# When is a real function orthogonal to its derivative?

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.

• 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
• @PeterT.off 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
• @PeterT.off: Aha. Yes, that's orthogonality in the room I'm considering. – Mark Feb 27 '12 at 14:49

## 1 Answer

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

• 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