Help on proof the Sturm-Liouville eigenfunctions are all real Now, there is a guiding question as follows

  
*
  
*Let $\lambda$ and $\phi$ be an eigenvalue and a corresponding eigenfunction respectively. Let $\phi(x)=U(x)+iV(x)$ and show that $U,V$ are eigenfunctions corresponding to $\lambda$.
  
*Using theorem $A$ or otherwise, show that $U,V$ are linearly dependent.
  
*Show $\phi$ must be real, apart from an arbitrary multiplicative constant that may be complex.
  
  
  Theorem $A$
  The eigenvalues of the Sturm-Liouville problem are all simple; to each eigenvalue corresponds only one linearly independent eigenfunction. Furthermore, the eigenvalues form an infinite sequence and can be ordered according to increasing magnitude.

I have managed $1.$ Namely, $U,V$ are indeed eigenfunctions that correspond to $\lambda$. But I don't get what theorem $A$ is saying; "to each eigenvalue corresponds only one linearly independent eigenfunction"
I mean, what's that supposed to mean? Linearly independent with...what?
Linear independence (over some field $K$ if I am going to be pedantic. Probably $\mathbb{C}$ here), if I understand is about a bunch of vectors/functions $\phi_n$ etc such that $a_1\phi_1++a_2\phi_2+...+a_n\phi_n=0$ can only be satisfied with $a_1=...=a_n=0$.
I have $\phi, U,V$ corresponding to $\lambda$ so far, and how does this relate here? is it saying $a_1\phi+a_2U+a_3U=0$ then $a_1=a_2=a_3 \neq 0$? or $=0$? I don't understand because "only one linearly independent" eigenfunction sounds bizarre. I can say $v_1,v_2$ are linearly independent vectors because there are two of them; this particular pair is linearly independent. Or this particular $3$ vectors are independent etc.
But in above, what is it linearly independent to? 
And even if I manage to show that $U,V$ are linearly dependent, how does that directly relate to stating that $\phi$ must be real? 
These are the $2$ main points I am confused about. Please help!
 A: Suppose you are working on a bounded interval $[a,b]$ with Sturm-Liouville operator $L$ given by
$$
     Lf = \frac{1}{w(x)}\left[-\frac{d}{dx}\left(p(x)\frac{df}{dx}\right)+q(x)f\right].
$$
Assume that $p,q,w$ are real functions, that $p$ is continuously differentiable, and that $q$, $w$ are continuous on $[a,b]$ with $w > 0$ and $p > 0$ on $[a,b]$. The Sturm-Liouville eigenvalue problem for $L$ is posed with the domain $\mathcal{D}(L)$ restricted to twice continuously differentiable funtions $f$ on $[a,b]$ that satisfy endpoint conditions
$$
                \cos\alpha f(a)+\sin\alpha f'(a) =0,\\
                \cos\beta f(b) + \sin\beta f'(b) = 0.
$$
The natural space to consider this problem is in $L^2_{w}[a,b]$, with inner product
$$
             (f,g)_w = \int_{a}^{b}f(x)\overline{g(x)}w(x)dx.
$$
For $f,g\in\mathcal{D}(L)$ (meaning both satisfy the endpoint conditions,) the Lagrange identity gives
$$
          (Lf,g)_w = (f,Lg)_w.
$$
Any time you have a symmetric operator on a complex inner product space, the eigenvalues must be real. To see why, suppose $Lf = \lambda f$. Then
$$
          \lambda (f,f)_w = (Lf,f)_w = (f,Lf)_w = (f,\lambda f)_w = \overline{\lambda}(f,f)_w.
$$
So, either (a) $\lambda=\overline{\lambda}$ or (b) $(f,f)_w=0$, which means $f=0$. The operator $L$ is real, meaning that $\overline{Lf}=L\overline{f}$. Therefore, if $Lf=\lambda f$, then $\lambda$ must be real, and
$$
          L\overline{f} = \overline{Lf} = \overline{\lambda f}=\lambda\overline{f}.
$$
In other words, if $f$ is an eigenfunction, then so is $\overline{f}$, which means that $\Re f$ and $\Im f$ are eigenfunctions, and both $f$, $\overline{f}$ satisfy the same endpoint conditions because the endpoint conditions are formulated with real coefficients. To see that $\{ \Re f,\Im f\}$ must be a linearly dependent set of solutions, start by noticing that if $g$ is a real eigenfunction with $g(a)=g'(a)=0$, then $g$ must be zero by classical uniqueness results for second order ODE's. Therefore, $\Re f$ must satisfy the following for some non-zero scalar $C$:
$$
                  (\Re f)(a) = -C\sin\alpha,\;\; (\Re f)'(a)=C\cos\alpha
$$
And the same must be true of $\Im f$, with a different constant $D \ne 0$. It follows that $g=D\Re f - C\Im f$ is a real eigenfunction with $g(a)=g'(a)=0$, which forces $g\equiv 0$. So $\{ \Re f,\Im f\}$ must be a linearly dependent set of functions. The final conclusion is that $f= \rho\Re f$ where $\rho$ is a non-zero complex number, or $f=\rho\Im f$ where $\rho$ is a non-zero complex number. So you can always assume the eigenfunctions are real after multiplying by the correct scalar.
