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Suppose $\Omega_s \subset \mathbb{R}^n$ is a compact subset for each $s \in [0,T]$. I have a linear operator $$p_t^s:H^1(\Omega_t) \to H^1(\Omega_s)$$ which maps functions on $\Omega_t$ to functions on $\Omega_s$, and $p_t^s$ is continuous with continuous inverse $p_s^t$. It does this via a diffeomorphism $P_t^s:\Omega_s \to \Omega_t$ with $p_t^s f = f \circ P_t^s$.

Question We can define (the Levi-Civita connection) a modified gradient on $\Omega_t$ by $$\nabla_{\Omega_t} f := (\nabla f)^T := \nabla f - (\nabla f \cdot N)N$$ where the superscript $T$ denotes projection onto the tangent space of $\Omega_t$ and $N$ is the normal vector on $\Omega_t$. My question is for functions $g \in C^1(\Omega_s)$, is it true that $$\nabla_{\Omega_t} (fg) =g\nabla_{\Omega_t} (f)?$$ So can I take it out as a constant?

Essentially, I want to use this condition in an integral over $\Omega_t$. I tried writing everything out in coordinates (eg. $x$ on $\Omega_t$ and $y$ on $\Omega_s$) and used the diffeomorphism but I am very confused.

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up vote 1 down vote accepted

The answer is no. For one thing, with $t=s$ you would be contradicting the Leibniz rule (unless $g$ is indeed constant). Even if you insist on $t\ne s$, I can consider the case of all $P_t^s$ being the identity map (and all compact subsets being the same), or translations, etc.

If you are trying to understand an argument in a paper you are reading, I recommend adding a reference to your question.

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Thanks for your answer. I think you're right, but when one is given, with no other knowledge, a $g \in C^1(\Omega_s)$ and a linear operator $T:H^1(\Omega_t) \to L^2(\Omega_t)$ then doesn't it seem natural that you can write $Tfg = gTf$. – HarryP Dec 27 '12 at 21:56
This question doesn't arise from any literature -- I have an integral that would simplify a lot with this "result". – HarryP Dec 27 '12 at 21:59
@HarryP It doesn't. "Linear" in the context of operators between vector spaces means "Linear over the base field" (here it's $\mathbb R$). Which means $T(\alpha f)=\alpha Tf$ for all $\alpha\in \mathbb R$ (plus additivity)... The differentiation operator $Tf=f'$ satisfies $T(fg)=gTf+fTg$, the Leibniz (product) rule that we know and love from calculus. – user53153 Dec 27 '12 at 22:01
OK, but isn't it the case that for $\Omega_s$ we use coordinates $y_i$, say, and for $\Omega_t$ we use coordinates $x_i$. So for example $g(y) = y_1 + y_2$. So the $y_i$ are just like real numbers to the operator $T$ which concerns itself with the $x_i$. $T$ doesn't need to know that there's a diffeomorphism that maps the $y$ to $x$. – HarryP Dec 27 '12 at 22:10
@Harry In order to multiply $f$ by $g$, you first have to pull $g$ to $\Omega_t$ via composition with some map $\phi:\Omega_t\to\Omega_s$. We cannot multiply functions defined on different sets. So now the product is not $f(x)g(y)$ but rather $f(x)g(\phi(x))$. – user53153 Dec 27 '12 at 23:39

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