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On page 325 of Stein's Functional Analysis, he writes "We consider the following vector field

$$ L = \frac{1}{i\lambda} \sum_{k=1}^d a_k \frac{\partial}{\partial x_k} = \frac{1}{i\lambda}(a \cdot \nabla) $$

... then the transpose $L^t$ of $L$ is given by

$$ L^t(f) = -\frac{1}{i\lambda} \sum_{k=1}^d \frac{\partial}{\partial x_k} (a_k f) = -\frac{1}{i\lambda} \nabla \cdot (af) $$

..." (Here, $f$ is a function $\mathbb R^d \to \mathbb R$, and $a$ is a function $\mathbb R^d \to \mathbb R^d$.)

Can someone explain to me where the expression for the transpose comes from?

If I understand correctly, $L^t$ and $L$ are related by

$$ \int_{\mathbb R^d} L(f) g = \int_{\mathbb R^d} f L^t(g) $$

In an attempt to show that the two sides are equal, I made the calculation

$$ \int_{\mathbb R^d} L(f) g - \int_{\mathbb R^d} f L^t(g) = \int_{\mathbb R^d} \nabla \cdot (afg) $$

However, I am not sure what to do from here. Can someone help?

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1 Answer 1

up vote 1 down vote accepted

Integration by parts! Also known as the divergence theorem.

The integral $$\int_{\mathbb{R}^d} \nabla \cdot (afg) = \int_{\text{sphere at infinity}} afg = 0$$ vanishes if we assume that the domain of definition of $L$ is suitably restricted to functions decaying strongly at infinity.

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Ah, thanks! I was thinking of using the divergence theorem, but I wasn't sure if it was applicable (partly because I didn't know what was the exact space of functions $L$ was acting on). –  Alan C Apr 11 '13 at 0:42
    
It's more or less a standard convention (at least outside of the most rigorous functional analysis) that such operators are defined on whatever space of functions they behave nicely on. I'm a physicist at heart, though, so be careful... Morally the divergence theorem is certainly the correct answer. –  Sharkos Apr 11 '13 at 0:46
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