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I have been recently studing Fourier transform and there is a proposition that says: If $\lim \limits_{x \to\infty}xf(x)=\lim\limits_{{x}\to -\infty}xf(x)=0$ then $$\hat{f'}(z)=-iz\hat{f}(z)$$

and this is proven like this:

$(f(x)e^{izx})'=f'(x)e^{izx}+izf(x)e^{izx}$

and intergrating by parts we have

$\displaystyle\int_{-\infty}^{\infty}(f(x)e^{izx})'dx=\int_{-\infty}^{\infty}f'(x)e^{izx}dx+\int_{-\infty}^{\infty}izf(x)e^{izx}dx$

or

$f(x)e^{izx}\vert_{x=-\infty}^{x=+\infty}=\hat{f'}(z)+iz\hat{f}(z)$

But $f(x)e^{izx}\vert_{x=-\infty}^{x=+\infty}=0$ , so we get the desired equality.

My quetion here is why we should suppose that $\lim\limits_{{x}\to \infty}xf(x)=\lim\limits_{{x}\to -\infty}xf(x)=0$ and not just that $$\lim\limits_{{x}\to \infty}f(x)=\lim\limits_{{x}\to -\infty}f(x)=0$$ ? Because then we would have $f(-\infty)e^{iz(-\infty)}=f(\infty)e^{iz(\infty)}=0$

Thanks in advance for your time and effort.

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    $\begingroup$ See the last half of this answer for a method to avoid integration to prove the theorem math.stackexchange.com/a/3088728/441161 $\endgroup$ – Andy Walls Jan 29 at 17:28
  • $\begingroup$ @AndyWalls, Thanks for the refference (+1) but what I need is to tell me if my statement is correct or wrong not another way to prove it. $\endgroup$ – dmtri Jan 29 at 19:40

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