I've got a question about the following exercise:
I'm supposed to prove that the functional $F(u):=\int_{-1}^1 x^2u'(x)^2 + xu'(x)^3 dx$ satisfies $\delta F(0,\zeta)=0 \; \forall\zeta\in C_0^1([-1,1])$ and $\delta^2 F(0,\zeta)>0 \; \forall 0 \neq \zeta\in C_0^1([-1,1])$ but that $u=0$ is no weak relative minimum of $F$.
$u$ is a weak relative minimum of $F$ in $C_0^1([-1,1])$ if there exists a $\delta>0$ so that $F(u)\leq F(v) \; \forall v\in C_0^1([-1,1])\cap B_\delta (u)$ with the $C^1$-norm.
I've succeeded to prove the assumptions about the first and the second variation of $F$. In a book, I've found the following hint regarding the proof that $u=0$ is no weak relative minimum of $F$: Consider the function $u_\epsilon(x) = \left\{ \begin{array}{ll} \epsilon cos(\frac{x}{\epsilon})^2 & |x| <\frac{1}{2} \pi \epsilon \\ 0 & |x| \geq \frac{1}{2} \pi \epsilon\\ \end{array} \right. $ for $\epsilon < \frac{2}{\pi}$ and prove that there exist non-negative constants $c_1$ and $c_2$ so that $F(u_\epsilon)=c_1 \epsilon^3 -c_2\epsilon^2 \; \forall \epsilon < \frac{2}{\pi}.$ This implies that $u=0$ is no weak relative minimum of $F$ if we define the distance between the functions with the $C^1$-norm.
I've proven that $F(u_\epsilon)=c_1 \epsilon^3 -c_2\epsilon^2$, but I don't see how I can prove that $u=0$ is no weak relative minimum of $F$ by knowing that $F(u_\epsilon)=c_1 \epsilon^3 -c_2\epsilon^2$.
I would appreciate if someone could explain the connection between these two statements to me.