I am considering the following equation:
$$ f(t):=\int_\Omega[(I-t\Delta)^{-1}\Delta(I-t\Delta)^{-1}u]\cdot u\,dx $$ where $u\in C_c^\infty(\Omega)$ and $t\geq 0$ a real number. $I$ is the identity operator and by $-1$ we denote the inverse of operator. For example, if $v=\Delta w$, we may write $w=\Delta^{-1}v$.
I am trying to study the positiveness of $f$. Clearly, when $t=0$, I have $$ f(0)=\int_\Omega\Delta u\cdot u = -\int_\Omega |\nabla u|^2<0 $$ My question: I wish to prove $f(t)\leq 0$ for $t>0$ and $\lim_{t\to\infty}f(t)=0$ as well. I am confused at as $t$ gets larger, how $(I-t\Delta)^{-1}u$ changes?
Update:
We have \begin{align*} f(t)&=\int_\Omega[(I-t\Delta)^{-1}\Delta(I-t\Delta)^{-1}u]\cdot u=\\ &=\int_\Omega(\Delta[(I-t\Delta)^{-1}(I-t\Delta)^{-1}u])\cdot u\\ &=-\int_\Omega(\nabla[(I-t\Delta)^{-1}(I-t\Delta)^{-1}u])\cdot\nabla u\\ &=-\int_\Omega((I-t\Delta)^{-1}(I-t\Delta)^{-1}\nabla u])\cdot\nabla u \end{align*}
But why I have $$ \int_\Omega((I-t\Delta)^{-1}(I-t\Delta)^{-1}\nabla u])\cdot\nabla u = \int_\Omega((I-t\Delta)^{-1}\nabla u])\cdot((I-t\Delta)^{-1}\nabla u) \tag 1 $$ I am not sure I can move $(I-t\Delta)^{-1}$ like this.
To answer your edit: once you have found the correct domain of definition for the Laplacian, you have that $(I-t\Delta)^{-1} = \sum \limits _{k=0} ^\infty t^k \Delta^k$. Since $\int (\Delta u) v = \int u (\Delta v)$ (on test functions, at least), you may use induction and prove that $\int (\Delta^k u) v = \int u (\Delta^k v)$, so $\int (\sum \limits _{k=0} ^\infty t^k \Delta^k u) v = \int u (\sum \limits _{k=0} ^\infty t^k \Delta^k v)$, which means $\int [(I-t\Delta)^{-1} u] v = \int u [(I-t\Delta)^{-1} v]$.