# What is $\left\| u \right\|_{ H_{0}^{k}, H_{0}^{k}}$ norm when $H_{0}^{k}=\left\{u \in H^{k, 2}(M) \mid \int_{M} u \operatorname{vol}_{g}=0\right\}$

What is $$\left\| f \right\|_{ H_{0}^{k}, H_{0}^{k}}$$ norm when $$H_{0}^{k}=\left\{u \in H^{k, 2}(M) \mid \int_{M} u \operatorname{vol}_{g}=0\right\}$$.

I'm reading a paper Chern-Yamabe flow which said that

Now for $$k$$ big enough $$(k>n)$$, the first eigenvalue of the operator $$-\Delta$$ on the space $$H_{0}^{k}=\left\{u \in H^{k, 2}(M) \mid \int_{M} u \mathrm{vol}_{g}=0\right\}$$ is strictly negative hence $$\left\|e^{-t \Delta}\right\|_{H_{0}^{k}, H_{0}^{k}} \leq$$ $$C_{0} e^{-c t}$$.

I have no idea what is $$\left\| - \right\|_{ H_{0}^{k}, H_{0}^{k}}$$ norm.

• Taking $K:f\to (\Delta)^{-1}f$ as a (compact) operator between $H^k_0$ and $H^k_0$, it is the usual linear operator norm, perhaps? Jun 27 at 7:16
• @username given the context, your interpretation is almost certainly the correct one (that $\| e^{-t\Delta}\|_{H_0^k,H_0^k}$ is the operator norm of $e^{-t\Delta}$ as a mapping from $H^k_0$ to itself). Can you post it as an answer? Jun 27 at 14:48
• It seems that the $\Delta$ in this paper is $-\Delta$. I add the link in the question. Jun 27 at 14:58
• Oof, I don't like this notation. Normally for operator norms I would write $\| - \|_{X\to Y}$ because the one they chose, $\| - \|_{X,Y}$, does not make clear whether they want it to be the operator norm of a mapping from $X\to Y$ or from $Y\to X$. This becomes a bit of an issue later on when they write $\| e^{-t\Delta} \|_{H^k_0, H^0_0}$. At a quick glance it is not obvious to me which direction they meant. Jun 27 at 19:13

A partial answer to your question. I would think it means: $$\sup\bigg(\| u(t,\cdot)\|_{H^k_0} : u(t,\cdot)=\exp(-t\Delta) u(0,\cdot),\textrm{ with } \| u(0,\cdot)\|_{H^k_0}=1\bigg).$$