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.

  • $\begingroup$ 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. $\endgroup$ May 6, 2013 at 6:53
  • $\begingroup$ How do you show that it is not symmetric for n odd in the space of test functions? $\endgroup$
    – user66906
    May 6, 2013 at 7:33
  • $\begingroup$ Integration by parts gives you a minus sign, doesn't it? $\endgroup$ May 6, 2013 at 10:59

2 Answers 2


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).


You must log in to answer this question.