# Transpose of a linear operator on functions

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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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.
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