# Given $F(u)=u''+\lambda_n u$ on $X=\{u \in C^2 | u'(0)=u'(l)=0\}$, prove Im$(F) = \{ \theta | \langle \theta, \phi_n \rangle = 0 \}$

I'm working on a problem on bifurcation, getting stuck in the following point. Let me explain the settings of my problem first: I'm trying to find the bifurcation values of the following system.

$$\theta''+\lambda \sin \theta=0\text{ ; }0<s<l$$ $$\theta'(0)=\theta'(l)=0$$

So I setup the spaces $$X=\{u \in C^2([0,l]); u'(0)=u'(l)=0\}, Y=C([0,l])$$

and the operator

$$F(\lambda,u)=u''+\lambda u$$

The Fréchet derivative is $F_u(\lambda,0)[\theta]=\theta''+\lambda \theta$, which gives $\lambda_n=\frac{n^2 \pi^2}{l^2}$ as the only possible bifurcation values. I have

$$\ker(F_u(\lambda_n,0))=\text{span}\langle \phi_n \rangle \text{ ; } \phi_n=\cos \sqrt{\lambda_n}s$$

The problem explicitly requires me to prove that

$$\text{Im}(F_u(\lambda_n,0))=\{\theta \in Y | \langle \theta, \phi_n \rangle =0\}$$

I've proven the "$\theta =u''+u, \exists u \in X\Rightarrow\langle \theta, \phi_n \rangle=0$" half by straightforward calculations, but how can I prove the other direction? Please help.