When is the following operator self-adjoint?(Is there a difference between self-adjoint and Hermitian?)

$O:= \sum_{n=0}^4 f_n(x){d^n\over dx^n}$ subjected to boundary conditions $y(0)=y'(0)=y(1)=y'(1)=0$ and where $f_n$'s are real functions.

Thanks.

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On which space is $y$? – Davide Giraudo Sep 27 '11 at 12:02
@DavideGiraudo: I beleive it is the space of functions that satisfy those boundary conditions? – M. M. Sep 27 '11 at 12:03
If I remember correctly, in the case of unbounded operators (as in here), self-adjoint is not the same as hermitian/symmetric. – Mark Sep 27 '11 at 12:19

In a first step, we need to define the base space we are talking about, where our operators are defined. In this case we could choose $$H := L^2[0, 1]$$ the complex (or real) Hilbert space of complex (or real) valued square integrable functions on the interval [0 ,1]. An operator like the differential operator $$A: f \mapsto f'$$ will be defined not on the whole space, but on a (dense) subspace $$D(A) \subset H$$ An important difference of symmetric and selfadjoint is that a symmetric operator agrees with its adjoint on the subspace where it is defined, i.e. $$A = A^* \; \text{on } D(A)$$ while an operator is selfadjoint, if in addition we have that $$D(A) = D(A^*)$$ i.e. its domain of definition is the same as the one of its adjoint. This is not a minor technicality, but as a consequence theorems that are true for selfadjoint operators generally fail for operators that are only symmetric.
So the real problem here is to find the domain of definition of the adjoint of the given operator. Here is a hint: If we further impose boundary conditions like the one you mention, we have as domain of definition $$D(A) := \{f \in H: f \text{ is four times differentiable and } f(0) = f(1) = 0 \text{ etc.} \}$$ then the problem with boundary conditions like that is that they usually lead to a situation where the adjoint has a bigger domain of definition. As an example let $A$ be the operator of multiplication by a real valued function $h$ with domain of definition as above. Then you have $$\langle Af, g \rangle = \langle h f, g \rangle = \langle f, h g \rangle$$ for arbitrary $g$. This shows that $$D(A^*) = H \neq D(A)$$ so that A is symmetric, but not selfadjoint.
Thanks, Tim. :-) But I need the conditions on $O$ in the form of constraints on the $f_n$'s such that $O$ is self-adjoint. Could you shed some light on that? – M. M. Sep 27 '11 at 13:40