Skip to main content
MathOverflow

Return to Question

Post Closed as "Not suitable for this site" by Deane Yang, Joonas Ilmavirta, Willie Wong, Alex Degtyarev, Stefan Kohl♦
edited body
Source Link
rook
rook
  • 11
  • 2

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$$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 $\Vert u\Vert \simeq \Vert u\Vert_{H^1}$?

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 $\Vert u\Vert \simeq \Vert u\Vert_{H^1}$?

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 $\Vert u\Vert \simeq \Vert u\Vert_{H^1}$?

Source Link
rook
rook
  • 11
  • 2

Norm equivalent to Sobolev norm?

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 $\Vert u\Vert \simeq \Vert u\Vert_{H^1}$?