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Is the differential operator ${d\over dx}$ antisymmetric? If so, what does it even mean to take it's transpose? Thank you.

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On what function space do you mean? –  Berci Nov 8 '12 at 11:57
@Berci ,"complex inner product space"? –  Henry Nov 8 '12 at 12:03

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This is actually a quite large topic. Typically, for the inner product space you pick $L^2(a,b)$ with the inner product $$\langle u,v\rangle=\int_a^b u(x)\overline{v(x)}\,dx,$$ and you will find that $d/dx$ is at least formally skew symmetric in that partial integration $$\langle u',v\rangle=-\langle u',v\rangle,$$ at least when $u$ and $v$ are suitably smooth.

But wait (you say), this is wrong, for partial integration also involves boundary terms! And herein lies the rub. You need homogenous boundary conditions to complete this argument, such as $u(a)=u(b)=0$, or you may require periodic boundary conditions on $u$ and $v$ both.

To complicate things further, differentiation is an unbounded operator on $L^2$, and you can extend the definition from the smooth functions in various ways, depending on the homogeneous boundary conditions you chose to begin with. In order to get a good theory, you need the resulting operator to be closed (in the sense of having a closed graph).

In general, the adjoint of a closed, densely defined operator $A$ is defined such that $$\langle u,A^*v\rangle=\langle Au,v\rangle,$$ for all $v$ such that $u\mapsto\langle Au,v\rangle,$ is a bounded linear functional. (Then $A^*v$ exists by the Riesz representation theorem.)

There is much more to be said, but I can't write an entire textbook into an answer!

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Thank you, Harald! –  Henry Nov 8 '12 at 13:20

The integration by parts rule can be interpreted as saying that the adjoint of $\frac{d}{dx}$ is $-\frac{d}{dx}$, which means $\frac{d}{dx}$ is anti-symmetric.

Here's the idea: let $V$ be the vector space of all $C^{\infty}$ functions from $\mathbb{R}$ to $\mathbb{R}$ with period $T$. $V$ is an inner produce space over $\mathbb{R}$ with inner product defined by \begin{equation} \langle f, g \rangle = \int_0^T f(x) g(x) \, dx. \end{equation}

$\frac{d}{dx}$ is a linear operator on $V$, and integration by parts tells us that if $u,v \in V$ then \begin{equation} \int_0^T \frac{du }{dx} v \, dx = - \int_0^T u \frac{dv}{dx} \, dx. \end{equation} (The boundary term is $0$ because $u$ and $v$ are periodic.) In other words, \begin{equation} \langle \frac{d}{dx} u, v \rangle = \langle u, - \frac{d}{dx} v \rangle \end{equation} which shows the adjoint of $\frac{d}{dx}$ is $-\frac{d}{dx}$.

This is a very interesting result. Because $\frac{d}{dx}$ is anti-symmetric, it is also "normal", so one could hope a version of the spectral theorem applies. This leads us to expect that there is a complete orthonormal basis of eigenfunctions for $\frac{d}{dx}$, and this is where Fourier series comes from.

Note: Somebody better at functional analysis may be able to improve on my explanation.

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Thank you, littleO! –  Henry Nov 8 '12 at 13:21

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