I am looking in the space of test functions $\{f \in C^\infty|f^{(n)}(a)=f^{(n)}(b)=0\};n \in \mathbb{N}_0\}$whether the n-th derivative is a self adjoint operator. the dot product is given by $(f,g)=\int_a^b f(x)g(x) dx$. I was able to show it for even n that the n-th derivative is self adjoint, just by doing integration by parts but for odd n this does not work as integration by parts would give me a minus sign here. Unfortunately I am neither able to produce a counterexample nor am I able to show that the derivative is self adjoint, which I highly doubt.

• What Hilbert space are you working over? This really is the key to whether or not $D^n$ is self-adjoint for $n$ even; in the case that $n$ is odd, $D^n$ of course cannot possibly be self-adjoint for any relevant choice of Hilbert space, given that it isn't even symmetric. May 6, 2013 at 6:53
• How do you show that it is not symmetric for n odd in the space of test functions?
– user66906
May 6, 2013 at 7:33
• Integration by parts gives you a minus sign, doesn't it? May 6, 2013 at 10:59

You already know the answer: derivatives of odd orders are anti-symmetric, not symmetric. But if you want a concrete example, take something like $f(x)=(x-a)^{2n+1}(b-x)^{2n+1}$ and $g=f^{(n)}$. (Check that these functions satisfy the boundary conditions). Then $\int_a^b f^{(n)}g>0$ and $\int_a^b fg^{(n)}=-\int_a^b f^{(n)}g<0$, so these integrals are not equal.
There are a number of comments which might be appropriate here. Firstly, the general rule of thumb is that you require $n$ boundary conditions to specify a self-adjoint operator for a differential operator of order $n$ (in the sense of unbounded operators in Hilbert space), not the $2 n$ given here. In fact, if one considers the formal differential operator of differentiating $2 n$ times rather than $n$ times, then it is essentially self-adjoint in your Hilbert space if one imposes the boundary conditions $f^{(n)}(a)=f^{(n)}(b)=0$ for $n$ even between $0$ and $2n-2$. This is a significant example---the corresponding eivenvectors are the functions $\sin n x$ and the expansion by eigenvectors is just the Fourier Sine expansion (we assume that $a=0$ and $b= \pi$ to simplify the notation). In order to obtain a genuinely self-adjoint operator one has to be more careful about the domain of definition (e.g., by using distributions or Sobolev spaces). A very thorough acount of the hard analysis involved in this and in many of the important Sturm-Liouville problems which arise in classical mathematical physics can be found in Hans Triebel's text "Hoehere Analysis (an english translation "Higher Analysis" is available).