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 $\sum_{|\alpha|=s} \| e^{-\gamma t}\partial^\alpha_{t,x} u\|_{L^2}.$