1

i want to prouve that if $f$ is continuously derivable and $f' \in L^1(\mathbb{R})$, then $F(f')=i(yF(f))$. where $F(f')$ denote the Fourier transformate of $f'$.

I try following solution: we have by definition: $F(f')(\xi)= \displaystyle\int_{-\infty}^{+\infty} e^{-ix\xi} f'(x) dx, \forall \xi \in \mathbb{R}$.

Then by integration by parts we have $$ F(f')(\xi)= [e^{-ix \xi} f(x)]_{-\infty}^{+\infty} + i \xi \displaystyle\int_{-\infty}^{+\infty} e^{-i x \xi} f(x) dx. $$ My question is what's the value of $ [e^{-ix \xi} f(x)]_{-\infty}^{+\infty}$? I think that the result is zero but how we justify it? Please.

Thank you in advance.

user487908
  • 235
  • My instinct is it's zero. Maybe you could reason like this: we're generally talking about $L^1$ functions in the first place, in which case at least one of those limits is zero. Which limit is zero depends on the sign of $\text{Im}[\xi]$. You'd like them both to be zero, but that might depend on $f$. – Adrian Keister Apr 05 '18 at 19:02
  • Wait a minute: where does $\xi$ live? Is it real or complex? If real, then the exponential is just oscillatory, and your limits are both zero. – Adrian Keister Apr 05 '18 at 19:05
  • yes, $\xi$ is an real but why the result is zeo? Please how we justify this result? – user487908 Apr 05 '18 at 19:15
  • It all has to do with the nature of $f$. You're talking about the Fourier Transform of $f$; necessary conditions for the existence of the FT are a bit complicated, but sufficient conditions are a bit easier. $f\in L^1$ is part of a set of sufficient conditions. If $f\in L^1$, then $$\lim_{x\to\pm\infty}|f(x)|=0,$$I think. Have to check... – Adrian Keister Apr 05 '18 at 19:28
  • Ok, $f\in L^1$ does not imply $$\lim_{x\to\pm\infty}|f(x)|=0,$$but I think if you examine physically realizable signals, they will have to vanish at infinity. That is, I think we're talking about a domain problem: exactly where does $f$ live? – Adrian Keister Apr 05 '18 at 19:37
  • no! $f \in L^1$ don't implies that $\lim_{|x| \to +\infty} |f(x)| =0$ – user487908 Apr 05 '18 at 20:11
  • I know! I just said that! But I think if you're dealing with physical signals, that'll be true for other reasons. – Adrian Keister Apr 05 '18 at 23:08

1 Answers1

1

Indeed, with these assumptions, the term $[e^{-ix \xi} f(x)]_{-\infty}^{+\infty}$ vanishes (but as discussed in the comments and in this thread or this one, this is not true in general). According to this thread, it suffices to prove that $f$ is uniformly continuous on $\mathbb R$. This follows from the fundamental theorem of analysis and the fact that for each positive $\varepsilon$, there exists a $\delta$ for which the integral on an interval of length smaller than $\delta$ of $\left\lvert f'\right\rvert$ is smaller than $\varepsilon.

Davide Giraudo
  • 172,925