The inverse of Laplacian operator for different orders

Question

I post this question in MSE 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!

Answer 1

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 $48 \lambda_h^2 < \lambda_g^2$.