Eigenfunctions of a second derivative operator Consider the operator $L :=\frac{-d^2}{dy^2}+ \alpha^2 - K(y)$ on the space of functions $f(y) $ on $H^2(-a,a) \cap H_0^1(-a,a)$. Here $K(y)$ is an even function and $\alpha >0$ is a positive real number. It is claimed that any eigenfunction of $L$ is either even or odd. Why is this so? Edit: OK, the claim any eigenfunction is odd or even is, according to my interpretation, saying that is $f$ is an eigenfunction such that $f=f_o + f_e$ where $f_o$ ($f_e$) is the odd (even) part of $f$, then one of them, either the odd part of the even part is necessarily zero. Is this the right interpretation?

If $f$ satisfies, $$ -f''+a^2f-K(y)f=\lambda f, \quad f(-1)=f(1)=0. $$ then so do $$ f_{even}=\frac{1}{2}\big(f(x)+f(-x)\big), \quad f_{odd}=\frac{1}{2}\big(f(x)-f(-x)\big). $$ Hence, the eigenspace of our operator is spanned by odd and even eigenfunctions.

We shall next show that only one of the two does not vanish identically. Clearly, if the both did not vanish identically, then they would be linearly independent - otherwise, $$ L_{even}=cL_{odd}, \quad\text{and thus}\quad L_{even}(-x)=cL_{odd}(-x)=-cL_{odd}(x)=-L_{even}(x). $$ But in such case, they would constitute a fundamental set of solutions (i.e., a basis of the set of solutions of $Ly=0$), as the set of solutions of $Ly=$ is 2-dimensional. This is impossible, since they both vanish at $x=-1$, and hence the Wronskian is equal to zero.