Self-Adjoint Operator If I have a self-adjoint operator which operates on twice differentiable functions defined by:
$$ Lx(t) = [k(t)x'(t)]' + g(t)x(t) $$ How can I show that $k(t)$ is real-valued given that $k \in C^1[a,b]$ and $g \in C[a,b]$? Not really sure where to begin other than using the self-adjoint condition; that is, $(Lu,v) = (u,Lv)$ where $u,v$ are twice differentiable functions and $$ (u,v) = \int_a^b u(t) \overline{v(t)} \, dt $$ Any ideas to help me out?
 A: As Stuart duly pointed out boundary conditions are essential here. Write out the scalar product explicitly and integrate by parts twice
$$\left(Lx,y\right)=\int_{a}^{b}\left(kx'\right)'\bar{y}dt+\int_{a}^{b}gx\bar{y}dt=\left.kx'\bar{y}\right|_{a}^{b}-\int_{a}^{b}kx'\bar{y}'dt+\int_{a}^{b}gx\bar{y}dt=\\\left.\left(kx'\bar{y}-kx\bar{y}'\right)\right|_{a}^{b}+\int_{a}^{b}x\left(k\bar{y}'\right)'dt+\int_{a}^{b}xg\bar{y}dt$$
For what follows it is necessary that the first term in the last expression vanishes. Now 
$$\left(x,Ly\right)=\int_{a}^{b}x\left(\bar{k}\bar{y}'\right)'dt+\int_{a}^{b}x\bar{g}\bar{y}dt$$
Since the difference $\left(Lx,y\right)-\left(x,Ly\right)$ vanishes identically for all admissible $x$ and $y$ one may deduce the desired result.
A: You're right, you can just use self-adjointness. Keep in mind that self-adjointness is more than just defining the operator, you need to define boundary conditions too. E.g. $u(a) = u(b) = 0$.
From self-adjointness, you get $k\bar{v}'' + k'\bar{v}' + g\bar{v} = \overline{kv'' + k'v' + gv}$. So $g$ is real too. 
In general, if $L$ is self-adjoint and  $L = p_0 u^{(n)} + p_1u^{(n-1)} + \cdots + p_nu$, then $L^* = (-1)^n (\overline{p_0}u)^{(n)} + (-1)^{n-1}(\overline{p_{1}}u)^{(n-1)} + \cdots + \overline {p_n}u$, where boundary conditions are such that $u(b)v(b) - u(a)v(a) = 0$
