# self-adjoint differential operator on $C^{0}([a,b])$?

I've got problems in understanding the way a special (self-adjoint) differential operator is acting on the domain and the range.

So, I try to explain my difficulties:

The differential operator is given by $L=\frac{d^2}{dx^2}-F'(\varphi(x))$ where $F$ is a $C^2$-Function and $\varphi$ is an element in $\mathcal{D}(L):=\{u \in C^{2}([a,b])|u_{x}(a)=u_{x}(b)=0\}$. Now it is said that

1. "$L$ is an operator on $C^0([a,b])$ with domain $\mathcal{D}(L)$" and
2. $L$ is self-adjoint.

Now concerning my problems: Without considering (2) L is an well-defined unbounded (linear differential) operator, and for that densly-defined because $\mathcal{D}(L)$ is dense in $C^{0}([a,b])$ with respect to the uniform-norm (right?). So we can see $L$ as an operator $L:C^{0}([a,b])\supseteq\mathcal{D}(L)\longrightarrow C^{0}([a,b])$.
In this sense I will agree to (1), mentioning that $L$ is a dense-defined linear Operator on the banach-space $C^{0}([a,b])$.

But what is with (2)? At first I notice that I necessarily need a hilbert-space for self-adjointness. So I need a scalar product and the suitable scalar-product will be the $L^{2}$-scalar-product $<.,.>_{L^{2}}$. But then $(C^{0}([a,b]),<.,.>_{L^{2}})$ is no hilbert-space. Probably I need a "better space" to understand the defined operator. Finally my idea is, that $L^2([a,b])$ will do it. It makes sense in the way that I will show $\mathcal{D}(L)$ is also dense in $L^{2}([a,b])$ and I obtain (nearly) the same operator
$L:L^{2}([a,b])\supseteq\mathcal{D}(L)\longrightarrow L^{2}([a,b])$
Then $L$ will be self-adjoint and well-defined. That is only my idea and I'm not sure to be right with it. The question that follows with that is :"Why does the author explicitly mention $C^0([a,b])$, where $L^{2}([a,b])$ makes much more sense (for me)?"

So, after the large text of blah blah I hope that,nevertheless, someone will read it and could help me :)

A lot of thanks in advance!

• We use the completeness of Hilbert spaces just to ensure the existence of adjoint operator. However, the definition of adjoint operators still make sense in pre-Hilbert spaces (as is the case for $(C^0[a,b],\langle\cdot,\cdot\rangle_{L^2})$): a operator $S$ is the adjoint of $T$ if $\langle Tx,y\rangle=\langle x,Sy\rangle\ \forall x,y$. In this case, if it exists, the adjoint is unique. There are the definitions and some results here, for example. – Luiz Cordeiro Jun 12 '14 at 9:22
• And one more comment: in the case of (linear) differential operators, you can usually use integration by parts to obtain an explicit formula for the adjoint. Here's the case for second order: mathworld.wolfram.com/Self-Adjoint.html – Luiz Cordeiro Jun 12 '14 at 9:26