^2 \\ &\ge C \u\^2_{H^1} \end{align*}$$ where $C = \min(\frac{r}{2}, \frac{r\lambda}{2})$.