Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.