I'm discussing this in another forum but I'd like to run it past you as well. I'm observing a periodic comb function $f(t_i)$ such that all its non-zero values within the period [0,T] are
$f(t_0)=−5$,
$f(t_1)=−2.5$,
$f(t_2)=0$,
$f(t_3)=−2.5$,
$f(t_4)=−5$,
$f(t_5)=−7.5$,
$f(t_6)=−10$,
$f(t_7)=−7.5$,
$f(t_8)=−5$
where $f(t_8)=f(T)$ and I'm looking for the first derivative of that function at each one of the non-zero points -- 9 altogether. Mathematicians, most likely, will not agree that this is a differentiable function (a function should be continuous in order to be differentiable) but it appears to me that I can consider the real change of that function around each point in this particular case as $f(t_{i+1})−f(t_{i−1})$ and, therefore, the first derivative over time at each point can be written as $\frac{f(t_{i+1})−f(t_{i−1})}{t_{i+1}−t_{i−1}}$. If that's not a first derivative then what would you call that ratio?
EDIT: I think the problem boils down to the following:
Let us denote the comb function by f(t) although in our case we know it consists of only 9 values. This is given. Now, the theorem that has to be proved is that no matter what way of obtaining the 9 outcomes from f(t) we may find (note I'm not calling these outcomes by the term derivatives), which we will denote by f'(t), there will always be a condition (offset) whereby the sum of the 9 products f(t)f'(t) will be negative.
EDIT 2:
I'll be trying to refine further the problem:
Let us denote the periodic comb function by $f(t)$, consisting of only $2n+1$ values within a period $[0.T]$, then no matter what way of obtaining the $2n + 1$ difference quotient outcomes from $f(t)$ we may find (for instance, backward, central or forward difference quotients), which we will denote by $f^{dq}(t)$, there will always be a condition (offset) whereby the sum of the $2n + 1$ products $f(t)f^{dq}(t)$ will be negative.
