Variational methods : Why i can't apply this theorem? Consider the following problem:
Find a weak solution for 
$$u'' + u  = \sin t  ,\,\,  0 < t < \pi$$
$$u(0) = u(\pi)=0 $$
the corresponding functional for the problem is $\varphi(u) = \displaystyle\int_{0}^{\pi}\biggl( \frac{|u^{'}|^2}{2} - \frac{u^2}{2} + (sin \ t )u\biggr)dt,\;\; u \in H^{1}_{0}(0,\pi)$.
My book says that we can't aply the following theorem(to show the existence of weak solution):
theorem: Let $\varphi \in C^{1}(X,R)$ ($X$ is Banach) with:
i) Exist $c$ such that $c = \inf_X \varphi$
ii) $\varphi$ satisfies the property:
For every sequence $u_n$ with $\varphi(u_n) \rightarrow c$ and $\varphi^{'}(u_n) \rightarrow 0$, then $c$ is a critical value of $\varphi$
Then exist $u_0 \in X$ such that $\varphi(u_0) = c = \inf_X \varphi.$
I believe that the first condition fails. I tried to check this without success. About the second condition of the theorem i dont know a sequence where this conditions fails..
Someone can give me a help to explain why i cant aply the theorem?
thanks in advance.
 A: The fact that the first condition does not hold is not the main reason why the theorem cannot be applied. It can never be applied, because it is basically wrong. Indeed, take $X=\mathbb{R}$ and consider a functional $\varphi\,\colon \mathbb{R}\to\mathbb{R}\,$ of the form $\,\varphi(x)=e^x$.  It is clear that $\,\inf\limits_{x\in\mathbb{R}}{e^x}=0$, so we have $c=0$,  while for any sequence $\,x_n\to -\infty\,$ we have $\,\varphi(x_n)\to c\,$ and $\,\varphi'(x_n)\to 0$.   Still, there is no $\,x_0\in\mathbb{R}\,$ such that $\,\varphi(x_0)=c=\inf\limits_{x\in\mathbb{R}}{e^x}$.  In fact,
possessing no zero, even on the complex plane, is the most important feature enjoyed by the exponent as an entire function.
A: Your problem has no solution. It is easy to see by the Riesz–Fredholm theory (see 
Theorem 1.14, Part (ii), page 10, http://books.google.de/books/about/Nonlinear_Analysis_and_Semilinear_Ellipt.html?id=4O6tfmZtxQAC&redir_esc=y ), since $\lambda_1 = 1$ is the first eigenvalue and $v_1 = \sin t$ is the first eigenfunction, but
$$
\int_0^\pi \sin t \sin t \ dx \neq 0.
$$
