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.

  • $\begingroup$ Probably both $f$ and $f'$ ought to lie in $L^2$. $\endgroup$ – Qiaochu Yuan Feb 23 '12 at 16:29
  • $\begingroup$ @Mark Are you considering only the case for $\omega(x) =1$? $\endgroup$ – Pedro Tamaroff Feb 23 '12 at 18:40
  • $\begingroup$ @PeterT.off I don't follow, what's $\omega(x)$? $\endgroup$ – Mark Feb 25 '12 at 8:51
  • $\begingroup$ @Mark The weighing function. Not all orthogonal functions have the same weighing function. $\endgroup$ – Pedro Tamaroff Feb 25 '12 at 14:22
  • $\begingroup$ @PeterT.off: Aha. Yes, that's orthogonality in the room I'm considering. $\endgroup$ – Mark Feb 27 '12 at 14:49

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. $$ \int_{-\infty}^\infty f(t)f^\prime(t)dt = \tfrac{1}{2}\int_{-\infty}^\infty \tfrac{d}{dt}f(t)^2 dt = \tfrac{1}{2}f(t)^2|_{-\infty}^\infty $$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.