Argue that $\gamma_0$ cannot be a maximum of the action
We consider once more the one-dimensional harmonic oscillator with the Lagrangian given by: $$L(x,v)=\frac{m}{2}v^2-\frac{k}{2}x^2,k>0$$ and the corresponding action on trajectories $\gamma$ given by the functional: $$S(\gamma)=\int_{t_1}^{t_2}L(\gamma(t),\dot{\gamma(t)})dt.$$
I have found the Euler-Lagrange equation of the system to: $$-kx=m \ddot{x}$$ And the solution to: $$x(t) = c_1 \cos(\sqrt{\frac{k}{m}} t ) + c_2 \sin(\sqrt{\frac{k}{m}} t)$$ We assume that $\gamma_0(t)$ solution of the Euler-Lagrange-equation satisfying $\gamma_0(t_1 = 0) = \xi$ and $\gamma_0(t_2 = T) = \eta$. And variation of the trajectory $\gamma_0$ given as $\gamma(t) = \gamma_0(t) + \nu(t)$ with $\nu$ some curve satisfying $\nu(0) = \nu(T) = 0$. We have that $S(\gamma)$ and $S(\gamma_0)$ is given by: $$\Delta S=S(\gamma)-S(\gamma_0)=\frac{1}{2} \int_0^T m \dot \nu^2(t)-k\nu^2(t)dt$$ We have to use this to argue that that $\gamma_0$ cannot be a maximum of the action. Can anyone help me what that argue could be? I'm a bit lost
Hints:
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Consider the sequence $$ \nu_n(t)~:=~A \sin\left( \frac{n\pi t}{T}\right) ,\qquad n~\in~\mathbb{N}, \qquad A~>~0. $$
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Show that $$\Delta S~=~S[\gamma_0+\nu_n]-S[\gamma_0]~=~S[\nu_n]~\to~ \infty\quad {\rm for}\quad n~\to~\infty. $$
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To develop a more systematic understanding of the problem, see e.g. this related Phys.SE post.