I post this question in MSE couple of days before and get no response. So I repost it here for better luck. Thank you!

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 $t>t_0$?

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...

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!

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