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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!

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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.

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  • $\begingroup$ Thanks for your answer! Just to clarify: Since the cut-off function is just a special case of a L2 function, i can always identify it's functional via Riesz-Fischer by the L2 scalar product? Could you elaborate a bit on your last estimate? By using Cauchy-Schwartz, setting t=0 in the inner integral and integrating in the outer from 0 to 1/2 (since the inner integral is 0 for t>1/2), the best estimate i can get is or order $\left\Vert \rho \right\Vert_{L^2}^2$ $\endgroup$ – Stefan Litzel Dec 18 '17 at 14:25
  • $\begingroup$ As you mentioned, there is still room for improvements of the constant, but it seems that it doesn't matter for the purposes of the article. In the very last part I did: $\int_0^1(\int_t^1\rho\,dx)^2\,dt\le |\rho|_2^2\int_0^1\int_t^1\,dxdt= \frac{1}{2}|\rho|_2^2$. Concerning your first question, $v\mapsto\int uv$ is the definition of the functional attached to $u$, it doesn't come from Riesz-Thorin. Even a function $u\in L^1$ defines an element in $H^{-1}$ with this pairing. $\endgroup$ – user90189 Dec 19 '17 at 16:20
  • $\begingroup$ Other way you can get this kind of estimates is by Fourier transform. Notice that $|\int u\bar{v}|=|\sum \hat{u}(n)\hat{\bar{v}}(n)|=|\sum \hat{u}(1+|n|^2)^{-1/2}(1+|n|^2)^{1/2}\hat{\bar{v}}|\le\Big( \sum |\hat{u}|^2(1+|n|^2)^{-1}\Big)^{1/2}\Big(\sum|\hat{v}|^2 (1+|n|^2)\Big)^{1/2}\le \Big( \sum |\hat{u}|^2(1+|n|^2)^{-1}\Big)^{1/2}|v|_{H_0^1}$. $\endgroup$ – user90189 Dec 19 '17 at 16:27

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