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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.

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  • $\begingroup$ 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? $\endgroup$
    – username
    Commented Jun 27, 2022 at 7:16
  • $\begingroup$ @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? $\endgroup$
    – Willie Wong
    Commented Jun 27, 2022 at 14:48
  • $\begingroup$ It seems that the $\Delta$ in this paper is $-\Delta$. I add the link in the question. $\endgroup$
    – Elio Li
    Commented Jun 27, 2022 at 14:58
  • $\begingroup$ 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. $\endgroup$
    – Willie Wong
    Commented Jun 27, 2022 at 19:13

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

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