I'd like to run past you also this problem, connected with the discrete function problems I posted earlier:
It would be interesting to look for conditions whereby the product of the non-zero part of a periodic function and its first derivative, integrated over a subinterval within the period might be less than zero. Thus, observe the skewed Gaussian function between $t = 0.5$ and $t = 5$
\begin{equation}
f(t)=-ae^{-\frac{(t - b)^2}{2c^2}} -ae^{-\frac{(0.5t - b)^2}{2c^2}}
\end{equation}
which for $a=1$, $b=1$ and $c=1$ is
\begin{equation}
f(t) = -e^{-\frac{(0.5t-1)^2}{2}} -e^{-\frac{(t-1)^2}{2}}
\end{equation}
The first derivative of the simplified $f(t)$ over $t$ is
\begin{equation}
\frac{df(t)}{dt} = 0.5e^{-\frac{(0.5t-1)^2}{2}} (0.5t-1) + e^{-\frac{(t-1)^2}{2}} (t-1)
\end{equation}
We now want to integrate the product $f(t)f'(t)$ for which we will get
\begin{equation}
\int\limits_{0.5}^{5} f(t)f'(t)dt = \frac{1}{2} \int\limits_{0.5}^{5} d (f(t))^2 = (f(t))^2|_{0.5}^{5} =
\end{equation}
\begin{equation}
\frac{1}{2}(-e^{-\frac{(0.5t-1)^2}{2}} -e^{-\frac{(t-1)^2}{2}})^2|_{0.5}^{5} = -1.28763 < 0
\end{equation}
Thus, we have obtained a non-zero value of $\int\limits_{0.5}^5 f(t)f'(t)dt$ for a part of the period $[0,10]$ of the periodic function $f(t)$ whose values are zero for $t>5$ and for $0 I'd like to hear objections to the above approach.