If $X$ is an eigenfunction satisfying the boundary conditions and $XX'|_a^b\le0$ , then there is no negative real eigenvalues.
Prove that this is valid for symmetric condition.
Attempt
Is symmetric condition if $f'g-fg'|_a^b=0$.
Then $(f'g-fg')(b)=(f'g-fg')(a).$
From here can I say that $a=b$ ?
If yes, then $ff'|_a^b\le0$ follows immediately.
If not, then what can I do to prove that symmetric condition satisfies $XX'|_a^b\le0$ ?
$X$ is an eigenfunction, i.e. a nonzero function satisfying $-X''=\lambda X$ for some $\lambda$, subject to a pair of boundary conditions. The boundary conditions are called symmetric if, for any pair of functions $f$ and $g$ which satisfy both boundary conditions, $(f'g-fg')|_a^b = 0$.
Assume $XX'|_a^b\leq 0$ for all eigenfunctions $X$ which satisfy the boundary conditions. By Green's First Identity,
$$\int_a^b X''Xdx = XX'|_a^b-\int_a^bX'^2dx$$
$$\Rightarrow XX'|_a^b = \int_a^b X''X + X'^2dx = \int_a^b -\lambda X^2 + X'^2dx \leq 0 $$
If $\lambda < 0$, then $-\lambda X^2+X'^2 > 0$, so $\int_a^b -\lambda X^2 + X'^2dx > 0$, a contradiction. Thus, $\lambda \geq 0$.