How can I show that the action of a SHO is a saddle-point solution if $t_{f} - t_{0} > T/2$? In this post and in this post, QMechanic claims that a simple harmonic oscillator with Dirichlet boundary conditions has saddle-point solutions if $t_{f} - t_{0} > T/2$ where $T$ is the characteristic period of the oscillator. I am curious, how can one prove this assertion?
I tried to consider $t_{0} = 0$ and $x(0) = 0$ for simplicity with Lagrangian $$L = \tfrac{1}{2}m\dot{x}^{2} - \tfrac{1}{2}m\omega^{2}x^{2}$$ where $T = 2\pi/\omega$. Going through the usual steps, I found the solution
$$ x(t) = \frac{x_{f}\sin\omega t}{\sin\omega t_{f}} $$
where $x_{f} = x(t_{f})$.
Integrating out the Lagrangian gives me the action
$$ S[x] = \frac{1}{2}m\omega x_{f}^{2} \cot\omega t_{f}. $$
How do I use this to show that $S[x]$ is neither a local minimum nor a local maximum if $t_{f} > T/2$?
 A: 
Theorem. Consider a simple harmonic oscillator (SHO)
$$ S[x]~=~\int_{t_i}^{t_f} \! dt~L, \qquad  
L~=~\frac{m}{2}\dot{x}^2 - \frac{k}{2}x^2,\tag{1} $$
with characteristic frequency
$$ \frac{2\pi}{T}~=~\omega~=~\sqrt{\frac{k}{m}},\tag{2} $$
and Dirichlet boundary conditions (BC)
$$ x(t_i)~=~x_i \quad\text{and}\quad x(t_f)~=~x_f.\tag{3} $$
Then the action $S$ has a minimum if
$$ \Delta t~:=~t_f-t_i~\leq~ \frac{T}{2},\tag{4}$$
(if $\Delta t=\frac{T}{2}$ there is a zeromode), and $S$ has a saddle point if $\Delta t > \frac{T}{2}$.

Sketched proof:

*

*The classical solution on the SHO is on the form
$$x_{\rm cl}(t)~=~A\cos(\omega t) + B\sin(\omega t).\tag{5}$$


*An arbitrary virtual path that satisfies the BC (3) is of the form
$$ x(t)~=~x_{\rm cl}(t)+ y(t),\tag{6} $$
where the fluctuation part is a Fourier series
$$ y(t)~=~\sum_{n\in\mathbb{N}} a_n\sin\left(n\pi \frac{t-t_i}{\Delta t}\right),\qquad a_n~\in~\mathbb{R}. \tag{7} $$


*The quadratic action (1) contains no cross-terms
$$ S[x]~=~S[x_{\rm cl}] + S[y]\tag{8}  $$
[because of the EL equation and the BC (3)].


*The fluctuation part doesn't get any cross-terms contributions:
$$\begin{align} S[y]~=~&\ldots \cr 
~=~&\frac{\Delta t}{4}\sum_{n\in\mathbb{N}} a_n^2 \left[m\left(\frac{n\pi}{\Delta t}\right)^2  -k \right]\cr
~=~&\frac{m\pi^2\Delta t}{4}\sum_{n\in\mathbb{N}} a_n^2 \left[\left(\frac{n}{\Delta t}\right)^2  -\left(\frac{2}{T}\right)^2 \right].
\end{align} \tag{9}$$


*Case $\Delta t\leq\frac{T}{2}$: Then $S[y]\geq 0$. If $S[y]=0$, then $y$ can at most contain an $n=1$ mode.


*Case $\Delta t<\frac{T}{2}$: Then $S[y]=0$ implies $y=0$. It's a minimum.


*Case $\Delta t=\frac{T}{2}$: $n=1$ mode is a zeromode.


*Case $\Delta t>\frac{T}{2}$: The $n=1$ mode is unbounded from below. It's a saddle point. $\Box$
