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I saw this proof that a function $f$ is orthogonal to its derivative $f'$:

$$ \int_{-\infty}^\infty f(t)f'(t)dt = \frac{1}{2\pi} \int_{-\infty}^\infty F(\Omega) (-j\Omega) F^*(\Omega) d\Omega = -\frac{1}{2\pi} \int_{-\infty}^\infty j\Omega |F(\Omega)|^2 d\Omega = 0 $$

where $F(\Omega)$ denotes the Fourier transform of $f(t)$.

This clearly isn't true for all functions, e.g. $f(t) = \max(0,t)$. Could anyone help me figure out which assumptions were made? The original text was not more specific than this.

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    Probably both $f$ and $f'$ ought to lie in $L^2$.2012-02-23
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    @Mark Are you considering only the case for $\omega(x) =1$?2012-02-23
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    @PeterT.off I don't follow, what's $\omega(x)$?2012-02-25
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    @Mark The weighing function. Not all [orthogonal functions](http://mathworld.wolfram.com/OrthogonalFunctions.html) have the same weighing function.2012-02-25
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    @PeterT.off: Aha. Yes, that's orthogonality in the room I'm considering.2012-02-27

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In addition to the assumptions that the integral even makes sense, this particular result is based on the assumption is that $f(t)^2$ is defined at $t = \pm\infty$ and that the two limiting values (at $t = \infty$ and $t = -\infty$) are equal. $$ \int_{-\infty}^\infty f(t)f^\prime(t)dt = \tfrac{1}{2}\int_{-\infty}^\infty \tfrac{d}{dt}f(t)^2 dt = \tfrac{1}{2}f(t)^2|_{-\infty}^\infty $$

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    By "$f(t)^2$ is defined at $t=\pm\infty$", are you referring to the limits at $\pm\infty$? And by making sense, do you mean that it converges?2012-02-23
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    Yes for both of them.2012-02-23