This question may not be at the research level, but it has really bothered me for a long time. The following space is used for handling initial boundary value problem for first order hyperbolic equation.

I am reading Multidimensional Hyperbolic Partial Differential Equations written by S.Benzoni-Gavage and D.Serre. There is a statement about equivalence of weighted Sobolev space norm in Page 240-241. I will first show this statement below.

For function $u=u(x),$ where $x=(t,x_1,\cdots,x_{n-1}) \in \mathbb{R}^n,$ and for parameter $\gamma \geq 1,$ and for $s\geq 1$ is a positive integer, we define the following norm $$ \| u \|_{H^s_{\gamma}}^2:=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} (\gamma^2+|\xi|^2)^s |\hat{u}(\xi)|^2 d\xi. $$ Here $\hat{u}$ denotes the Fourier transform of $u$, and $\xi$ is the dual variable of $x$. The norm $\| \cdot \|_{H^s_{\gamma}}$ is clearly the normal Sobolev norm of $H^s$ when $\gamma=1.$

Now we want to define a norm of the space $\mathcal{H}_\gamma^s:=\{u:e^{-\gamma t} u \in H_\gamma^s\}. $ A direct idea is to let $$ \| u\|_{\mathcal{H}_\gamma^s}^2=\| e^{-\gamma t} u\|_{H_\gamma^s}^2. $$ The authors here state that since we note that $$ \| u\|_{H_\gamma^s}^2= \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} (\gamma^2+|\xi|^2)^s |\hat{u}(\xi)|^2 d\xi\approx \sum_{|\alpha|\leq s} \gamma^{2(s-|\alpha|)} \| \partial^\alpha_{t,x} u\|_{L^2}^2. $$ The $\approx$ here stands for two-sided inequalities with constants independent of $\gamma$ and $u.$(It is clearly true just expanding the binomial $(\gamma^2+|\xi|^2)^s$ and using property of Fourier transform.) Then they say that the following relation holds true: $$ \| e^{-\gamma t}u\|_{H_\gamma^s}^2 \approx \sum_{|\alpha|\leq s} \gamma^{2(s-|\alpha|)} \| e^{-\gamma t}\partial^\alpha_{t,x} u\|_{L^2}^2. $$ But I don't think the above equality holds since both $e^{-\gamma t}$ and $u$ depend on the variable $t$. To be exact, let's denote the dual variable $\xi$ by $(\sigma,\eta)$ where $\sigma \in \mathbb{R}$ is the dual variable of $t$, and $\eta\in \mathbb{R}^{n-1}$ is the dual variable of $(x_1,\cdots,x_{n-1}).$ Since we have $$ (\gamma^2+\sigma^2+|\xi|^2)^s \approx \sum_{l=0}^s (\gamma^2+\sigma^2)^l |\xi|^{2(s-l)} \approx \sum_{l=0}^s |\gamma+i\sigma|^{2l} \sum_{|\beta|=s-l}|\xi^\beta |^2 =\sum_{l+|\beta|=s} |(\gamma+i\sigma)^l\xi^{\beta}|^2. $$ All the $\approx$ notations are independent of $\gamma,\sigma,\eta.$

Therefore \begin{equation*} \begin{aligned} \| e^{-\gamma t} u\|_{H_\gamma^s}=&\frac{1}{(2\pi)^{n}}\int (\gamma^2+\sigma^2+|\xi|^2)^s|\mathcal{F}(e^{-\gamma t}u)(\sigma, \xi)|^2 d\sigma d\xi \\\approx& \frac{1}{(2\pi)^{n}}\sum_{l+|\beta|=s} \int |(\gamma+i\sigma)^l\xi^{\beta}\mathcal{F}(e^{-\gamma t}u)(\sigma, \xi)|^2 d\sigma d\xi \\ =&\frac{1}{(2\pi)^{n}}\sum_{l+|\beta|=s} \int |\mathcal{F}(e^{-\gamma t}\partial_t^l \partial _x^\beta u)(\sigma, \xi)|^2 d\sigma d\xi \\ =& \sum_{|\alpha|=s} \| e^{-\gamma t}\partial^\alpha_{t,x} u\|_{L^2}. \end{aligned} \end{equation*} In the above calculation, we have used the fact that $$ \mathcal{F}(e^{-\gamma t}\partial_t^l u)=(-1)^l(\gamma+i\sigma)^l\mathcal{F}(e^{-\gamma t}u). $$ Hence, an equivalent norm of the weighted Sobolev space $\mathcal{H}_\gamma^s=e^{\gamma t}H^s_\gamma$ should be $$ \sum_{|\alpha|=s} \| e^{-\gamma t}\partial^\alpha_{t,x} u\|_{L^2} $$ instead of $$ \sum_{|\alpha|\leq s} \gamma^{2(s-|\alpha|)} \| e^{-\gamma t}\partial^\alpha_{t,x} u\|_{L^2}^2. $$ Am I right?

Remark: The definition of norm of $\mathcal{H}_\gamma^s$ is quite important. In Page 251-254, the authors spend lots of time discussing estimate in $\mathcal{H}_\gamma^s$ under their norm $\sum_{|\alpha|\leq s} \gamma^{2(s-|\alpha|)} \| e^{-\gamma t}\partial^\alpha_{t,x} u\|_{L^2}^2$ via some techniques in Harmonic analysis, but I think the estimate could be more easily deduced as in J.Rauch's old paper" $\mathcal{L}^2$ is a Continuable Initial Condition for Kreiss' Mixed Problems" (see the discussions in the beginning of Section 1 in this paper) under the norm $\sum_{|\alpha|=s} \| e^{-\gamma t}\partial^\alpha_{t,x} u\|_{L^2}.$


1 Answer 1


I am Denis Serre, one of the authors, and am happy to give you an answer. My impression is that everything is OK. The matter is to see the equivalence between the expressions $$A:=\sum_{|\alpha|\le s}\gamma^{2(s-|\alpha|)}\left|\nabla^\alpha(e^{-\gamma t}u)\right|^2$$ and $$B:=\sum_{|\alpha|\le s}\gamma^{2(s-|\alpha|)}\left|e^{-\gamma t}\nabla^\alpha u\right|^2.$$

Let us just consider the first non-trivial case $s=1$. Then \begin{eqnarray} A & = & \gamma^{2s}\left|e^{-\gamma t}u\right|^2+\gamma^{2s-2}\left|\nabla(e^{-\gamma t}u\right|^2 \\ & = & \gamma^{2s}\left|e^{-\gamma t}u\right|^2+\gamma^{2s-2}\left(\left|e^{-\gamma t}\nabla_yu\right|^2 +e^{-2\gamma t}(\partial_tu-\gamma u)^2\right). \end{eqnarray} That $A\sim B$ follows from the equivalence of quadratic forms (in $(p,Q,r)$) $$\gamma^{2s}p^2+\gamma^{2s-2}(|Q|^2+(r-\gamma p)^2)\sim \gamma^{2s} p^2+\gamma^{2s-2}(|Q|^2+r^2),$$ and this equivalence s uniform in $\gamma$.

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    $\begingroup$ I am very honored to receive the answer from the author himself. Thank you so much! $\endgroup$ Sep 14, 2023 at 13:30

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