# Norm equivalent to Sobolev norm? [closed]

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 off-topic by Deane Yang, Joonas Ilmavirta, Willie Wong, Alex Degtyarev, Stefan KohlJun 3 '15 at 8:37

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• Am I missing something, or isn't this just integration by parts? – Nate Eldredge Jun 3 '15 at 1:27
• @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. – 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 $\lambda = \frac{(n-1)^2}{4}$ 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 $C = \min(\frac{r}{2}, \frac{r\lambda}{2})$.