A necessary condition for the differentiable functional $J[y]$ to have an extremum for $y=\hat{y}$ is that variation vanish for $y=\hat y$, i,e., that $$\delta J[h]=0$$ for $y=\hat y$ and for all admissible $h$.
I was referring the following proof given in The Textbook Calculus of Variation, Gelfand Fomin.
I understood that for sufficient small $||h||$, I understood that $\Delta J[h]$ and $\delta J[h]$ must have same sign in the small neighbourhood of $\hat{y}.$
I understood that author was trying to prove by the method of contradiction.
Then suppose there exist an admissible function such that $\delta J[h]\neq 0$ Without loss of generality I can assume $\delta J[h]> 0$ or $\delta J[h]< 0$
Question:-Then why did the author took $\alpha>0$? How does the proof evolve from there? Could you explain? I understood the intermediary step $\delta J[-\alpha h_0]=-\delta[\alpha h_0](\because$ variation is linear.)
Could you explain the proof after that? How does it help to create contradiction? Please help me...
$\alpha$ is taken to be positive since this means that $- \alpha < 0$ so that $\delta J[ - \alpha h_0]$ has a different sign to $\delta J[\alpha h_0]$ as long as $\delta J[h_0] \neq 0$ (by linearity).
The reason you need an arbitrary $\alpha$ and not just $\alpha = 1$ here is then that you want to use $(9)$ to conclude from this that $\Delta J[-\alpha h_0]$ and $\Delta J[\alpha h_0]$ have different signs and this requires $\|\alpha h_0\|$ to be sufficiently small.