I post this question in [MSE](https://math.stackexchange.com/questions/1629349/the-inverse-of-laplacian-for-different-orders) couple of days before and get no response. So I repost it here for better luck. Thank you!




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Let $u,v\in C\_c^\infty(\Omega)$ and $\Omega\subset \mathbb R^N$ is open bounded with smooth boundary. Let $\Delta$ denote the Laplacian operator, $I$ denotes the identity operator and $t\in\mathbb R^+$ is a positive real number.


Let 
$$
f(t):=\|(I-t\Delta)^{-1}\nabla u\|\_{L^2}^2-\|(I-t\Delta)^{-\frac32}\nabla v\|\_{L^2}^2.
$$
It is given that $\|\nabla u\|\_{L^2}^2<\|\nabla v\|\_{L^2}^2$, that is, $f(0)<0$; and I know that there exists $t\_0>0$ such that $f(t\_0)=0$. I also know that 
$$
\int u = \int v.
$$



> 
> My question: Can I prove that $f(t)<0$ for $0<t<t\_0$ and $f(t)>0$ for
>  $t>t\_0$?
> 
> 
> 




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My try: I compute that 
$$
\frac{d}{dt}(\|(I-t\Delta)^{-1}\nabla u\|\_{L^2}^2)=-\|(I-t\Delta)^{-\frac32}\Delta u\|\_{L^2}^2<0
$$
so I know $\|(I-t\Delta)^{-1} \nabla u\|\_{L^2}^2$ is decreasing as $t$ increasing, so is $\|(I-t\Delta)^{-\frac32} \nabla v\|\_{L^2}^2$. But I can't prove that the later one decreasing faster... I guess the order $3/2$ would do sth but I am not sure...




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Also, I am wondering that how may I write



> 
> $$ \|(I-t\Delta)^{-s}u\|\_{L^2}^2=\sum\_{k=0}?? $$ where $t$ and $s$ are
>  real numbers, based on Fourier transform.
> 
> 
> 


I was trying to look for Bessel potential but had no luck...


Please advise!



The inverse of Laplacian operator for different orders
**Disclaimer**


The below argument applies directly to the previous version of the question. For the current version the basic argument still hold, using that 
$$ \| \nabla u\|\_2 = \langle u, -\Delta u\rangle $$
and so behaves nicely for eigenfunctions of the Laplacian. 


To make the argument also work for $C^\infty\_c$ functions simply apply a cut-off near the boundary. This is a small perturbation and so you just need to add some small $\epsilon$ terms to the whole thing. 


The added requirement that $\int u = \int v$ is of no consequence since the constructed example below has no zero-frequency component, and that the insertion of $\nabla$ in the new version of the question kills this component anyway.


**End disclaimer**


With the given information it seems unlikely. 


Let $b,g,h$ be three eigenfunctions of $-\Delta$ with eigenvalues $\lambda\_b \gg \lambda\_g \gg \lambda\_h$, normalized so $\|b\|\_2 = \|g\|\_2 = \|h\|\_2 = 1$. Let $\tau\_0 = 0$, $\tau\_1 = 1/\lambda\_g$, and $\tau\_2 = 1 / \lambda\_h$. 


Let $u = a\_g g$ and $v = b + a\_h h$ where $a\_g$ and $a\_h$ are scalars. We have that 
$$ \begin{align}
f(\tau\_0) &= a\_g^2 - (1 + a\_h^2) \\
f(\tau\_1) &= \frac{a\_g^2}{4} - \frac{1}{(1 + \frac{\lambda\_b}{\lambda\_g})^3} - \frac{a\_h^2}{(1 + \frac{\lambda\_h}{\lambda\_g})^3} > \frac{a\_g^2}{4} - \frac{\lambda\_g^3}{\lambda\_b^3} - a\_h^2 \\
f(\tau\_2) &= - \frac{a\_h^2}{8} - \frac{1}{(1 + \frac{\lambda\_b}{\lambda\_h})^3} + \frac{a\_g^2}{(1 + \frac{\lambda\_g}{\lambda\_h})^2} < - \frac{a\_h^2}{8} + \frac{a\_g^2 \lambda\_h^2}{\lambda\_g^2} 
\end{align}$$
It is clear that for very large $t$, $f(t) > 0$ (since $t^{-1}$ decays slower than $t^{-3/2}$). So to produce an almost counterexample (only thing missing is that $u$ and $v$ are not in $C^\infty\_0(\Omega)$), we just need to arrange


$$\begin{align}
a\_g^2 &< a\_h^2 + 1 \\
a\_g^2 & \geq 4 a\_h^2 + \frac{4 \lambda\_g^3}{\lambda\_b^3} \\
a\_g^2 & \leq \frac{\lambda\_g^2}{8 \lambda\_h^2} a\_h^2 
\end{align} $$


This can be satisfied by, for example, $a\_h^2 = 1/6$ and $a\_g^2 = 1$ provided the ratio of eigenvalues $24\lambda\_g^3 < \lambda\_b^3$ and $48 \lambda\_h^2 < \lambda\_g^2$. 


