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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 $L^2$. – Qiaochu Yuan Feb 23 '12 at 16:29
@Mark Are you considering only the case for $\omega(x) =1$? – Pedro Tamaroff Feb 23 '12 at 18:40 I don't follow, what's $\omega(x)$? – Mark Feb 25 '12 at 8:51
@Mark The weighing function. Not all orthogonal functions have the same weighing function. – Pedro Tamaroff Feb 25 '12 at 14:22 Aha. Yes, that's orthogonality in the room I'm considering. – Mark Feb 27 '12 at 14:49
up vote 4 down vote accepted

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? – Mark Feb 23 '12 at 16:33
Yes for both of them. – josh Feb 23 '12 at 17:50

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