Good evening,
In the paper "Asymptotic analysis of the linearized Navier-Stokes equations in a channel" (Differential Integral Equations 8, No. 7(1995), 1591-1618) the authors define a cut-off function as follows:
$\rho \in C^{\infty}\left[0, \infty\right), \rho(0) = 1, \text{supp}\space \rho \subset \left[0,1/2\right]$
$\rho^{\eta}\left(y\right) = \rho\left(y/\eta^{1/2}\right), \space y \geq 0 $
$\widetilde{\rho}^\eta = \rho^\eta(y) + \rho^\eta(1-y), \space 0 \leq y \leq 1$
They state some straightforward estimates in the following in the $L^p$ Norm. Afterwards, however, they postulate an estimate in the $H^{-1}$ Norm for $\eta < 1$:
$\left|\widetilde{\rho}^\eta\right|_{H^{-1}\left(0,1\right)} \leq 2 \eta^{3/4} \left|\rho\right|_{L^2\left(\Omega\right)}$
I am not an expert in functional analysis, but isnt $H^{-1}$ the dual space of $H^1_0$? So why does that statement even make sense? The cut-off function is a function and not a functional. So how is the pairing defined? Do they mean the $L^2$ scalar product in that case? If yes, I still dont know how they arrive at that estimate since I only get $\eta^{1/4}$ by using Cauchy-Schwartz and an integral transformation.
Thank you for your help!
I don't have access to the paper, and I'm not quite confident about I'm to write, but I hope it could help you. I think that the pairing you're referring to is $v\mapsto \int uv\,dt$, for $v\in H^1_0$, where $u$ is a function, in this case $u=\tilde{\rho}^\eta$, or for simplicity we can take $u=\rho^\eta$.
Suppose $v$ is smooth, since a limiting argument prove the general statement. Since $v$ is compactly supported in $(0,1)$ we have
$\int \rho^\eta v\,dt = -\int(\int_t^1\rho(\frac{x}{\eta^{1/2}})\,dx)'v\,dt=\int(\int_t^1\rho(\frac{x}{\eta^{1/2}})\,dx)v'\,dt\le \lVert \int_t^1\rho(\frac{x}{\eta^{1/2}})\,dx\rVert_2\lVert v\rVert_{H_0^1}$.
The $L^2$-norm of $\int_t^1\rho(\frac{x}{\eta^{1/2}})\,dx$, which equals $\eta^{1/2}\int_{t/\eta^{1/2}}^1 \rho(x)\,dx$ for $t<\eta^{1/2}$ and zero otherwise, is
$\eta^{1/2}\Big[\int_0^{\eta^{1/2}}(\int_{t/\eta^{1/2}}^1 \rho(x)\,dx)^2\,dt\Big]^{1/2}=\eta^{3/4}\Big[\int_0^1(\int_t^1 \rho(x)\,dx)^2\,dt\Big]^{1/2}\le \frac{1}{\sqrt{2}}\eta^{3/4}\lVert\rho\rVert_2$.
The other side of $\tilde{\rho}^\eta$ is treated similarly. The term $1/\sqrt{2}$ is superfluous, the autor drops it from the inequality.