I've got a problem in the following computation.
I am looking for the general solution of periodic boundary value problem $$ Ly(t):=y''(t)+y(t)=f(t),t\in(0,\pi/2)\quad y(0)=y(\pi/2),y'(0)=y'(\pi/2). $$ I tried to find the corresponding Green's function in the following form $$ G(s,t)=\begin{cases} A(s)y_1(t),& s>t,\\B(s)y_2(t),& s\leq t\end{cases}$$ where $y_{1,2}$ are some solutions of the homogeneous problem $y''(t)+y(t)=0$ satisfying $y_1(0)=y_2(\pi/2)$ and $y_1'(0)=y_2'(\pi/2)$. I chose $y_1(t)=\sin t+\cos t$ and $y_2(t)=\sin t-\cos t$. I used the basic properties of Green's functions to determine $A$ and $B$ and I got that $A(s)=y_2(s)/2$ and $B(s)=y_1(s)/2$. The solution $y$ of the problem is then given by $$ y(t)=y_2(t)/2\int_0^ty_1f(s)\mbox{ d}s+y_1(t)/2\int_t^{\pi/2}y_2f(s)\mbox{ d}s.$$
The problem started when I tried to verify the condition $y(0)=y(\pi/2)$. From this, I found that necessarily it holds $\int_0^{\pi/2}\cos sf(s)\mbox{ d}s=0$. I do not understand this since the operator $L:L^2(0,\pi/2)\to L^2(0,\pi/2) $ is self-adjoint and so the closure of the range of $L$ is equal to the orthogonal complement of the kernel of $L$, that is, the range of $L$ should be dense in $L^2(0,\pi/2)$. However the condition $\int_0^{\pi/2}\cos s\,f(s)\mbox{ d}s=0$ indicates that this density is not true.
There should be a mistake in my computation but I have no idea what could be wrong. I appreciate any help.