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For $f(x)$ we have that

$$ \widehat{f}(\xi) = \int_\mathbb{R} e^{-2 \pi i x \xi} f(x) $$

Now let $s \in \mathbb{R}$ and consider

$$ \widehat{f(sx)}(\xi) = \int_\mathbb{R} e^{-2 \pi i (sx) \xi} f(sx) = \int_\mathbb{R} e^{-2 \pi i (x) \xi} f(x)*{d \over ds}(1/s) = -{1 \over s^2}\int_\mathbb{R} e^{-2 \pi i (x) \xi} f(x) = -{1 \over s^2} \widehat{f}(\xi) $$

through a change in variable $x \mapsto (1/s)x$.

Question: Is this calculation correct?

user1770201
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1 Answers1

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No. Let $\varphi(x)=sx$, so the question is addressing the relationship between $\widehat{f}$ and $\widehat{f\circ\varphi}$. We have \begin{align*} \widehat{f\circ\varphi}(\xi)&=\int f\circ\varphi(x)e^{-2\pi ix\xi}dx\\ &=\int f(sx)e^{-2\pi ix\xi}dx\\ &=\int f(sx)e^{-2\pi i(sx)(\xi/s)}\dfrac{d(sx)}{s}\\ &=\dfrac{\text{sgn}(s)}{s}\int f(x)e^{-2\pi ix\xi/s}dx\\ &=\dfrac{\text{sgn}(s)}{s}\widehat{f}(\xi/s). \end{align*}

user284331
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  • Do you mean $d(sx)/x$ instead of $d(sx)/s$ (since $s$ is constant)? – user1770201 Mar 28 '18 at 01:52
  • Doesn't $e^{-2 \pi i (sx)( \xi /s)} = e^{-2 \pi i x \xi}$? – user1770201 Mar 28 '18 at 01:53
  • No, I really mean that $d(sx)/s=dx$, actually this is a bad notation, one should write formally by substituting $u=sx$. – user284331 Mar 28 '18 at 01:54
  • No, the $x$ is still in the $e^{-2\pi i\xi x}=e^{-2\pi i(sx)(\xi/s)}$. – user284331 Mar 28 '18 at 01:55
  • I'm confused what's going on in line 3. You're changing the variable $x$ in the integral? – user1770201 Mar 28 '18 at 01:57
  • Oh sorry, it is me that made the typo, thanks for reminding. – user284331 Mar 28 '18 at 01:58
  • Yes, that is why I said that, it is better to use $u=sx$ to do the job, and then change to $x$ back as the dummy variable. – user284331 Mar 28 '18 at 01:59
  • Why do we have $\text{sgn}(s)/s = |1/s|$ and not $1/s$ as our final factor in the final two steps of your reasoning? – user1770201 Mar 28 '18 at 02:28
  • $s$ may be negative, when you do the change of variable, the domain of integration swipes the $-\infty$ and $\infty$. – user284331 Mar 28 '18 at 02:29
  • But if $s < 0$, we will also have that $1/s$ is negative, thus canceling the effect of $\int_{\infty}^{-\infty}$ already, right? In any event the Wikipediage page for integration by substitution doesn't directly mention domain of integration swiping; but isn't this because the domain of integration will swap if and only if the derivative of the change in variable function is negative (so that the effects cancel out)? – user1770201 Mar 28 '18 at 04:19
  • The wikipedia is not correct, it actually requires $\varphi'>0$, but luckily the examples provided meet this requirement. Right now just think of how do we deal with the basic calculus like $\displaystyle\int_{-\infty}^{\infty}f(x)dx$. If $s=-1$ and $u=-x$, then $du=-dx$, the integral is $\displaystyle\int_{\infty}^{-\infty}f(-u)(-du)=\displaystyle\int_{-\infty}^{\infty}f(-u)du=-s\displaystyle\int_{-\infty}^{\infty}f(-x)dx$. – user284331 Mar 28 '18 at 04:32