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This is a continuation of the question asked in second-order divided differenc.

Suppose the periodic function $f(t)$ has only one value, $f(t_0) = -10$ within the period $[0,T]$, all other values of the function being zero within that period. We can interpolate $f(t)$ by $asin(t) - 10$ which for $t_0 = 0$ will give the value $-10$. Now, if we accept the interpolation then $f'(t) = acos(t)$ and therefore for $t_0=0$ we will have $f'(t_0) = acos(0) = a$. Thus, at $t_0 =0$ we will obtain $f(t_0)f'(t_0) = -10a$. At the end of the period the value of $f(t)f'(t)$ will be also $-10a$. Therefore, the average value of $f(t)f'(t)$ for the period is $\frac{2(-10a)}{2} = -10a$. Do you agree with this?

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If I understand your question correctly, you have an $f: [0,T] \to \mathbb{R}$, $$ f(t) = \begin{cases} -10 &\text{if } t=t_0 \\ 0 &\text{otherwise} \end{cases} $$ and are asking if for the value of $$ \frac{1}{T}\int_0^T f(t)f'(t) dt \text{ .} $$

That question doesn't make a lot of sense then, because $f$ is not differentable, hence $f'(t)$ isn't even defined. And if you go on to define $f'(t)$ somehow, the integral will be $0$, because the integrand is zero everywhere except at $t_0$.

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No, if $g$ is zero on $A$ except at finitely many points, then $\int_A g(t)dt = 0$. Or you do actually envision $f$ to be a measure? I.e. do you have a measure $\mu$ with $\mu([a,b]) = f(b) - f(a)$ and want to compute $\int_A f(t)d\mu(t)$? In that case, $\int_A f(t)d\mu(t) = \int_A f(t)f'(t)dt$ only if $f$ is differentiable, which your $f$ is not. – fgp Oct 20 '12 at 12:21
So? Pick any other interpolation, and you'll get a different result. In fact, I bet that you can get every possible result just by picking a suitable interpolation for $f$. So what's the point? – fgp Oct 20 '12 at 12:31
Pick the value of $f$ at any $n$ points. There always is a polynomial of degree $n$ which coincides with $f$ at those $n$ points. If you add additional conditions, e.g. about $f'$, you'll need to pick a polynomial with a larger degree, but again, you will find a polynomial which fulfills all of your conditions. I really don't know what you're trying to do there, but you need to assume some pretty strong property of $f$ to be able to make assumptions about $f'$ purely from samples values of $f$. Limiting the bandwidth of $f$ seems like it might work, but I didn't check... – fgp Oct 20 '12 at 12:50
Because you can choose arbitrary values of $f'(t)$, and always find an interpolating polynomial, so you can get any answer you want. Which makes the whole thing completely devoid of any meaning. I'll say it one last time: You cannot deduce anythinga bout $f'$ from samples of $f$ without some strong additional constraint on $f$. Nothing. Nada. Zilch. – fgp Oct 20 '12 at 13:06
Sure, yeah, if you know that $f(t) = a\sin(t) - b$, you can do that. But you never stated that you know that, you basically said "I'm going to just pick $a\sin(t) - b$ 'cause it seems easy to do". There's a huge difference there. And btw, you can easily find polynomial interpolations which make $f$ periodic - just add constraints $p(0) = f(T)$.... – fgp Oct 20 '12 at 13:14

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